8.4 Matrix Representations of Transformations and Change of Basis

8.4 Matrix Representations of Transformations and Change of Basis

Every linear transformation between finite-dimensional vector spaces can be faithfully represented as a matrix once ordered bases are specified for both the domain and codomain. The GRE Subject Test tests your ability to construct transformation matrices, relate map composition to matrix multiplication, perform similarity transformations under a change of basis, and exploit similarity invariants (trace, determinant, characteristic polynomial, and rank).

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Construction of Matrix Representations

Let $V$ and $W$ be finite-dimensional vector spaces over $\mathbb{F}$ with ordered bases $\mathcal{B} = (v_1, \dots, v_n)$ and $\mathcal{C} = (w_1, \dots, w_m)$, respectively. Let $T: V \to W$ be a linear transformation. The matrix representation of $T$ with respect to bases $\mathcal{B}$ and $\mathcal{C}$, denoted $[T]_{\mathcal{C}, \mathcal{B}}$, is the $m \times n$ matrix whose columns are the $\mathcal{C}$-coordinate vectors of the images of the $\mathcal{B}$-basis vectors: $[T]_{\mathcal{C}, \mathcal{B}} = \begin{bmatrix} [T(v_1)]_\mathcal{C} & [T(v_2)]_\mathcal{C} & \dots & [T(v_n)]_\mathcal{C} \end{bmatrix} \in M_{m \times n}(\mathbb{F}).$

Fundamental Action Formula

For any vector $v \in V$, matrix multiplication reproduces transformation evaluation: $[T(v)]_\mathcal{C} = [T]_{\mathcal{C}, \mathcal{B}} [v]_\mathcal{B}.$

Linear Operators on a Single Space

When $T: V \to V$ is a linear operator and $\mathcal{B}$ is a single ordered basis for $V$, the operator matrix is denoted: $[T]_\mathcal{B} = [T]_{\mathcal{B}, \mathcal{B}} \in M_{n \times n}(\mathbb{F}).$

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Matrix Composition and Operator Invertibility

Matrix representations establish an algebraic isomorphism between the algebra of linear maps $\mathcal{L}(V, W)$ and the algebra of matrices $M_{m \times n}(\mathbb{F})$.

Composition as Matrix Multiplication

Let $T: V \to W$ and $S: W \to U$ be linear transformations with ordered bases $\mathcal{B}$ for $V$, $\mathcal{C}$ for $W$, and $\mathcal{D}$ for $U$. Then: $[S \circ T]_{\mathcal{D}, \mathcal{B}} = [S]_{\mathcal{D}, \mathcal{C}} [T]_{\mathcal{C}, \mathcal{B}}.$

• For linear operators on $V$, powers of transformations correspond directly to matrix powers: $[T^k]\mathcal{B} = ([T]\mathcal{B})^k$.
• A linear operator $T: V \to V$ is invertible if and only if $[T]_\mathcal{B}$ is an invertible matrix, with: $[T^{-1}]_\mathcal{B} = ([T]_\mathcal{B})^{-1}.$

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Change of Basis for Linear Operators and Matrix Similarity

How does the matrix representation of an operator $T: V \to V$ change when we replace ordered basis $\mathcal{B}$ with a new ordered basis $\mathcal{C}$?

The Similarity Transformation Formula

Let $P = P_{\mathcal{B} \leftarrow \mathcal{C}}$ be the transition matrix from $\mathcal{C}$ to $\mathcal{B}$, whose columns are $[w_j]_\mathcal{B}$ (where $\mathcal{C} = (w_1, \dots, w_n)$). Then: $[T]_\mathcal{C} = P^{-1} [T]_\mathcal{B} P = P_{\mathcal{C} \leftarrow \mathcal{B}} [T]_\mathcal{B} P_{\mathcal{B} \leftarrow \mathcal{C}}.$

Matrix Similarity

Two square matrices $A, B \in M_{n \times n}(\mathbb{F})$ are similar (written $A \sim B$) if there exists an invertible matrix $P \in M_{n \times n}(\mathbb{F})$ such that: $B = P^{-1} A P.$

• Geometric Meaning: $A$ and $B$ are similar if and only if they represent the exact same linear operator $T: V \to V$ under two different choices of basis.
• Similarity is an equivalence relation (reflexive, symmetric, and transitive).

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Similarity Invariants

Because similar matrices represent the same geometric transformation, any intrinsic coordinate-free property of the operator must be shared by all similar matrices. These shared quantities are similarity invariants.

Catalog of Preserved Invariants

If $B = P^{-1} A P$:

1. Trace: $\operatorname{tr}(B) = \operatorname{tr}(P^{-1}(AP)) = \operatorname{tr}((AP)P^{-1}) = \operatorname{tr}(A)$.
2. Determinant: $\det(B) = \det(P^{-1}) \det(A) \det(P) = \det(A)$.
3. Characteristic Polynomial: $p_B(\lambda) = \det(\lambda I - P^{-1}AP) = \det(P^{-1}(\lambda I - A)P) = \det(\lambda I - A) = p_A(\lambda).$
4. Eigenvalues and Algebraic Multiplicities: Identical roots of the characteristic polynomial.
5. Geometric Multiplicities: $\dim(\ker(B - \lambda I)) = \dim(\ker(A - \lambda I))$ for every eigenvalue $\lambda$.
6. Minimal Polynomial: $m_B(\lambda) = m_A(\lambda)$.
7. Rank and Nullity: $\operatorname{rank}(B) = \operatorname{rank}(A)$ and $\operatorname{nullity}(B) = \operatorname{nullity}(A)$.
8. Jordan Canonical Form: Identical up to permutation of Jordan blocks.

What Is NOT Invariant Under Similarity

• Individual Matrix Entries: The numbers in the arrays generally differ.
• Eigenvectors: If $v$ is an eigenvector of $A$ ($A v = \lambda v$), then $P^{-1} v$ is the corresponding eigenvector of $B$: $B(P^{-1} v) = (P^{-1} A P)(P^{-1} v) = P^{-1} A v = P^{-1} (\lambda v) = \lambda (P^{-1} v).$
• Row Space and Column Space: $\operatorname{col}(B) \neq \operatorname{col}(A)$ in general.

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Invariant vs Non-Invariant Quantities

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Step-by-Step Worked Problems

Problem 1: Matrix Representation of a Polynomial Operator

Let $T: P_2(\mathbb{R}) \to P_2(\mathbb{R})$ be defined by $T(p(x)) = (x + 1) p'(x)$. Find $[T]_\mathcal{B}$ with respect to the standard basis $\mathcal{B} = (1, x, x^2)$.

Solution:

1. Apply $T$ to each basis vector:
   • $T(1) = (x + 1)(0) = 0 = 0(1) + 0(x) + 0(x^2) \implies [T(1)]_\mathcal{B} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}$.
   • $T(x) = (x + 1)(1) = 1 + x = 1(1) + 1(x) + 0(x^2) \implies [T(x)]_\mathcal{B} = \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix}$.
   • $T(x^2) = (x + 1)(2x) = 2x + 2x^2 = 0(1) + 2(x) + 2(x^2) \implies [T(x^2)]_\mathcal{B} = \begin{pmatrix} 0 \ 2 \ 2 \end{pmatrix}$.
2. Assemble columns: $[T]_\mathcal{B} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix}.$
3. Notice that $[T]_\mathcal{B}$ is upper triangular! Its eigenvalues are immediately readable from the diagonal: $\lambda = 0, 1, 2$.

Problem 2: Exploiting Similarity Invariants

Let $A = \begin{pmatrix} 1 & 4 \ 2 & 3 \end{pmatrix}$. Compute $\operatorname{tr}(A^{10})$ without computing $A^{10}$ explicitly.

Solution:

1. Find eigenvalues of $A$:
   • $\operatorname{tr}(A) = 1 + 3 = 4$.
   • $\det(A) = (1)(3) - (4)(2) = 3 - 8 = -5$.
   • Characteristic equation: $\lambda^2 - 4\lambda - 5 = 0 \implies (\lambda - 5)(\lambda + 1) = 0$.
   • Eigenvalues are $\lambda_1 = 5$ and $\lambda_2 = -1$.
2. Since $A$ has distinct eigenvalues, $A$ is diagonalizable: $A = P D P^{-1}$ with $D = \begin{pmatrix} 5 & 0 \ 0 & -1 \end{pmatrix}$.
3. Then $A^{10} = P D^{10} P^{-1}$, which is similar to $D^{10}$.
4. By similarity invariance of trace: $\operatorname{tr}(A^{10}) = \operatorname{tr}(D^{10}) = 5^{10} + (-1)^{10} = 5^{10} + 1.$

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GRE Exam Traps & Pitfalls

Trap 1: The Necessary vs. Sufficient Condition of Similarity Having identical trace, determinant, and characteristic polynomial is necessary for similarity, but not sufficient. For example, $A = \begin{pmatrix} 3 & 0 \ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 1 \ 0 & 3 \end{pmatrix}$ have identical eigenvalues (3, 3), trace (6), determinant (9), and characteristic polynomial $(\lambda - 3)^2$, yet they are NOT similar because $A$ is scalar and only similar to itself.

Trap 2: Inverting Basis Transition Direction In the change-of-basis formula $[T]\mathcal{C} = P^{-1} [T]\mathcal{B} P$, $P$ must be the transition matrix from $\mathcal{C}$ to $\mathcal{B}$ ($P = P_{\mathcal{B} \leftarrow \mathcal{C}}$). Inverting the placement of $P$ and $P^{-1}$ yields the matrix with respect to a different basis.

Trap 3: Mixing Input and Output Bases When computing $[T]_{\mathcal{C}, \mathcal{B}}$, remember that inputs come from basis $\mathcal{B}$ while output coordinates must be decomposed along target basis $\mathcal{C}$.