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Matrix Representation of Linear Transformations

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Matrix Representation of Linear Transformations

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Matrix Representation of Linear Transformations

🎯 Learning Objectives

After this topic you will be able to:

Find the matrix of a linear transformation relative to given ordered bases
Compute the image of a vector using the matrix representation
Understand how the matrix changes when bases change
Compute change-of-basis matrices
Determine transition matrices between bases

📋 Prerequisites

Linear Transformations (Week 5) — definition and properties
Basis & Dimension (Week 4) — coordinates relative to a basis
Matrix Multiplication (Week 1) — matrices as linear maps

1. Intuition: Translating Between Languages

A linear transformation

T: V \to W

is an abstract map. To compute with it, we need to translate it into a matrix. But matrices are defined on

\mathbb{R}^n

, not on abstract vector spaces like

P_2

. The solution: choose bases. A basis lets us:

Represent any vector $\mathbf{v} \in V$ by its coordinate vector $[\mathbf{v}]_B \in \mathbb{R}^n$
Represent $T$ by a matrix $M$ so that $[T(\mathbf{v})]_C = M [\mathbf{v}]_B$ (Diagram)

2. Coordinates Relative to a Basis

2.1 Definition

B = \{\mathbf{b}_1, \dots, \mathbf{b}_n\}

is an ordered basis of

V

, then every

\mathbf{v} \in V

has a unique representation:

\mathbf{v} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + \cdots + c_n\mathbf{b}_n

The coordinate vector of

\mathbf{v}

relative to

B

is:

[\mathbf{v}]_B = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix} \in \mathbb{R}^n

Example 1: Coordinates

$B = \{1-x, 1+x, x^2\}$ is a basis for $P_2$ .

For $p(x) = 3 + x + 2x^2$ , find $[p]_B$ .

Solve $c_1(1-x) + c_2(1+x) + c_3 x^2 = 3 + x + 2x^2$ . Group: $(c_1 + c_2) + (-c_1 + c_2)x + c_3 x^2 = 3 + x + 2x^2$ .

> \begin{cases} c_1 + c_2 = 3 \\ -c_1 + c_2 = 1 \\ c_3 = 2 \end{cases} \Rightarrow c_1 = 1, c_2 = 2, c_3 = 2 >

> [p]_B = \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} >

.

3. Matrix of a Linear Transformation

Definition (Matrix of $T$ ). Let $T: V \to W$ be linear, $B = \{\mathbf{b}_1, \dots, \mathbf{b}_n\}$ an ordered basis of $V$ , and $C = \{\mathbf{c}_1, \dots, \mathbf{c}_m\}$ an ordered basis of $W$ . The matrix of $T$ relative to $B$ and $C$ is:

> [T]_{C \leftarrow B} = \begin{bmatrix} | & | & & | \\ [T(\mathbf{b}_1)]_C & [T(\mathbf{b}_2)]_C & \dots & [T(\mathbf{b}_n)]_C \\ | & | & & | \end{bmatrix} >

It satisfies: $[T(\mathbf{v})]_C = [T]_{C \leftarrow B} [\mathbf{v}]_B$ .

3.1 Algorithm

Compute $T(\mathbf{b}_j)$ for each basis vector $\mathbf{b}_j$ of $V$
Express $T(\mathbf{b}_j)$ in the basis $C$ of $W$ to get $[T(\mathbf{b}_j)]_C$
These coordinate vectors become the columns of $[T]_{C \leftarrow B}$

Example 2: Matrix of a transformation

$T: P_2 \to P_1$ , $T(p) = p'$ (derivative). $B = \{1, x, x^2\}$ (basis for $P_2$ ) $C = \{1, x\}$ (basis for $P_1$ )

> T(1) = 0 = 0\cdot1 + 0\cdot x \Rightarrow [T(1)]_C = \begin{bmatrix}0\\0\end{bmatrix} >

> T(x) = 1 = 1\cdot1 + 0\cdot x \Rightarrow [T(x)]_C = \begin{bmatrix}1\\0\end{bmatrix} >

> T(x^2) = 2x = 0\cdot1 + 2\cdot x \Rightarrow [T(x^2)]_C = \begin{bmatrix}0\\2\end{bmatrix} >

> [T]_{C \leftarrow B} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} >

Check: For $p(x) = a + bx + cx^2$ , $p'(x) = b + 2cx$ . $[p]_B = (a,b,c)^T$ .

> [T] [p]_B = \begin{bmatrix} b \\ 2c \end{bmatrix} >

= $[p']_C$ ✓ Example 3: Standard matrix = matrix relative to standard bases

$T: \mathbb{R}^3 \to \mathbb{R}^2$ , $T(x,y,z) = (2x-y, x+3y+z)$ . Using standard bases $B = \{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}$ , $C = \{\mathbf{e}_1,\mathbf{e}_2\}$ :

> [T]_{C \leftarrow B} = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 3 & 1 \end{bmatrix} >

(the standard matrix we found in Week 5).

4. Change of Basis

4.1 Change-of-Basis Matrix

If we have two bases

B

and

B'

of the same space

V

, how are coordinates related?

Definition (Change-of-Basis Matrix). The change-of-basis matrix from $B$ to $B'$ is:

> P_{B' \leftarrow B} = \begin{bmatrix} | & | & & | \\ [\mathbf{b}_1]_{B'} & [\mathbf{b}_2]_{B'} & \dots & [\mathbf{b}_n]_{B'} \\ | & | & & | \end{bmatrix} >

It satisfies: $[\mathbf{v}]_{B'} = P_{B' \leftarrow B} [\mathbf{v}]_B$ . Key fact: $P_{B' \leftarrow B} = (P_{B \leftarrow B'})^{-1}$ . Example 4: Change of basis in ℝ²

Standard basis $S = \{\mathbf{e}_1, \mathbf{e}_2\}$ , new basis $B = \{(1,2), (3,4)\}$ . Find $P_{B \leftarrow S}$ (from standard to $B$ ).

$\mathbf{e}_1 = (1,0)$ in $B$ : solve $c_1(1,2) + c_2(3,4) = (1,0)$ .

> \begin{cases} c_1 + 3c_2 = 1 \\ 2c_1 + 4c_2 = 0 \end{cases} \Rightarrow c_1 = -2, c_2 = 1 >

So

> [\mathbf{e}_1]_B = \begin{bmatrix} -2 \\ 1 \end{bmatrix} >

.

Similarly, $\mathbf{e}_2 = (0,1)$ in $B$ : solve $c_1(1,2) + c_2(3,4) = (0,1)$ . $c_1 = \frac{3}{2}, c_2 = -\frac{1}{2}$ .

> [\mathbf{e}_2]_B = \begin{bmatrix} 3/2 \\ -1/2 \end{bmatrix} >

.

> P_{B \leftarrow S} = \begin{bmatrix} -2 & 3/2 \\ 1 & -1/2 \end{bmatrix} >

.

4.2 Effect on the Matrix of a Transformation

T: V \to W

has matrix

[T]_{C \leftarrow B}

relative to bases

B, C

, and we change to bases

B', C'

, then:

[T]_{C' \leftarrow B'} = P_{C' \leftarrow C} \; [T]_{C \leftarrow B} \; P_{B \leftarrow B'}

where

P_{C' \leftarrow C}

is the change-of-basis in

W

and

P_{B \leftarrow B'}

is the change-of-basis in

V

. Special case:

T: V \to V

(operator) with

B = B'

and

C = C' = B'

[T]_{B'} = P^{-1} [T]_B P

where

P = P_{B' \leftarrow B}

. This is called a similarity transformation (Week 7).

5. Edge Cases & Gotchas

Situation	What Happens
Same basis for domain and codomain	$[T]_B = [T]_{B \leftarrow B}$ is a square matrix
Standard basis	Makes everything simpler — coordinates are just the vector itself
$V = W$ but different bases	$[T]_{B' \leftarrow B}$ is possible

6. Common Pitfalls

❌ Pitfall 1: Wrong order in $[T]_{C \leftarrow B}$

The notation means "from

B

C

[T(\mathbf{v})]_C = [T]_{C \leftarrow B} [\mathbf{v}]_B

❌ Pitfall 2: Confusing change-of-basis direction

P_{B' \leftarrow B}

converts from

B

B'

. It's

P_{B' \leftarrow B} [\mathbf{v}]_B = [\mathbf{v}]_{B'}

❌ Pitfall 3: Forgetting to convert the output basis

After computing

[T] [\mathbf{v}]_B

, the result is

[T(\mathbf{v})]_C

, not

T(\mathbf{v})

directly. Convert back using basis

C

7. Formula Summary Table

Concept	Formula
Matrix of $T$	$[T]_{C \leftarrow B} = [[T(\mathbf{b}_1)]_C \; \dots \; [T(\mathbf{b}_n)]_C]$
Coordinate transformation	$[T(\mathbf{v})]_C = [T]_{C \leftarrow B} [\mathbf{v}]_B$
Change-of-basis	$P_{B' \leftarrow B} = [[\mathbf{b}_1]_{B'} \; \dots \; [\mathbf{b}_n]_{B'}]$
Inverse relationship	$P_{B \leftarrow B'} = (P_{B' \leftarrow B})^{-1}$
Change of bases for $T$	$[T]_{C' \leftarrow B'} = P_{C' \leftarrow C} [T]_{C \leftarrow B} P_{B \leftarrow B'}$

8. 📝 Practice Questions

Q1: Matrix relative to non-standard bases

$T: \mathbb{R}^2 \to \mathbb{R}^2$ , $T(x,y) = (x+2y, 3x+4y)$ . $B = \{(1,1), (1,-1)\}$ , $C = \{(1,0), (0,1)\}$ . Find $[T]_{C \leftarrow B}$ .

Solution: $T(1,1) = (3,7)$ , $[T(1,1)]_C = (3,7)$ (standard basis). $T(1,-1) = (-1,-1)$ , $[T(1,-1)]_C = (-1,-1)$ .

> [T]_{C \leftarrow B} = \begin{bmatrix} 3 & -1 \\ 7 & -1 \end{bmatrix} >

. Q2: Matrix of a derivative operator with different bases

$T: P_2 \to P_1$ , $T(p) = p'$ . $B = \{1, 1+x, x+x^2\}$ , $C = \{1, x\}$ . Find $[T]_{C \leftarrow B}$ .

Solution: $T(1) = 0$ , $[T(1)]_C = (0,0)$ . $T(1+x) = 1$ , $[T(1+x)]_C = (1,0)$ . $T(x+x^2) = 1+2x$ , $[T(x+x^2)]_C = (1,2)$ .

> [T]_{C \leftarrow B} = \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 2 \end{bmatrix} >

. Q3: Change of basis

Find $P_{B \leftarrow S}$ where $B = \{(1,0,0), (1,1,0), (1,1,1)\}$ , $S$ = standard basis.

Solution: $\mathbf{e}_1 = 1\cdot(1,0,0)$ , so $[\mathbf{e}_1]_B = (1,0,0)$ . $\mathbf{e}_2 = -(1,0,0) + 1\cdot(1,1,0)$ , so $[\mathbf{e}_2]_B = (-1,1,0)$ . $\mathbf{e}_3 = 0\cdot(1,0,0) + (-1)(1,1,0) + 1\cdot(1,1,1)$ , so $[\mathbf{e}_3]_B = (0,-1,1)$ .

> P_{B \leftarrow S} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} >

. Q4: Using the matrix to compute

Using $[T]_{C \leftarrow B}$ from Q1, find $T(2,3)$ .

Solution: Find $[(2,3)]_B$ : solve $c_1(1,1)+c_2(1,-1) = (2,3)$ ⇒ $c_1+c_2=2$ , $c_1-c_2=3$ ⇒ $c_1=2.5$ , $c_2=-0.5$ .

> [(2,3)]_B = \begin{bmatrix}2.5 \\ -0.5\end{bmatrix} >

.

> [T(2,3)]_C = [T]_{C \leftarrow B} [(2,3)]_B = \begin{bmatrix} 3 & -1 \\ 7 & -1 \end{bmatrix} \begin{bmatrix}2.5 \\ -0.5\end{bmatrix} = \begin{bmatrix} 7.5+0.5 \\ 17.5+0.5 \end{bmatrix} = \begin{bmatrix} 8 \\ 18 \end{bmatrix} >

.

$T(2,3) = (8,18)$ (in standard coordinates). Check directly: $T(2,3) = (2+6, 6+12) = (8,18)$ ✓ Q5: Similarity transformation

$T: \mathbb{R}^2 \to \mathbb{R}^2$ has matrix

> A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} >

in standard basis. Find $[T]_B$ for $B = \{(1,1), (1,-1)\}$ .

Solution: $[T]_B = P^{-1} A P$ where $P = P_{B \leftarrow S}$ .

> P = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} >

(columns are basis vectors).

> P^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} >

.

> [T]_B = P^{-1}AP = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} >

> = \begin{bmatrix} 2 & 3 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 5 & -1 \\ -2 & 0 \end{bmatrix} >

. Q6: Finding transition matrix

$B = \{(1,0), (1,1)\}$ , $B' = \{(2,1), (1,2)\}$ . Find $P_{B' \leftarrow B}$ .

Solution: For each vector in $B$ , find coordinates in $B'$ .

$(1,0) = c_1(2,1) + c_2(1,2)$ : $2c_1 + c_2 = 1$ , $c_1 + 2c_2 = 0$ ⇒ $c_2 = -\frac{1}{3}$ , $c_1 = \frac{2}{3}$ .

> [(1,0)]_{B'} = \begin{bmatrix} 2/3 \\ -1/3 \end{bmatrix} >

.

$(1,1) = c_1(2,1) + c_2(1,2)$ : $2c_1 + c_2 = 1$ , $c_1 + 2c_2 = 1$ ⇒ $c_1 = \frac{1}{3}$ , $c_2 = \frac{1}{3}$ .

> [(1,1)]_{B'} = \begin{bmatrix} 1/3 \\ 1/3 \end{bmatrix} >

.

> P_{B' \leftarrow B} = \begin{bmatrix} 2/3 & 1/3 \\ -1/3 & 1/3 \end{bmatrix} >

. Q7: Matrix of identity

What is $[I]_{C \leftarrow B}$ where $I: V \to V$ is the identity map?

Solution: $I(\mathbf{b}_j) = \mathbf{b}_j$ . So $[I]_{C \leftarrow B}$ has columns $[\mathbf{b}_j]_C$ . This is exactly the change-of-basis matrix $P_{C \leftarrow B}$ ! Q8: Composition

$T: V \to W$ , $S: W \to U$ are linear. Prove $[S \circ T]_{D \leftarrow B} = [S]_{D \leftarrow C} [T]_{C \leftarrow B}$ .

Proof: For any $\mathbf{v} \in V$ : $[(S \circ T)(\mathbf{v})]_D = [S(T(\mathbf{v}))]_D = [S]_{D \leftarrow C} [T(\mathbf{v})]_C = [S]_{D \leftarrow C} [T]_{C \leftarrow B} [\mathbf{v}]_B$ .

So $[S \circ T]_{D \leftarrow B} = [S]_{D \leftarrow C} [T]_{C \leftarrow B}$ . ✓ Q9: Matrix of a projection

$T: \mathbb{R}^3 \to \mathbb{R}^3$ , $T(x,y,z) = (x,y,0)$ . Find $[T]_B$ where $B = \{(1,0,0), (1,1,0), (1,1,1)\}$ .

Solution: $T(1,0,0) = (1,0,0)$ . In $B$ : solve $(1,0,0) = c_1(1,0,0)+c_2(1,1,0)+c_3(1,1,1)$ ⇒ $c_1=1$ , $c_2=0$ , $c_3=0$ . $[T(1,0,0)]_B = (1,0,0)$ .

$T(1,1,0) = (1,1,0)$ = $0(1,0,0)+1(1,1,0)+0(1,1,1)$ . $[(1,1,0)]_B = (0,1,0)$ .

$T(1,1,1) = (1,1,0)$ = $0(1,0,0)+1(1,1,0)+0(1,1,1)$ . $[(1,1,1)]_B = (0,1,0)$ .

> [T]_B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} >

. Q10: Going back and forth

If

> [T]_{C \leftarrow B} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} >

and

> [\mathbf{v}]_B = \begin{bmatrix} 5 \\ 6 \end{bmatrix} >

, find $[T(\mathbf{v})]_C$ .

Solution:

> [T(\mathbf{v})]_C = [T]_{C \leftarrow B} [\mathbf{v}]_B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 \\ 6 \end{bmatrix} = \begin{bmatrix} 17 \\ 39 \end{bmatrix} >

.

🔗 Cross-References

Next topic: Equivalent & Similar Matrices
Week 7 (Similarity): $A$ and $P^{-1}AP$ represent the same operator in different bases
Week 8 (Orthogonal Transformations): Orthogonal matrices represent rotations/reflections
BSCS3004 (Deep Learning): Weight matrices in neural networks are linear transformations between feature spaces Join Discord Previous6.1 Kernel & ImageNext7.1 Equivalent & Similar Matrices