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Change of Basis






Same Transformation, Different Matrix

The same linear transformation has different matrix representations in different bases. Changing the basis changes the numbers but not the map itself. The relationship between two matrix representations of the same transformation is similarity — and choosing the right basis is how difficult matrices become simple ones.



The Problem

A linear transformation T:VVT: V \to V is a fixed geometric object — it sends each vector to a definite image regardless of how coordinates are assigned. But the matrix that represents TT depends on the choice of basis. Different bases assign different coordinates to the same vectors, and the matrix that converts input coordinates to output coordinates changes accordingly.

This raises a natural question: if TT has matrix AA in one basis and matrix AA' in another, how are AA and AA' related? The answer is the similarity relation A=P1APA' = P^{-1}AP, where PP is the change-of-basis matrix. Understanding this relation is the key to choosing bases strategically — picking the basis that makes the matrix as simple as possible.

The Change-of-Basis Matrix

If B\mathcal{B} and C\mathcal{C} are two bases for VV, the change-of-basis matrix PCBP_{\mathcal{C} \leftarrow \mathcal{B}} converts B\mathcal{B}-coordinates to C\mathcal{C}-coordinates:

[v]C=PCB[v]B[\mathbf{v}]_\mathcal{C} = P_{\mathcal{C} \leftarrow \mathcal{B}} \, [\mathbf{v}]_\mathcal{B}


Column jj of PP is the C\mathcal{C}-coordinate vector of the jj-th basis vector of B\mathcal{B}. The reverse conversion uses the inverse: PBC=P1P_{\mathcal{B} \leftarrow \mathcal{C}} = P^{-1}.

Worked Example


In R2\mathbb{R}^2, let B={(1,1),(1,1)}\mathcal{B} = \{(1, 1), (1, -1)\} and let C\mathcal{C} be the standard basis. The C\mathcal{C}-coordinates of the B\mathcal{B}-basis vectors are just their components: (1,1)(1, 1) and (1,1)(1, -1). So

P=(1111)P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}


To find the B\mathcal{B}-coordinates of v=(5,1)\mathbf{v} = (5, 1): solve Pc=(5,1)P\mathbf{c} = (5, 1). Using P1=12(1111)P^{-1} = \frac{1}{-2}\begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix}, we get c=(3,2)\mathbf{c} = (3, 2). So v=3(1,1)+2(1,1)\mathbf{v} = 3(1, 1) + 2(1, -1).

The Similarity Relation

If T:VVT: V \to V has matrix AA in basis B\mathcal{B} and matrix AA' in basis C\mathcal{C}, then

A=P1APA' = P^{-1}AP


where P=PCBP = P_{\mathcal{C} \leftarrow \mathcal{B}} is the change-of-basis matrix from B\mathcal{B} to C\mathcal{C}.

The derivation is direct. For any vector v\mathbf{v}, the transformation in B\mathcal{B}-coordinates reads [T(v)]B=A[v]B[T(\mathbf{v})]_\mathcal{B} = A[\mathbf{v}]_\mathcal{B}. Converting to C\mathcal{C}-coordinates: [T(v)]C=P1[T(v)]B=P1A[v]B=P1AP[v]C[T(\mathbf{v})]_\mathcal{C} = P^{-1}[T(\mathbf{v})]_\mathcal{B} = P^{-1}A[\mathbf{v}]_\mathcal{B} = P^{-1}AP[\mathbf{v}]_\mathcal{C}. Since this holds for every v\mathbf{v}, the matrix of TT in basis C\mathcal{C} is P1APP^{-1}AP.

Two matrices related by A=P1APA' = P^{-1}AP for some invertible PP are called similar. Similarity is an equivalence relation: every matrix is similar to itself (P=IP = I), similarity is symmetric (A=P1APA' = P^{-1}AP implies A=PAP1A = PA'P^{-1}), and it is transitive.

Properties Preserved by Similarity

Similar matrices represent the same transformation, so they share every property that is intrinsic to the transformation rather than to a particular coordinate system.

The determinant is preserved: det(P1AP)=det(P1)det(A)det(P)=det(A)\det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \det(A).

The trace is preserved: tr(P1AP)=tr(APP1)=tr(A)\text{tr}(P^{-1}AP) = \text{tr}(APP^{-1}) = \text{tr}(A) by the cyclic property.

The eigenvalues are preserved: det(P1APλI)=det(P1(AλI)P)=det(AλI)\det(P^{-1}AP - \lambda I) = \det(P^{-1}(A - \lambda I)P) = \det(A - \lambda I), so the characteristic polynomial — and therefore all eigenvalues with their multiplicities — is the same.

The rank is preserved: multiplying by invertible matrices cannot change the rank.

Individual matrix entries, symmetry, and sparsity are generally not preserved. A symmetric matrix AA can become non-symmetric under P1APP^{-1}AP if PP is not orthogonal.

Diagonalization as a Change of Basis

If TT has nn linearly independent eigenvectors v1,,vn\mathbf{v}_1, \dots, \mathbf{v}_n with eigenvalues λ1,,λn\lambda_1, \dots, \lambda_n, use them as the basis B\mathcal{B}. In this eigenvector basis, TT acts by scaling each basis vector:

T(vi)=λiviT(\mathbf{v}_i) = \lambda_i \mathbf{v}_i


The matrix of TT in this basis is diagonal: D=diag(λ1,,λn)D = \text{diag}(\lambda_1, \dots, \lambda_n).

The change-of-basis matrix PP has the eigenvectors as columns: P=[v1    vn]P = [\mathbf{v}_1 \; \cdots \; \mathbf{v}_n]. The similarity relation gives A=PDP1A = PDP^{-1}, or equivalently D=P1APD = P^{-1}AP.

Diagonalization is the most powerful application of basis change. It reduces matrix powers to diagonal powers: Ak=PDkP1=Pdiag(λ1k,,λnk)P1A^k = PD^kP^{-1} = P\,\text{diag}(\lambda_1^k, \dots, \lambda_n^k)\,P^{-1}. It simplifies differential equations, recurrence relations, and any computation involving repeated application of the same transformation.

When Diagonalization Fails

Not every matrix is diagonalizable. A transformation may not have nn linearly independent eigenvectors — this happens when the geometric multiplicity of some eigenvalue is strictly less than its algebraic multiplicity.

For example, A=(2102)A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} has eigenvalue λ=2\lambda = 2 with algebraic multiplicity 22, but the eigenspace is one-dimensional (spanned by (1,0)(1, 0)). There is no basis of eigenvectors, so AA cannot be diagonalized.

In such cases, the best achievable form under similarity is the Jordan normal form: a block-diagonal matrix where each block is an upper triangular matrix with a single eigenvalue on the diagonal and ones on the superdiagonal. The Jordan form is unique up to ordering of blocks and is the canonical representative of the similarity class. Its full development belongs to advanced linear algebra.

Orthogonal Similarity

When the change-of-basis matrix PP is orthogonal (P1=PTP^{-1} = P^T), the similarity relation becomes A=PTAPA' = P^TAP. This is called orthogonal similarity.

Orthogonal similarity preserves more than ordinary similarity. If AA is symmetric, then PTAPP^TAP is also symmetric — a property that ordinary similarity does not guarantee.

The Spectral Theorem states that every real symmetric matrix is orthogonally similar to a diagonal matrix. The eigenvectors of a symmetric matrix can be chosen orthonormal, and the columns of PP form an orthonormal basis. This is a stronger conclusion than ordinary diagonalizability — the diagonalizing basis is not just independent but orthonormal, which simplifies projections, least squares, and numerical computation.

Worked Example: Full Basis Change

Let A=(4123)A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}. Find a diagonalization A=PDP1A = PDP^{-1}.

The characteristic polynomial is det(AλI)=(4λ)(3λ)2=λ27λ+10=(λ2)(λ5)\det(A - \lambda I) = (4 - \lambda)(3 - \lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda - 2)(\lambda - 5). Eigenvalues: λ1=2\lambda_1 = 2, λ2=5\lambda_2 = 5.

For λ1=2\lambda_1 = 2: (A2I)v=0(A - 2I)\mathbf{v} = \mathbf{0} gives (2121)v=0\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}, so v1=(1,2)\mathbf{v}_1 = (1, -2).

For λ2=5\lambda_2 = 5: (A5I)v=0(A - 5I)\mathbf{v} = \mathbf{0} gives (1122)v=0\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0}, so v2=(1,1)\mathbf{v}_2 = (1, 1).

P=(1121),D=(2005)P = \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}


Verification: P1=13(1121)P^{-1} = \frac{1}{3}\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}, and PDP1=(1121)(2005)13(1121)=(4123)=APDP^{-1} = \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}\frac{1}{3}\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} = A.

Application: A10=PD10P1=P(21000510)P1=P(1024009765625)P1A^{10} = PD^{10}P^{-1} = P\begin{pmatrix} 2^{10} & 0 \\ 0 & 5^{10} \end{pmatrix}P^{-1} = P\begin{pmatrix} 1024 & 0 \\ 0 & 9765625 \end{pmatrix}P^{-1}.

Why Basis Choice Matters

The standard basis is the default, but it is rarely the best choice for a given problem.

An eigenvector basis diagonalizes the matrix, reducing powers and exponentials to operations on diagonal entries. A system of differential equations x=Ax\mathbf{x}' = A\mathbf{x} decouples into independent scalar equations when AA is diagonal.

An orthonormal basis simplifies projections and least-squares computations. Coordinates relative to an orthonormal basis are computed by dot products rather than by solving systems, and numerical errors are minimized because the change-of-basis matrix has condition number 11.

A Jordan basis achieves the simplest possible form for non-diagonalizable matrices, isolating the defective eigenvalues into small blocks.

Choosing the right basis is often the key insight that converts a hard problem into an easy one. The transformation does not change — only its numerical description does — but the right description can make all the difference between a tractable computation and an intractable one.