Quick Example: The system

has augmented matrix:

REF example:

Pivots are boxed. Three non-zero rows → rank = 3.

Why it matters — a concrete example:

System: $0.0001x + y = 1$ and $x + y = 2$.

Exact solution (for reference): $x = \dfrac{10000}{9999} \approx 1.0001$, $\;y = \dfrac{19998}{9999} \approx 1.9999$.

Without Pivoting (4-digit arithmetic)

Augmented matrix:

Step 1 — Eliminate below pivot $a_{11} = 0.0001$:

Multiplier: $m = \dfrac{1}{0.0001} = 10000$

In 4-digit floating point, $-9999$ rounds to $-10000$ and $-9998$ rounds to $-10000$:

Step 2 — Back substitution:

Result without pivoting: $x = 0, \; y = 1$ — wildly wrong! (true $x \approx 1$)

The tiny pivot caused the multiplier $10000$ to amplify the rounding error and obliterate the true value of $x$.

With Partial Pivoting (4-digit arithmetic)

Before elimination, compare $\lvert a_{11} \rvert = 0.0001$ vs $\lvert a_{21} \rvert = 1$. Since $1 > 0.0001$, swap $R_1 \leftrightarrow R_2$:

Step 1 — Eliminate below pivot $a_{11} = 1$:

Multiplier: $m = \dfrac{0.0001}{1} = 0.0001$ (small — no error amplification!)

In 4-digit floating point, $0.9999$ and $0.9998$ are stored exactly (4 significant digits):

Step 2 — Back substitution:

Result with pivoting: $x = 1.000, \; y = 1.000$ — excellent accuracy! ✓

Takeaway: Partial pivoting keeps multipliers $\lvert m \rvert \leq 1$, preventing catastrophic amplification of rounding errors. Always use the largest available pivot.

Problem: Solve

Step 1 — Augmented Matrix:

Step 2 — Forward Elimination: Pivot = $a_{11} = 1$. Multiplier: $m_{21} = 3/1 = 3$. Operation: $R_2 \leftarrow R_2 - 3R_1$

Step 3 — Back Substitution:

• Row 2: $-2y = -6 \Rightarrow y = 3$
• Row 1: $x + 2(3) = 8 \Rightarrow x = 2$

Solution: $x = 2, \; y = 3$ ✓

Verification: $2 + 2(3) = 8$ ✓, $3(2) + 4(3) = 18$ ✓

Problem: Solve

Augmented Matrix:

Eliminate column 1: $R_2 \leftarrow R_2 - 2R_1$, $R_3 \leftarrow R_3 - R_1$

Eliminate column 2: $R_3 \leftarrow R_3 - R_2$

Back Substitution:

• Row 3: $3z = 9 \Rightarrow z = 3$
• Row 2: $y - 3 = 1 \Rightarrow y = 4$
• Row 1: $x + 4 + 3 = 6 \Rightarrow x = -1$

Solution: $x = -1, \; y = 4, \; z = 3$ ✓

Problem: Solve

Augmented Matrix:

Eliminate column 1: $R_2 \leftarrow R_2 - 2R_1$, $R_3 \leftarrow R_3 - R_1$

Row 2 is all zeros — equations are dependent. Swap $R_2 \leftrightarrow R_3$:

2 equations, 3 unknowns — there are more unknowns than pivot rows.

Here column 3 ($z$) has no pivot, so $z$ is a free variable. Let $z = t$ (free parameter):

• Row 2: $-y - 2t = -8 \Rightarrow y = 8 - 2t$
• Row 1: $x + 2(8 - 2t) + 3t = 14 \Rightarrow x = -2 + t$

Solution:

Problem: Solve

The pivot $0.0001$ is very small — swap rows first!

$R_2 \leftarrow R_2 - 0.0001 R_1$:

Back substitution: $y \approx 1$, $x = 2 - 1 = 1$.

Solution: $x = 1, \; y = 1$ (with improved numerical accuracy!)

Problem: Investigate for what values of $\lambda, \mu$ the system

has (i) no solution, (ii) a unique solution, (iii) infinitely many solutions.

Augmented matrix:

$R_2 \leftarrow R_2 - 2R_1$, $R_3 \leftarrow R_3 - 3R_1$:

$R_3 \leftarrow R_3 - R_2$:

Analysis of Row 3:

Problem: Solve the homogeneous system:

Coefficient matrix:

Swap $R_1 \leftrightarrow R_4$ for unit pivot, then eliminate:

$\rho(A) = 2 < 4$ (unknowns) $\Rightarrow$ non-trivial solutions exist.

Number of free variables $= 4 - 2 = 2$. Let $z = s$, $w = t$ (free).

From Row 2: $-3y - 24s - 9t = 0 \Rightarrow y = -8s - 3t$

From Row 1: $x + 3(-8s - 3t) + 13s + 3t = 0 \Rightarrow x = 11s + 6t$

Solution:

Problem: Discuss for all values of $k$, whether the system has only the trivial solution or also non-trivial solutions:

Coefficient matrix and determinant:

Computing $\det(A)$:

After expansion and simplification:

Analysis:

• Trivial solution only: $\det(A) \neq 0 \Rightarrow k \neq \pm 2$
• Non-trivial solutions: $\det(A) = 0 \Rightarrow k = 2$ or $k = -2$

Problem: Find values of $\lambda$ for which the following homogeneous system has non-trivial solutions:

This system can be written as $(A - \lambda I)\vec{x} = \vec{0}$ where:

For non-trivial solutions, we need $\det(A - \lambda I) = 0$:

Expanding the determinant:

This is the characteristic equation. The values of $\lambda$ satisfying this equation are called eigenvalues — a topic explored fully in Unit II.

Example 1: Let $\vec{u} = (1,2,3)$ and $\vec{v} = (4,5,6)$.

• Addition: $\vec{u} + \vec{v} = (1+4, \; 2+5, \; 3+6) = (5, 7, 9)$
• Scalar multiplication: $2\vec{u} = (2 \cdot 1, \; 2 \cdot 2, \; 2 \cdot 3) = (2, 4, 6)$
• Subtraction: $\vec{u} - \vec{v} = \vec{u} + (-1)\vec{v} = (1-4, \; 2-5, \; 3-6) = (-3, -3, -3)$
• Linear combination: $3\vec{u} + 2\vec{v} = (3,6,9) + (8,10,12) = (11, 16, 21)$

Example 2 (Real-world): You walk 3 blocks east and 4 blocks north: $\vec{d}_1 = (3, 4)$. Then 1 block west and 2 blocks north: $\vec{d}_2 = (-1, 2)$.

Total displacement: $\vec{d}_1 + \vec{d}_2 = (3 + (-1), \; 4 + 2) = (2, 6)$ — 2 blocks east, 6 blocks north.

Example 1: Is $\vec{w} = (5, 8)$ a linear combination of $\vec{v}_1 = (1, 2)$ and $\vec{v}_2 = (2, 3)$?

We need scalars $c_1, c_2$ such that: $c_1(1,2) + c_2(2,3) = (5,8)$

Writing component-wise:

From the first equation: $c_1 = 5 - 2c_2$. Substituting into the second:

Verification: $1 \cdot (1,2) + 2 \cdot (2,3) = (1,2) + (4,6) = (5,8)$ ✓

So $\vec{w} = \vec{v}_1 + 2\vec{v}_2$. ✓

Example 2: Is $\vec{w} = (1, 0, 0)$ a linear combination of $\vec{v}_1 = (1, 1, 0)$ and $\vec{v}_2 = (0, 1, 1)$?

We need $c_1(1,1,0) + c_2(0,1,1) = (1, 0, 0)$:

From Eq.1: $c_1 = 1$. From Eq.3: $c_2 = 0$. But Eq.2: $1 + 0 = 1 \neq 0$. Contradiction!

No — $\vec{w}$ is NOT a linear combination of $\vec{v}_1$ and $\vec{v}_2$. The system is inconsistent.

Quick example: Is $S = \lbrace(x, y) \in \mathbb{R}^2 : x \geq 0, y \geq 0\rbrace$ (the first quadrant) a vector space?

• Check closure under scalar multiplication: Take $\vec{v} = (1, 2) \in S$ and $c = -1$. Then $c\vec{v} = (-1, -2) \notin S$ (negative components!).
• Fails axiom 2 → NOT a vector space. Done!

We didn’t need to check the other 9 axioms. One failure is enough.

Claim: $\mathbb{R}^3$ is a vector space over $\mathbb{R}$.

What is the set? All triples of real numbers: $(a, b, c)$ where $a, b, c \in \mathbb{R}$.

What are the operations?

• Addition: $(a_1, b_1, c_1) + (a_2, b_2, c_2) = (a_1+a_2, \; b_1+b_2, \; c_1+c_2)$
• Scalar multiplication: $k(a, b, c) = (ka, kb, kc)$

Verification of all 10 axioms:

• Axiom 1 (Closure +): If $\vec{u}, \vec{v} \in \mathbb{R}^3$, then $\vec{u} + \vec{v} = (u_1+v_1, u_2+v_2, u_3+v_3)$. Since the sum of two real numbers is real, every component is in $\mathbb{R}$, so the result is in $\mathbb{R}^3$ ✓
• Axiom 2 (Closure ·): $c\vec{v} = (cv_1, cv_2, cv_3)$. Product of reals is real, so the result is in $\mathbb{R}^3$ ✓
• Axiom 3 (Commutativity): $(u_1+v_1, u_2+v_2, u_3+v_3) = (v_1+u_1, v_2+u_2, v_3+u_3)$ because real number addition is commutative ✓
• Axiom 4 (Associativity): follows from associativity of real addition ✓
• Axiom 5 (Zero vector): $\vec{0} = (0, 0, 0)$ works because $(a,b,c) + (0,0,0) = (a,b,c)$ ✓
• Axiom 6 (Inverse): $-\vec{v} = (-v_1, -v_2, -v_3)$, and $\vec{v} + (-\vec{v}) = (0,0,0) = \vec{0}$ ✓
• Axioms 7–10: Follow from properties of real number arithmetic (distribution, associativity of multiplication, $1 \cdot a = a$). ✓

Conclusion: All 10 axioms hold. $\mathbb{R}^3$ is a vector space. ✓

Claim: $P_n = \lbrace a_0 + a_1x + a_2x^2 + \cdots + a_nx^n : a_i \in \mathbb{R}\rbrace$ is a vector space over $\mathbb{R}$.

What is the set? All polynomials with real coefficients and degree at most $n$.

What are the operations?

• Addition: add polynomials in the usual way (add corresponding coefficients)
• Scalar multiplication: multiply every coefficient by the scalar

Example with $P_2$: Let $p(x) = 3 + 2x - x^2$ and $q(x) = 1 - 4x + 5x^2$.

• $p + q = 4 - 2x + 4x^2$ (degree $\leq 2$ ✓, still in $P_2$)
• $2p = 6 + 4x - 2x^2$ (degree $\leq 2$ ✓, still in $P_2$)

Key axioms:

• Closure +: Sum of two polynomials of degree $\leq n$ has degree $\leq n$ ✓
• Closure ·: Scalar times a polynomial of degree $\leq n$ has degree $\leq n$ ✓
• Zero vector: The zero polynomial $0 + 0x + \cdots + 0x^n = 0$ ✓
• Inverse: $-(a_0 + a_1 x + \cdots) = (-a_0) + (-a_1)x + \cdots$ ✓
• Remaining axioms follow from polynomial arithmetic ✓

Claim: The set of all $2 \times 2$ real matrices is a vector space.

What is the set? All matrices of the form:

Operations: The usual matrix addition and scalar multiplication:

• Closure +:

✓ (sum of reals is real, so each entry is real)

• Zero vector:

✓ (the zero matrix plays the role of $\vec{0}$)

Claim: $C(\mathbb{R}) = \lbrace f : \mathbb{R} \to \mathbb{R} \mid f \text{ is continuous}\rbrace$ is a vector space.

What is the set? All continuous functions from $\mathbb{R}$ to $\mathbb{R}$ (e.g., $\sin x$, $e^x$, $x^3 + 1$, …).

Operations:

• Addition: $(f + g)(x) = f(x) + g(x)$ — add the function values at each point
• Scalar multiplication: $(cf)(x) = c \cdot f(x)$ — scale every output by $c$

Key axioms:

• Closure +: Sum of continuous functions is continuous (from calculus!) ✓
• Closure ·: Scalar multiple of a continuous function is continuous ✓
• Zero vector: $f(x) = 0$ for all $x$ (the function that is always zero) ✓

Note: This is an infinite-dimensional vector space! Unlike $\mathbb{R}^n$, you can’t describe every continuous function with finitely many numbers.

In $\mathbb{R}^3$, any vector is fully described by 3 numbers: $(a, b, c)$. But a continuous function like $\sin x$ has an output for every real number — infinitely many “coordinates.” No finite list of basis functions can span all of $C(\mathbb{R})$. That’s what “infinite-dimensional” means: you’d need infinitely many basis vectors to express every element.

Analogy: Think of $\mathbb{R}^n$ as a box with $n$ dials — turn each dial to build any vector. For $C(\mathbb{R})$, you’d need a box with infinitely many dials — one for each “direction” in function space.

Not a vector space: The set of all polynomials of degree exactly 2.

Why? Let $p(x) = x^2$ and $q(x) = -x^2 + 1$. Then $p + q = 1$ (degree 0, not 2). Fails closure under addition.

Not a vector space: $\mathbb{Z}^n$ (vectors with integer components) over $\mathbb{R}$.

Why? $\frac{1}{2} \cdot (1, 0) = (0.5, 0)$ has non-integer component. Fails closure under scalar multiplication.

Not a vector space: $\lbrace(x, y) : x + y = 1\rbrace$ (a line not through origin).

Why? $\vec{0} = (0,0)$ doesn’t satisfy $0 + 0 = 1$. Fails zero vector axiom.

Not a vector space: The set of all invertible matrices of order $n \times n$.

Why? The zero matrix $O$ is not invertible (it has no inverse), so the set has no additive identity. Fails the zero vector axiom.

Claim: $W$ is a subspace of $\mathbb{R}^2$.

1. Zero vector: $(0, 0)$? Check: $0 = 2(0)$? Yes. ✓
2. Closed under +: If $(x_1, 2x_1), (x_2, 2x_2) \in W$, then sum = $(x_1+x_2, 2x_1+2x_2) = (x_1+x_2, 2(x_1+x_2)) \in W$. ✓
3. Closed under ·: $c(x, 2x) = (cx, 2cx) \in W$. ✓

Conclusion: YES, $W$ is a subspace. Geometrically, it’s the line $y = 2x$ through the origin.

Claim: $W$ is NOT a subspace.

1. Zero vector: $(0, 0) \in W$? Check: $0 = 2(0) + 1 = 1$? NO!

Conclusion: NOT a subspace — fails immediately at condition 1. The line $y = 2x + 1$ doesn’t pass through the origin.

1. Zero: $0 + 0 - 0 = 0$ ✓
2. Closed +: $(x_1+x_2) + (y_1+y_2) - (z_1+z_2) = (x_1+y_1-z_1) + (x_2+y_2-z_2) = 0 + 0 = 0$ ✓
3. Closed ·: $c(x+y-z) = c \cdot 0 = 0$ ✓

Conclusion: Subspace — it’s the plane $x + y - z = 0$ through the origin.

1. $A\vec{0} = \vec{0}$, so $\vec{0} \in N(A)$. ✓
2. If $A\vec{u} = \vec{0}$ and $A\vec{v} = \vec{0}$, then $A(\vec{u} + \vec{v}) = A\vec{u} + A\vec{v} = \vec{0}$. ✓
3. $A(c\vec{v}) = cA\vec{v} = c\vec{0} = \vec{0}$. ✓

This is a very important result in linear algebra! The solution space of any homogeneous system is always a subspace.

1. Zero polynomial: $p(x) = 0$ satisfies $p(0) = 0$. ✓
2. If $p(0) = 0$ and $q(0) = 0$, then $(p+q)(0) = p(0)+q(0) = 0$. ✓
3. $(cp)(0) = c \cdot p(0) = c \cdot 0 = 0$. ✓

These are polynomials with no constant term: $p(x) = a_1x + a_2x^2 + a_3x^3$.

$\text{Span}\lbrace(1,2), (2,4)\rbrace$

Since $(2,4) = 2(1,2)$, both point in the same direction:

Result: A line through origin with direction $(1,2)$ — a 1D subspace.

$\text{Span}\lbrace(1,0), (0,1)\rbrace$

For any $(x, y) \in \mathbb{R}^2$, choose $c_1 = x, c_2 = y$.

Result: $\text{Span}\lbrace(1,0), (0,1)\rbrace = \mathbb{R}^2$ — the entire plane!

$\text{Span}\lbrace(1,0,0), (0,1,0), (1,1,0)\rbrace$

Since $(1,1,0) = (1,0,0) + (0,1,0)$, the third vector is redundant.

All combinations give $(c_1+c_3, c_2+c_3, 0)$ — always has $z = 0$.

Result: The $xy$-plane — a 2D subspace of $\mathbb{R}^3$.

$\text{Span}\lbrace 1, x, x^2\rbrace$

Result: $\text{Span}\lbrace 1, x, x^2\rbrace = P_2$ — all polynomials of degree $\leq 2$.

Problem: Express $(3, 7, -4)$ as a linear combination of $\vec{u}_1 = (1, 2, 3)$, $\vec{u}_2 = (2, 3, 7)$, $\vec{u}_3 = (3, 5, 6)$.

Set up $\alpha_1 \vec{u}_1 + \alpha_2 \vec{u}_2 + \alpha_3 \vec{u}_3 = (3, 7, -4)$:

Form the augmented matrix and row-reduce:

$\rho(A) = \rho(A \mid B) = 3$ — unique solution exists.

Back-substitution: $\alpha_3 = 3$, $\alpha_2 = -4$, $\alpha_1 = 2$.

Remark: This problem is equivalent to solving a system $A\vec{x} = \vec{b}$ where the given vectors are the columns of $A$ and the target vector is $\vec{b}$. No solution means $\vec{b}$ is not a linear combination of the columns.

Problem: Show that $\lbrace\vec{w}_1, \vec{w}_2, \vec{w}_3\rbrace = \lbrace(1, 1, 1), (1, 1, 0), (1, 0, 0)\rbrace$ spans $\mathbb{R}^3$.

For any $(a, b, c) \in \mathbb{R}^3$, we need to find $x, y, z$ such that:

This gives:

Solving: $x = c$, $y = b - c$, $z = a - b$.

Since $x, y, z$ exist for every $(a, b, c) \in \mathbb{R}^3$, the set spans $\mathbb{R}^3$. $\checkmark$

$c_1(1,2) + c_2(2,4) = (0,0)$

System: $c_1 + 2c_2 = 0$ and $2c_1 + 4c_2 = 0$

From first: $c_1 = -2c_2$. Second equation: $2(-2c_2) + 4c_2 = 0$ ✓ for all $c_2$.

Non-trivial solution: Choose $c_2 = 1 \Rightarrow c_1 = -2$.

Check: $-2(1,2) + 1(2,4) = (-2,-4)+(2,4) = (0,0)$ ✓

Linearly dependent — indeed $(2,4) = 2(1,2)$.

$c_1(1,0) + c_2(0,1) = (0,0) \Rightarrow c_1 = 0, c_2 = 0$.

Only the trivial solution. Linearly independent!

Test $\lbrace(1,0,0), (0,1,0), (1,1,0)\rbrace$:

$c_1(1,0,0) + c_2(0,1,0) + c_3(1,1,0) = (0,0,0)$

System: $c_1 + c_3 = 0$, $c_2 + c_3 = 0$, $0 = 0$

Two equations, three unknowns — $c_3$ is free. Let $c_3 = 1$: $c_1 = -1, c_2 = -1$.

Dependent: $(1,1,0) = (1,0,0) + (0,1,0)$ — third vector is sum of first two!

Test $\lbrace(1,0,0), (0,1,0), (0,0,1)\rbrace$:

$c_1 = 0, c_2 = 0, c_3 = 0$ — only trivial solution.

Linearly independent — these are the standard basis vectors of $\mathbb{R}^3$.

Test $\lbrace1, x, 2+2x\rbrace$ in $P_1$:

$c_1(1) + c_2(x) + c_3(2+2x) = 0$

$(c_1 + 2c_3) + (c_2 + 2c_3)x = 0 + 0 \cdot x$

Comparing: $c_1 + 2c_3 = 0$ and $c_2 + 2c_3 = 0$.

Choose $c_3 = 1$: $c_1 = -2, c_2 = -2$.

Dependent: $2 + 2x = 2(1) + 2(x)$

Example: Test columns of

Rank = 2 (two pivots), but 3 vectors → $2 < 3$ → Dependent.

$B = \lbrace(1,0), (0,1)\rbrace$

Spanning: $(x, y) = x(1,0) + y(0,1)$ for any $(x,y)$. ✓

Independent: $c_1(1,0) + c_2(0,1) = (0,0) \Rightarrow c_1 = c_2 = 0$. ✓

Coordinates: $(3, 5)$ has coordinates $(3, 5)$ relative to $B$.

$B = \lbrace(1,1), (1,-1)\rbrace$

Spanning: Solve $c_1(1,1) + c_2(1,-1) = (a,b)$:

Unique solution for any $(a,b)$. ✓

Independent: Set $a = b = 0 \Rightarrow c_1 = c_2 = 0$. ✓

Example coordinates: $(3,1)$ has coordinates $c_1 = 2, c_2 = 1$, i.e., $(3,1) = 2(1,1) + 1(1,-1)$.

Why $c_1 = 2, c_2 = 1$ and not $(3,1)$?
In the standard basis $\lbrace(1,0),(0,1)\rbrace$, the vector $(3,1)$ is $3(1,0) + 1(0,1)$ — its coordinates are simply $3$ and $1$. But with respect to basis $B = \lbrace(1,1),(1,-1)\rbrace$, we must express it as a combination of those vectors. Plugging into the formulas: $c_1 = \frac{3+1}{2} = 2$, $c_2 = \frac{3-1}{2} = 1$. Verify: $2(1,1) + 1(1,-1) = (2,2) + (1,-1) = (3,1)$ ✓. The same vector has different coordinates depending on which basis you use — $c_1, c_2$ are the weights (scalars) you multiply each basis vector by to reconstruct the original vector.

$B = \lbrace1, x, x^2\rbrace$

Any $p(x) = a_0 + a_1x + a_2x^2 = a_0(1) + a_1(x) + a_2(x^2)$. ✓

Independent: $c_1 + c_2x + c_3x^2 = 0$ for all $x$ forces $c_1 = c_2 = c_3 = 0$. ✓

$(3 + 2x - x^2)$ has coordinates $(3, 2, -1)$.

Problem: Is $B = \lbrace(1,0,1), (0,1,1), (1,1,0)\rbrace$ a basis of $\mathbb{R}^3$?

Form the matrix with these as columns and compute the determinant:

Non-zero determinant → independent → spans $\mathbb{R}^3$ (3 independent vectors in 3D).

Yes, $B$ is a basis of $\mathbb{R}^3$.

Problem: Find a basis for $W = \lbrace(x,y,z) \in \mathbb{R}^3 : x + y + z = 0\rbrace$.

Parametrize: $z = -x - y$, so every element is:

$(x, y, -x-y) = x(1, 0, -1) + y(0, 1, -1)$

Basis: $B = \lbrace(1, 0, -1), (0, 1, -1)\rbrace$

These are independent (not proportional). They span $W$.

Two non-zero rows → $\text{rank}(A) = 2$. Both columns are independent (full column rank).

One non-zero row → $\text{rank}(B) = 1$. All rows are scalar multiples of Row 1 ($R_2 = 2R_1$, $R_3 = 1 \cdot R_1$) — lots of redundancy!

Already in REF. Two non-zero rows → $\text{rank}(C) = 2$. Since $\det(C) = 3 \neq 0$, the matrix is invertible (full rank).

Already in REF. Three non-zero rows → $\text{rank}(D) = 3$. Full row rank ($3 \times 4$ matrix, max rank is 3).

Example: For the following matrix, row reduce:

• $\text{rank}(A) = 1$ (one pivot)
• $\text{nullity}(A) = 3 - 1 = 2$ (two free variables)
• Verification: $1 + 2 = 3 = n$ ✓

Full Column Rank example ($3 \times 2$, rank $= 2 = n$):

Both columns are pivot columns → no free variables → at most one solution. For $b = (1, 2, 0)^T$: unique solution $x = (1, 2)^T$. For $b = (1, 2, 5)^T$: no solution (last row gives $0 = 5$).

Full Row Rank example ($2 \times 3$, rank $= 2 = m$):

Every row has a pivot → system is consistent for every $b$. But column 3 is free → infinitely many solutions. E.g., $b = (4, 6)^T$: $x_3 = t$, $x_1 = 4 - 2t$, $x_2 = 6 - 3t$.

Full Rank example ($2 \times 2$, rank $= 2 = m = n$):

$\det(A) = 7 - 6 = 1 \neq 0$, so $A$ is invertible. For any $b$, there is exactly one solution: $x = A^{-1}b$.

Problem: Find basis of row space and column space of:

Row reduce:

$\rho(A) = 3$.

Basis of row space: $\lbrace(1, 3, 1, 2),\; (0, -5, -4, -1),\; (0, 0, 0, 2)\rbrace$ — the non-zero rows of the echelon form.

Basis of column space: Pivot columns are $C_1, C_2, C_4$ (columns 1, 2, 4). Take these columns from the original matrix $A$:

Problem: Let $W$ be the subspace of $\mathbb{R}^4$ spanned by $\vec{u}_1 = (1, -2, 5, -3)$, $\vec{u}_2 = (2, 3, 1, -4)$, $\vec{u}_3 = (3, 8, -3, -5)$.

(i) Find a basis and dimension of $W$.

(ii) Extend the basis of $W$ to a basis of $\mathbb{R}^4$.

Part (i): Form the matrix with these as rows and row-reduce:

Two non-zero rows $\Rightarrow$ $\dim(W) = 2$.

Basis of $W$: $\lbrace(1, -2, 5, -3),\; (0, 7, -9, 2)\rbrace$.

Part (ii): To extend to a basis of $\mathbb{R}^4$, we need 2 more vectors. Append the standard basis vectors $\vec{e}_1, \vec{e}_2, \vec{e}_3, \vec{e}_4$ and check which ones are independent of $W$:

Row-reduce to verify this has rank 4. Indeed it does.

Basis of $\mathbb{R}^4$: $\lbrace(1, -2, 5, -3),\; (0, 7, -9, 2),\; (0, 0, 1, 0),\; (0, 0, 0, 1)\rbrace$.

Example: $\vec{u} = (1, 2, 3), \vec{v} = (4, -1, 2)$

$\lVert(3, 4)\rVert = \sqrt{9 + 16} = 5$

$\lVert(1, 1, 1)\rVert = \sqrt{3}$

Example:

• $(1, 0)$ and $(0, 1)$ are orthonormal: dot product = 0, both have norm 1. ✓
• $(1, 1)$ and $(1, -1)$ are orthogonal: $1(1) + 1(-1) = 0$. ✓ (But not orthonormal — need to normalize.)
• To normalize $(1,1)$: $\frac{(1,1)}{\lVert(1,1)\rVert} = \frac{(1,1)}{\sqrt{2}} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

Concrete Example — Messy vs. Orthonormal Basis

Consider $\mathbb{R}^2$. Suppose you have the basis:

These vectors are not perpendicular ($\vec{v}_1 \cdot \vec{v}_2 = 6 + 2 = 8 \neq 0$) and not unit length ($\lVert \vec{v}_1 \rVert = \sqrt{5}$, $\lVert \vec{v}_2 \rVert = \sqrt{13}$).

Task: Express $\vec{v} = (5, 3)$ in this basis.

With the messy basis $B$, you must solve the system:

Row reducing: $c_1 = 1, \, c_2 = 1$. That took real work (setting up equations, elimination).

Now compare with an orthonormal basis. The standard basis is:

which is orthonormal.

The coordinates are instant — just dot products:

No equations to solve. This is the power of orthonormal bases.

Now imagine doing this not in $\mathbb{R}^2$ but in $\mathbb{R}^{100}$ or an infinite-dimensional function space — the messy approach requires solving 100×100 systems, while the orthonormal approach is still just dot products!

Problem: Apply Gram-Schmidt to $\lbrace\vec{v}_1 = (1,0,0), \; \vec{v}_2 = (1,1,0), \; \vec{v}_3 = (1,1,1)\rbrace$.

Step 1:

Step 2:

Step 3:

Orthonormal basis: $\lbrace(1,0,0), (0,1,0), (0,0,1)\rbrace$ — the standard basis!

(This makes sense: the input vectors were already “staircase-shaped.”)

Problem: Apply Gram-Schmidt to $\lbrace\vec{v}_1 = (1, 1, 0), \; \vec{v}_2 = (1, 0, 1)\rbrace$.

Step 1:

Step 2:

Verification:

• $\langle \vec{u}_1, \vec{u}_2 \rangle = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{6}} + \frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{6}}\right) + 0 = 0$ ✓ (orthogonal)
• $\lVert\vec{u}_1\rVert = 1, \lVert\vec{u}_2\rVert = 1$ ✓ (unit length)

Problem: Apply Gram-Schmidt to $\lbrace1, x\rbrace$ in $P_1$ with inner product $\langle f, g \rangle = \int_0^1 f(x)g(x) \, dx$.

Why are $1$ and $x$ not already orthogonal on $[0,1]$?

So $1$ and $x$ are NOT orthogonal — the constant function and the linear function have “overlap” on $[0,1]$.

Step 1 — Start with $f_1(x) = 1$, normalize:

Step 2 — Make $x$ orthogonal to $p_1 = 1$:

Compute the projection of $x$ onto $p_1 = 1$:

Subtract the projection:

What is $q(x) = x - \frac{1}{2}$ geometrically? It’s the part of $x$ that the constant function $1$ cannot capture. It’s a line that passes through zero at $x = \frac{1}{2}$ — the “purely varying” part of $x$ after removing its average value on $[0,1]$.

Verify orthogonality: $\langle q, p_1 \rangle = \int_0^1 (x - \frac{1}{2}) \cdot 1 \, dx = \left[\frac{x^2}{2} - \frac{x}{2}\right]_0^1 = (\frac{1}{2} - \frac{1}{2}) - 0 = 0 \; \checkmark$

Normalize:

Orthonormal basis for $P_1$ on $[0,1]$: $\left\lbrace 1, \; 2\sqrt{3}\left(x - \frac{1}{2}\right) \right\rbrace$

Problem: Apply Gram-Schmidt to $\lbrace1, x, x^2\rbrace$ in $P_2$ with $\langle f, g \rangle = \int_0^1 f(x)g(x) \, dx$.

This produces the (shifted) Legendre polynomials on $[0,1]$ — one of the most important polynomial families in mathematics.

Step 1: $p_1(x) = 1$ (same as before, $\lVert 1 \rVert = 1$)

Step 2: $p_2(x) = 2\sqrt{3}\left(x - \frac{1}{2}\right)$ (same as Example 3a)

Step 3 — Make $x^2$ orthogonal to both $p_1$ and $p_2$:

Compute projections of $x^2$ onto each existing basis vector:

Subtract both projections:

Wait — let’s be more careful:

Verify: $\langle q, p_1 \rangle = \int_0^1 (x^2 - x + \frac{1}{6}) dx = \frac{1}{3} - \frac{1}{2} + \frac{1}{6} = 0 \; \checkmark$

Normalize:

Orthonormal basis for $P_2$ on $[0,1]$:

Why would we want this?

• Computation: Polynomials are trivial to evaluate (just multiply and add). Sine, exponentials, and other complex functions are expensive. Calculators internally use polynomial approximations.
• Analysis: Polynomials are easy to differentiate, integrate, and manipulate algebraically.
• Data fitting: Given measured data points, we often want a simple formula (polynomial) that captures the overall trend.

What does “close enough” mean?

We measure closeness using an inner product. On an interval $[a, b]$:

A smaller integral means a better approximation. The best degree-$n$ approximation is the one that minimizes this integral — and orthogonal polynomials give us an elegant way to find it term by term.

What to notice in Graph 1:

• The blue line ($x$) and green parabola ($x^2$) both start at $0$ and rise to $1$ — they move in the “same direction” over most of $[0,1]$.
• If you were approximating some function $f$ and first used $1$ and $x$, a big chunk of what $x^2$ does is redundant — it partially duplicates the upward trend that $x$ already described.
• Mathematically: $\langle x, x^2 \rangle = \int_0^1 x \cdot x^2\, dx = \frac{1}{4} \neq 0$ — these are not orthogonal. Their inner product is nonzero, meaning they overlap.
• Consequence: When you add $x^2$ to an approximation $a + bx$, the coefficients $a$ and $b$ must change to redistribute what each term is responsible for. The terms are entangled.

What to notice in Graph 2:

• Red ($q_1 = 1$): A flat horizontal line. It can only shift an approximation up or down — it captures the average height of a function.

• Blue ($q_2 = x - \frac{1}{2}$): A line that goes from $-\frac{1}{2}$ to $+\frac{1}{2}$. Notice it’s perfectly balanced around zero — it spends equal time above and below the $x$-axis. This means $\int_0^1 q_2\, dx = 0$, so it’s orthogonal to the constant $q_1$. It captures whether a function tilts left-to-right — something a constant can never do.

• Green ($q_3 = x^2 - x + \frac{1}{6}$): A U-shaped curve that dips below zero in the middle and curves up at both ends. Crucially:
  • $\int_0^1 q_3 \cdot 1\, dx = 0$ — it has zero average (orthogonal to $q_1$)
  • $\int_0^1 q_3 \cdot (x - \frac{1}{2})\, dx = 0$ — it has zero tilt (orthogonal to $q_2$)
  • It captures pure bowing/curvature — the U-shape that no combination of flat lines and tilted lines can produce.
• Key observation: Compare this graph with Graph 1. In Graph 1, the blue and green curves both rise together. Here, each curve oscillates in a fundamentally different pattern. They are geometrically perpendicular in function space.

How to read Graph 3 — point by point:

At every $x$-value:

Pick a concrete point to verify — say $x = 0.5$:

• Blue: $(0.5)^2 = 0.25$
• Red: $0.5 - \frac{1}{6} \approx 0.333$
• Green: $0.25 - 0.5 + 0.167 \approx -0.083$
• Check: $0.333 + (-0.083) = 0.25$ ✓

What does this decomposition mean?

• The red line is the “shadow” (projection) of $x^2$ onto the space of degree-$\leq 1$ polynomials. It’s the best approximation of $x^2$ you can get using only constants and straight lines. Think of it like compressing a photo — you keep the rough trend but lose the fine details.

• The green curve is the “error” or “residual” — everything about $x^2$ that a straight line simply cannot express. This is genuine degree-2 behavior: the curvature, the bowing. No matter what constants $a, b$ you choose, $a + bx$ will never look like that U-shape.

• This is exactly the “variation that no lower-degree polynomial can capture.”

What Graph 4 shows:

We approximate $f(x) = \sin(\pi x)$ by building up one orthogonal polynomial at a time:

1. Red (constant only): $c_1 q_1 = 0.637$ — just the average value of $\sin(\pi x)$ on $[0,1]$. A flat line at $\approx \frac{2}{\pi}$.

2. Blue (constant + linear): Adding $c_2 q_2$ tilts the line. Since $\sin(\pi x)$ is symmetric on $[0,1]$, the tilt is nearly zero — the blue line barely differs from the red.

3. Green (constant + linear + quadratic): Adding $c_3 q_3$ introduces curvature. Now the approximation starts to bend and follow the hump of the sine curve.

The crucial point: When we went from red → blue, the coefficient $c_1 = 0.637$ did not change. When we went from blue → green, neither $c_1$ nor $c_2$ changed. Each new term adds a layer without disturbing previous layers. This is only possible because the basis is orthogonal.

With the standard basis $\lbrace1, x, x^2\rbrace$, all coefficients would reshuffle at every step — making the approximation harder to interpret and compute.

Quick Example: Let $B = \lbrace(1,0), (0,1)\rbrace$ be the standard basis of $\mathbb{R}^2$ and $\vec{w} = (3, 7)$.

We need $c_1, c_2$ such that $c_1(1,0) + c_2(0,1) = (3,7)$. Clearly $c_1 = 3, \; c_2 = 7$.

In the standard basis, the coordinate vector is just the vector itself. But with a different basis (say $B’ = \lbrace(1,1), (1,-1)\rbrace$), the same vector $(3,7)$ would have different coordinates — you’d solve $c_1(1,1) + c_2(1,-1) = (3,7)$ to get $c_1 = 5, \; c_2 = -2$, so $[\vec{w}]_{B’} = \begin{pmatrix}5 \ -2\end{pmatrix}$.

Same vector, different “address” depending on the basis you use — just like the same location has different GPS coordinates depending on the map projection.

Basis: $B = \lbrace(1,0), (0,1)\rbrace$ (standard basis of $\mathbb{R}^2$)

Vector: $\vec{w} = (5, 3)$

Since $(5,3) = 5(1,0) + 3(0,1)$:

Observation: In the standard basis, the coordinate vector is just the vector itself! This is why we usually don’t write $[\vec{w}]_B$ when using the standard basis — coordinates and components coincide.

Basis: $B = \lbrace\vec{v}_1 = (1,1), \vec{v}_2 = (1,-1)\rbrace$

Vector: $\vec{w} = (5, 3)$

Find $c_1, c_2$ such that $c_1(1,1) + c_2(1,-1) = (5,3)$:

Adding: $2c_1 = 8 \Rightarrow c_1 = 4$. Subtracting: $2c_2 = 2 \Rightarrow c_2 = 1$.

Verification: $4(1,1) + 1(1,-1) = (4,4) + (1,-1) = (5,3)$ ✓

Key point: The same vector $(5,3)$ has coordinates $(5,3)$ in the standard basis but $(4,1)$ in basis $B$. The vector hasn’t changed — only the “language” we use to describe it.

Basis: $B = \lbrace(1,0,1), (0,1,1), (1,1,0)\rbrace$

Vector: $\vec{w} = (3, 2, 4)$

Solve $c_1(1,0,1) + c_2(0,1,1) + c_3(1,1,0) = (3,2,4)$:

From equations 1 and 2: $c_1 = 3 - c_3$ and $c_2 = 2 - c_3$.

Substituting into equation 3: $(3 - c_3) + (2 - c_3) = 4 \Rightarrow 5 - 2c_3 = 4 \Rightarrow c_3 = \frac{1}{2}$.

Then $c_1 = \frac{5}{2}$ and $c_2 = \frac{3}{2}$.

Verification: $\frac{5}{2}(1,0,1) + \frac{3}{2}(0,1,1) + \frac{1}{2}(1,1,0) = (\frac{5}{2},0,\frac{5}{2}) + (0,\frac{3}{2},\frac{3}{2}) + (\frac{1}{2},\frac{1}{2},0) = (3,2,4)$ ✓

Basis: $B = \lbrace1, x, x^2\rbrace$ (standard basis of $P_2$)

Vector: $p(x) = 3 - 2x + 5x^2$

Non-standard basis: $B’ = \lbrace1, (x-1), (x-1)^2\rbrace$

Expanding $p(x) = 3 - 2x + 5x^2$ in terms of $B’$:

Expanding the right side: $a_0 + a_1 x - a_1 + a_2 x^2 - 2a_2 x + a_2$

Comparing with $3 - 2x + 5x^2$:

• $a_2 = 5$
• $a_1 - 2(5) = -2 \Rightarrow a_1 = 8$
• $a_0 - 8 + 5 = 3 \Rightarrow a_0 = 6$

Verification: $6 + 8(x-1) + 5(x-1)^2 = 6 + 8x - 8 + 5x^2 - 10x + 5 = 3 - 2x + 5x^2$ ✓

Old basis: $B = \lbrace(1,1), (1,-1)\rbrace$

New basis: $B’ = \lbrace(1,0), (0,1)\rbrace = $ standard basis

Find $P_{B \to B’}$:

Since $B’$ is the standard basis, $[\vec{v}]_{B’} = \vec{v}$ for any $\vec{v}$. So:

(Just the old basis vectors as columns!)

Check: If $[\vec{x}]_B = (4, 1)^T$, then:

This matches Example 2 above: $(5,3)$ in standard coordinates ↔ $(4,1)$ in basis $B$ ✓

Basis $B$: $\lbrace(1,0), (1,1)\rbrace$

Basis $B’$: $\lbrace(1,2), (2,1)\rbrace$

Find $P_{B \to B’}$: Express each vector in $B$ as a linear combination of vectors in $B’$.

Set up $[B’ \mid B]$ and row reduce:

$R_2 \leftarrow R_2 - 2R_1$:

$R_2 \leftarrow -\frac{1}{3}R_2$:

$R_1 \leftarrow R_1 - 2R_2$:

Problem: Show that $f(t) = \sin t$, $g(t) = \cos t$, $h(t) = t$ are linearly independent in the vector space of functions from $\mathbb{R}$ to $\mathbb{R}$.

Suppose $x \sin t + y \cos t + z \cdot t = 0$ for all $t \in \mathbb{R}$.

Evaluate at specific values of $t$:

• $t = 0$: $x(0) + y(1) + z(0) = 0 \Rightarrow y = 0$
• $t = \pi/2$: $x(1) + y(0) + z(\pi/2) = 0 \Rightarrow x + z\pi/2 = 0$
• $t = \pi$: $x(0) + y(-1) + z\pi = 0 \Rightarrow z\pi = 0 \Rightarrow z = 0$

From the second equation with $z = 0$: $x = 0$.

Therefore $x = y = z = 0$ — the functions are linearly independent. $\checkmark$

Problem: Find a homogeneous system $AX = 0$ whose solution set is spanned by:

Approach: We need $\vec{v} = (x, y, z, t)$ to be in the solution space if and only if $\vec{v}$ is a linear combination of $\vec{u}_1, \vec{u}_2, \vec{u}_3$.

Form the matrix $M = [u_1^T \mid u_2^T \mid u_3^T \mid v^T]$ and require $\rho(M) = \rho$ of the left part:

Row-reduce and impose consistency. The conditions give the equations of the homogeneous system for which $\lbrace\vec{u}_1, \vec{u}_2, \vec{u}_3\rbrace$ is the solution basis.

After row reduction, the constraints on $(x, y, z, t)$ yield:

These are the two equations of the required homogeneous system.

Problem: In $M_{2 \times 2}(\mathbb{R})$, let $S = \left\lbrace E_1 = \begin{pmatrix}1&1\1&1\end{pmatrix}, E_2 = \begin{pmatrix}1&-1\1&0\end{pmatrix}, E_3 = \begin{pmatrix}1&1\0&0\end{pmatrix}, E_4 = \begin{pmatrix}1&0\0&0\end{pmatrix} \right\rbrace$ be a basis. Find $[A]_S$ where $A = \begin{pmatrix}3&-5\3&1\end{pmatrix}$.

Set $A = aE_1 + bE_2 + cE_3 + dE_4$:

Entry-wise:

Solving: $a = 1$, $b = 2$, $c = -8$, $d = 8$.

Problem: In $M_{2 \times 3}(\mathbb{R})$, determine whether $A = \begin{pmatrix}1&2&3\4&0&1\end{pmatrix}$, $B = \begin{pmatrix}1&3&-4\6&5&4\end{pmatrix}$, $C = \begin{pmatrix}3&8&-11\16&10&9\end{pmatrix}$ are linearly dependent.

Strategy: Convert each matrix to its coordinate vector relative to the standard basis of $M_{2 \times 3}$:

Form the matrix with these as rows and row-reduce:

$\rho = 2 < 3$ (number of vectors), so the coordinate vectors are LD.

Since the coordinate mapping is an isomorphism (preserves linear dependence): $A, B, C$ are linearly dependent. $\checkmark$

Problem: Express $v(t) = t^2 + 4t - 3$ as a linear combination of $p_1(t) = t^2 - 2t + 5$, $p_2(t) = 2t^2 - 3t$, $p_3(t) = t + 1$ in $P_2(t)$.

Convert to coordinate vectors: $(1, 4, -3)$ vs $(1, -2, 5)$, $(2, -3, 0)$, $(0, 1, 1)$.

Set up: $v = x \cdot p_1 + y \cdot p_2 + z \cdot p_3$, which gives:

$z = 2$, $y = 4$, $x = -7$.

Every matrix defines a linear transformation. If $A$ is an $m \times n$ matrix, then:

is linear.

Proof:

• $T(\vec{u} + \vec{v}) = A(\vec{u} + \vec{v}) = A\vec{u} + A\vec{v} = T(\vec{u}) + T(\vec{v})$ ✓
• $T(c\vec{u}) = A(c\vec{u}) = cA\vec{u} = cT(\vec{u})$ ✓

Example: Let $A = \begin{bmatrix} 2 & -1 \ 0 & 3 \end{bmatrix}$.

$T(1, 0) = (2, 0)$, $\;T(0, 1) = (-1, 3)$, $\;T(1, 1) = (1, 3) = T(1,0) + T(0,1)$ ✓

Counter-clockwise rotation by angle $\theta$:

Derivation: $R_\theta(\vec{e}1) = R\theta(1,0) = (\cos\theta, \sin\theta)$ and $R_\theta(\vec{e}2) = R\theta(0,1) = (-\sin\theta, \cos\theta)$.

$\theta = 90^\circ$:

$R_{90^\circ}(1,0) = (0,1)$ ✓, $R_{90^\circ}(0,1) = (-1,0)$ ✓, $R_{90^\circ}(3,4) = (-4,3)$.

Linearity check: Rotation preserves addition and scaling (rotating a sum equals the sum of rotations).

Projection onto the $x$-axis:

Matrix: $P = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}$

Linearity:

• $P(x_1+x_2, y_1+y_2) = (x_1+x_2, 0) = (x_1,0) + (x_2,0) = P(x_1,y_1)+P(x_2,y_2)$ ✓
• $P(cx, cy) = (cx, 0) = c(x, 0) = cP(x,y)$ ✓

Interesting property: $P^2 = P$ (projecting twice gives the same result — called idempotent).

Reflection about the $x$-axis:

Matrix: $T = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix}$

This is linear: $T(1,0) = (1,0)$, $T(0,1) = (0,-1)$.

Reflection about the line $y = x$:

Matrix: $T = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}$

Uniform scaling by factor $k$:

Matrix: $T = kI$ (scalar multiple of the identity).

For $k = 2$ in $\mathbb{R}^2$: $T = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}$. Every vector is doubled.

Non-uniform scaling: $T(x,y) = (2x, 3y)$ — stretches differently in $x$ and $y$.

Matrix: $T = \begin{bmatrix} 2 & 0 \ 0 & 3 \end{bmatrix}$.

Claim: $D: P_n \to P_{n-1}$ defined by $D(p) = p’$ (the derivative) is linear.

Proof:

• $D(p + q) = (p+q)’ = p’ + q’ = D(p) + D(q)$ ✓ (sum rule of differentiation)
• $D(cp) = (cp)’ = cp’ = cD(p)$ ✓ (constant multiple rule)

Examples:

• $D(3x^2 + 5x + 1) = 6x + 5$
• $D(x^3) = 3x^2$
• $D(7) = 0$ (constants map to the zero polynomial)

Matrix representation (with respect to standard bases $\lbrace 1, x, x^2\rbrace$ → $\lbrace 1, x\rbrace$):

$D(1) = 0 = 0 \cdot 1 + 0 \cdot x$, $\;D(x) = 1 = 1 \cdot 1 + 0 \cdot x$, $\;D(x^2) = 2x = 0 \cdot 1 + 2 \cdot x$.

$I: P_2 \to P_3$ defined by $I(p) = \int_0^x p(t)\,dt$ is linear.

Proof: Integral of a sum = sum of integrals, and integral of $cp$ = $c$ times integral of $p$.

$I(1) = x$, $\;I(x) = \frac{x^2}{2}$, $\;I(x^2) = \frac{x^3}{3}$.

NOT linear — Translation: $T(x,y) = (x+1, y+2)$

$T(0,0) = (1,2) \neq (0,0)$. Fails immediately!

NOT linear — Quadratic: $T(x) = x^2$

$T(2) = 4$ but $2T(1) = 2$. $T(2) \neq 2T(1)$. Fails homogeneity.

$T(-1 + 1) = T(0) = 0$ but $T(-1) + T(1) = 1 + 1 = 2$. Fails additivity.

NOT linear — Norm: $T(\vec{v}) = \lVert\vec{v}\rVert$

$T(2\vec{e}_1) = 2$ but $2T(\vec{e}_1) = 2$… wait, this works for this case. But try $T(\vec{e}_1 + \vec{e}_2) = \sqrt{2} \neq 1+1 = T(\vec{e}_1)+T(\vec{e}_2)$. Fails additivity.

Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ defined by $T(x,y,z) = (x+y, y+z)$.

Matrix: $A = \begin{bmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \end{bmatrix}$

Kernel: Solve $A\vec{x} = \vec{0}$:

$y = -x$, $z = -y = x$. So $\ker(T) = \lbrace t(1, -1, 1) : t \in \mathbb{R}\rbrace$.

$\dim(\ker(T)) = 1$. Basis: $\lbrace(1,-1,1)\rbrace$.

Image: The column space of $A$.

Columns: $(1,0)$ and $(1,1)$ are independent → $\text{Im}(T) = \mathbb{R}^2$ (full image).

$\dim(\text{Im}(T)) = 2$.

Rank-Nullity check: $\dim(\ker(T)) + \dim(\text{Im}(T)) = 1 + 2 = 3 = \dim(\mathbb{R}^3)$ ✓

$T: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $T(x,y) = (2x+y, x-y)$.

Standard bases: $B = B’ = \lbrace(1,0), (0,1)\rbrace$

$T(1,0) = (2,1)$, $\;T(0,1) = (1,-1)$

Non-standard basis example: Let $B = \lbrace(1,1), (1,-1)\rbrace$ and $B’ = \lbrace(1,0), (0,1)\rbrace$ (standard).

$T(1,1) = (2+1, 1-1) = (3, 0)$. In $B’$: $[T(1,1)]_{B’} = (3,0)^T$.

$T(1,-1) = (2-1, 1+1) = (1, 2)$. In $B’$: $[T(1,-1)]_{B’} = (1,2)^T$.

This is a different matrix representing the same transformation — just in different coordinates!

Example: Find $T: \mathbb{R}^2 \to \mathbb{R}^2$ if $T(1,0) = (3,1)$ and $T(0,1) = (2,-1)$.

Since $\lbrace(1,0), (0,1)\rbrace$ is a basis: for any $(x,y)$:

Matrix: $A = \begin{bmatrix} 3 & 2 \ 1 & -1 \end{bmatrix}$

Find $T(5,3)$: $T(5,3) = (15+6, 5-3) = (21, 2)$ ✓

Example (non-standard basis): Let $B = \lbrace(1,1), (1,-1)\rbrace$ be a basis of $\mathbb{R}^2$. Suppose $T(1,1) = (2,0)$ and $T(1,-1) = (0,4)$.

Find $T(3,1)$.

First, express $(3,1)$ in basis $B$: $(3,1) = 2(1,1) + 1(1,-1)$.

Then: $T(3,1) = 2T(1,1) + 1T(1,-1) = 2(2,0) + 1(0,4) = (4,4)$.

(i) $F: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $F(x, y) = (x - y, x - 2y)$.

Find $\ker(F)$: Set $F(x, y) = (0, 0)$:

$\ker(F) = \lbrace\vec{0}\rbrace$ $\Rightarrow$ $F$ is non-singular.

Find $F^{-1}$: Set $F(x, y) = (a, b)$:

Subtracting: $y = a - b$. Then $x = a + y = 2a - b$.

(ii) $G: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $G(x, y) = (2x - 4y, 3x - 6y)$.

Find $\ker(G)$: $2x = 4y$ and $3x = 6y$ $\implies$ $x = 2y$.

$\ker(G) = \lbrace(2y, y) : y \in \mathbb{R}\rbrace \neq \lbrace\vec{0}\rbrace$ $\Rightarrow$ $G$ is singular.

Non-zero vector in kernel: e.g., $(2, 1)$.

Let $F: \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $F(x, y, z) = (2x, y + z)$.

Let $G: \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $G(x, y, z) = (x + y, x - z)$.

(i) $(F + G)(x, y, z) = F(x,y,z) + G(x,y,z) = (2x, y+z) + (x+y, x-z) = (3x + y, x + y)$

(ii) $(3F)(x, y, z) = 3(2x, y+z) = (6x, 3y + 3z)$

(iii) $(2F - 5G)(x, y, z) = 2(2x, y+z) - 5(x+y, x-z) = (4x - 5x - 5y, 2y + 2z - 5x + 5z) = (-x - 5y, -5x + 2y + 7z)$

Let $F: \mathbb{R}^3 \to \mathbb{R}^2$ defined by $F(x, y, z) = (2x, y + z)$.

Let $G: \mathbb{R}^2 \to \mathbb{R}^3$ defined by $G(x, y) = (y, x, x + y)$.

Is $F \circ G$ defined? $G: \mathbb{R}^2 \to \mathbb{R}^3$ and $F: \mathbb{R}^3 \to \mathbb{R}^2$. So $F \circ G: \mathbb{R}^2 \to \mathbb{R}^2$. Yes, it is defined.

$(F \circ G)(x, y) = F(G(x, y)) = F(y, x, x+y) = (2y, x + x + y) = (2y, 2x + y)$

Is $G \circ F$ defined? $F: \mathbb{R}^3 \to \mathbb{R}^2$ and $G: \mathbb{R}^2 \to \mathbb{R}^3$. So $G \circ F: \mathbb{R}^3 \to \mathbb{R}^3$. Yes, also defined.

$(G \circ F)(x, y, z) = G(F(x,y,z)) = G(2x, y+z) = (y+z, 2x, 2x + y + z)$

Note: Even though both compositions exist, $F \circ G: \mathbb{R}^2 \to \mathbb{R}^2$ and $G \circ F: \mathbb{R}^3 \to \mathbb{R}^3$ have different domain/codomain, so they are different types of maps entirely!

$G: \mathbb{R}^2 \to \mathbb{R}^3$ defined by $G(x, y) = (x + y, x - 2y, 3x + y)$.

Find $\ker(G)$: $x + y = 0$, $x - 2y = 0$, $3x + y = 0$.

From $x + y = 0$ and $x - 2y = 0$: $y = -x$ and $x = 2y = -2x$, so $3x = 0$, hence $x = y = 0$.

$\ker(G) = \lbrace\vec{0}\rbrace$ $\Rightarrow$ $G$ is non-singular.

However, since $\dim(\mathbb{R}^2) = 2 \neq 3 = \dim(\mathbb{R}^3)$, $G$ cannot be an isomorphism (not surjective — image is a 2D subspace of $\mathbb{R}^3$).