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Module 3/Eigenvalues & Diagonalization

Diagonalization

A square matrix is diagonalizable when it is similar to a diagonal matrix, A=SDS−1A=SDS^{-1}. This lesson shows how an eigenbasis supplies the columns of SS and the diagonal of DD, how to decide whether a matrix is diagonalizable by matching geometric and algebraic multiplicities, and why the factorization makes powers AkA^k cheap.

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

A 3×33\times3 matrix AA has characteristic polynomial cA(λ)=−(λ−4)2(λ−1)c_A(\lambda)=-(\lambda-4)^2(\lambda-1), and its eigenspace E4=ker⁡(A−4I)E_4=\ker(A-4I) is a single line (dimension 11). Which conclusion is correct?

Suppose the columns of SS, in order, are eigenvectors v1,v2,v3v_1,v_2,v_3 of AA with Av1=3v1Av_1=3v_1, Av2=3v2Av_2=3v_2, and Av3=−1 v3Av_3=-1\,v_3. Enter the diagonal of the matrix D=S−1ASD=S^{-1}AS as the vector (d1,d2,d3)(d_1,d_2,d_3).

What you’ll be able to do

  • State what it means for a square matrix to be diagonalizable and write the factorization A=SDS−1A=SDS^{-1} (equivalently S−1AS=DS^{-1}AS=D) from Definition 7.5.
  • Define an eigenbasis and assemble the diagonalizing matrix S=[ v1  ⋯  vn ]S=[\,v_1\;\cdots\;v_n\,] and D=diag⁡(λ1,…,λn)D=\operatorname{diag}(\lambda_1,\dots,\lambda_n) from it (Definition 7.6), keeping the column order aligned with the eigenvalue order.
  • Decide whether a matrix is diagonalizable by checking gemu⁡(λ)=almu⁡(λ)\operatorname{gemu}(\lambda)=\operatorname{almu}(\lambda) for every eigenvalue (Theorem 7.8), or by finding nn distinct eigenvalues (Corollary 7.3).
  • Carry out the Strategy for Diagonalization end to end: eigenvalues, a basis of each eigenspace Eλ=ker⁡(A−λI)E_\lambda=\ker(A-\lambda I), then build SS and DD.
  • Use the factorization to compute matrix powers via Ak=SDkS−1A^k=SD^kS^{-1} with Dk=diag⁡(λ1k,…,λnk)D^k=\operatorname{diag}(\lambda_1^k,\dots,\lambda_n^k) (Remark 7.5).

In your course

· MATH2015 · Linear Algebra & Probability
§7.2 Matrix diagonalization
  • Definition 7.5Diagonalizable matrix
    AA is diagonalizable iff it is similar to a diagonal matrix: there is an invertible SS with S−1AS=DS^{-1}AS=D diagonal, i.e. A=SDS−1A=SDS^{-1}.
  • Definition 7.6Eigenbasis and diagonalizing matrix
    A basis {v1,…,vn}\{v_1,\dots,v_n\} of Rn\mathbb{R}^n of eigenvectors of AA; then S=[ v1  ⋯  vn ]S=[\,v_1\;\cdots\;v_n\,] gives S−1AS=diag⁡(λ1,…,λn)S^{-1}AS=\operatorname{diag}(\lambda_1,\dots,\lambda_n).
  • Theorem 7.7Diagonalizable ⟺ eigenbasis
    A square matrix is diagonalizable if and only if it has an eigenbasis.
  • Proposition 7.2Eigenvectors for distinct eigenvalues are independent
  • Theorem 7.8Eigenbasis ⟺ geometric = algebraic multiplicity
    AA has an eigenbasis iff gemu⁡(λ)=almu⁡(λ)\operatorname{gemu}(\lambda)=\operatorname{almu}(\lambda) for every eigenvalue λ\lambda; concatenate the eigenspace bases.
  • Corollary 7.3n distinct eigenvalues ⟹ diagonalizable
  • Remark 7.5Matrix powers via diagonalization
    If A=SDS−1A=SDS^{-1} then Ak=SDkS−1A^k=SD^kS^{-1} with Dk=diag⁡(λ1k,…,λnk)D^k=\operatorname{diag}(\lambda_1^k,\dots,\lambda_n^k).
See the Strategy for Diagonalization (find eigenvalues; a basis of each Eλ=ker⁡(A−λI)E_\lambda=\ker(A-\lambda I); diagonalizable iff the eigenspace dimensions sum to nn; then build SS and DD) and the worked cases in Example 7.7. Remark 7.4 advises starting with the eigenvalue of highest algebraic multiplicity. Example 7.7(3) notes a matrix with complex eigenvalues need not be diagonalizable over R\mathbb{R} even when it is over C\mathbb{C}.
1

Diagonalizable means “similar to diagonal”

A square matrix AA is diagonalizable when it is similar to a diagonal matrix (Definition 7.5): there is an invertible matrix SS with S−1AS=D⟺A=SDS−1,S^{-1}AS=D\quad\Longleftrightarrow\quad A=SDS^{-1}, where DD is diagonal. To diagonalize AA is to actually produce such an SS and DD. Think of SS as a change of coordinates into a basis where AA acts one axis at a time—the off-diagonal coupling disappears.

2

The eigenbasis builds $S$ and $D$

An eigenbasis for an n×nn\times n matrix AA is a basis {v1,…,vn}\{v_1,\dots,v_n\} of Rn\mathbb{R}^n whose vectors are all eigenvectors, Avi=λiviAv_i=\lambda_i v_i (Definition 7.6). Place these eigenvectors as the columns of SS and the matching eigenvalues on the diagonal of DD:

S=[∣∣v1⋯vn∣∣],S−1AS=diag⁡(λ1,…,λn).S=\begin{bmatrix} | & & |\\ v_1 & \cdots & v_n\\ | & & | \end{bmatrix},\qquad S^{-1}AS=\operatorname{diag}(\lambda_1,\dots,\lambda_n).

Order matters: the ii-th column of SS must pair with the ii-th diagonal entry of DD. The λi\lambda_i need not be distinct—a repeated eigenvalue simply contributes several columns.

3

When it works: multiplicities must match

Finding an eigenbasis is the whole game (Theorem 7.7). By Theorem 7.6 every eigenvalue satisfies 1≤gemu⁡(λ)≤almu⁡(λ)1\le\operatorname{gemu}(\lambda)\le\operatorname{almu}(\lambda), and the algebraic multiplicities sum to nn. So the eigenspace dimensions add up to nn—giving an eigenbasis—iff gemu⁡(λ)=almu⁡(λ)\operatorname{gemu}(\lambda)=\operatorname{almu}(\lambda) for every eigenvalue (Theorem 7.8). A matrix that fails this for some λ\lambda (so gemu⁡(λ)<almu⁡(λ)\operatorname{gemu}(\lambda)<\operatorname{almu}(\lambda)) is called defective and is not diagonalizable. Only repeated eigenvalues can cause trouble, which is why nn distinct eigenvalues always suffice (Corollary 7.3).

4

Why bother: cheap powers and decoupling

Once A=SDS−1A=SDS^{-1}, the middle terms telescope: Ak=SDS−1 SDS−1⋯SDS−1⏟k factors=SDkS−1,Dk=diag⁡(λ1k,…,λnk).A^k=\underbrace{SDS^{-1}\,SDS^{-1}\cdots SDS^{-1}}_{k\text{ factors}}=SD^kS^{-1},\qquad D^k=\operatorname{diag}(\lambda_1^k,\dots,\lambda_n^k). Raising a diagonal matrix to a power is just raising each diagonal entry to that power (Remark 7.5). The same idea powers up solving linear systems, decoupling linear transformations, differential equations, and PCA—anywhere repeated application of AA appears.

Theorem 7.7 — Diagonalizable ⟺ eigenbasis

A square matrix AA is diagonalizable if and only if AA has an eigenbasis. In that case S=[ v1  ⋯  vn ]S=[\,v_1\;\cdots\;v_n\,] (eigenvectors as columns) and S−1AS=diag⁡(λ1,…,λn)S^{-1}AS=\operatorname{diag}(\lambda_1,\dots,\lambda_n).

Intuition. Diagonalizing is nothing more than finding enough independent eigenvectors to form a basis. In that basis each coordinate is just scaled by its eigenvalue, so the matrix becomes diagonal.
Proposition 7.2 — Distinct eigenvalues give independence

If λ1,…,λk\lambda_1,\dots,\lambda_k are distinct eigenvalues of AA, then any corresponding eigenvectors v1,…,vkv_1,\dots,v_k are linearly independent.

Intuition. Eigenvectors from different eigenvalues point in genuinely different directions; one cannot be built from the others. This is the engine that lets separate eigenspaces be concatenated into a basis.
Theorem 7.8 — Eigenbasis ⟺ matched multiplicities

An n×nn\times n matrix AA has an eigenbasis if and only if gemu⁡(λ)=almu⁡(λ)\operatorname{gemu}(\lambda)=\operatorname{almu}(\lambda) for every eigenvalue λ\lambda. In that case an eigenbasis is obtained by concatenating bases of the eigenspaces Eλ=ker⁡(A−λI)E_\lambda=\ker(A-\lambda I).

Intuition. The algebraic multiplicities always total nn, and each eigenspace dimension can only fall short of its algebraic multiplicity, never exceed it. So the dimensions reach nn exactly when none falls short—no eigenvalue is “defective.”
Corollary 7.3 — Distinct eigenvalues suffice

An n×nn\times n matrix with nn distinct (real) eigenvalues is diagonalizable.

Intuition. With nn distinct eigenvalues each eigenspace is a line, and Proposition 7.2 makes the nn eigenvectors independent—instantly an eigenbasis. Sufficient, not necessary: a matrix can repeat eigenvalues and still be diagonalizable (e.g. the identity).

Worked examples

Example 1

Diagonalize A=[0−1−1121112]A=\begin{bmatrix} 0&-1&-1\\ 1&2&1\\ 1&1&2 \end{bmatrix} (Example 7.7). Its characteristic polynomial is cA(λ)=−(λ−1)2(λ−2)c_A(\lambda)=-(\lambda-1)^2(\lambda-2).

  1. 1

    Read off the eigenvalues and algebraic multiplicities from cA(λ)=−(λ−1)2(λ−2)c_A(\lambda)=-(\lambda-1)^2(\lambda-2): λ1=1\lambda_1=1 with almu⁡(1)=2\operatorname{almu}(1)=2 and λ2=2\lambda_2=2 with almu⁡(2)=1\operatorname{almu}(2)=1.

  2. 2

    Per Remark 7.4, start with the highest-multiplicity eigenvalue. For λ1=1\lambda_1=1 solve (A−I)v=0(A-I)v=0: A−I=[−1−1−1111111]A-I=\begin{bmatrix} -1&-1&-1\\ 1&1&1\\ 1&1&1 \end{bmatrix} has rank 11, so E1=ker⁡(A−I)=span⁡{[−110],[−101]}E_1=\ker(A-I)=\operatorname{span}\left\{\begin{bmatrix}-1\\1\\0\end{bmatrix},\begin{bmatrix}-1\\0\\1\end{bmatrix}\right\}, giving gemu⁡(1)=2\operatorname{gemu}(1)=2.

  3. 3

    Since gemu⁡(1)=2=almu⁡(1)\operatorname{gemu}(1)=2=\operatorname{almu}(1), this eigenvalue is fine. For λ2=2\lambda_2=2 solve (A−2I)v=0(A-2I)v=0: E2=span⁡{[−111]}E_2=\operatorname{span}\left\{\begin{bmatrix}-1\\1\\1\end{bmatrix}\right\}, so gemu⁡(2)=1=almu⁡(2)\operatorname{gemu}(2)=1=\operatorname{almu}(2).

  4. 4

    The eigenspace dimensions add up to 2+1=3=n2+1=3=n, so by Theorem 7.8 AA is diagonalizable. Concatenate the eigenspace bases into SS and record the matching eigenvalues in DD (same order).

  5. 5

    Check the pairing: columns 1,21,2 come from E1E_1 so carry λ=1\lambda=1; column 33 comes from E2E_2 so carries λ=2\lambda=2. A direct check gives S−1AS=DS^{-1}AS=D (here det⁡S=1≠0\det S=1\neq0, so SS is invertible).

Answer. AA is diagonalizable with S=[−1−1−1101011],S−1AS=D=[100010002].S=\begin{bmatrix} -1&-1&-1\\ 1&0&1\\ 0&1&1 \end{bmatrix},\qquad S^{-1}AS=D=\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&2 \end{bmatrix}.
Example 2

Diagonalize A=[2112]A=\begin{bmatrix} 2&1\\ 1&2 \end{bmatrix} and use it to compute A3A^3 via Ak=SDkS−1A^k=SD^kS^{-1}.

Example 3

Is A=[2102]A=\begin{bmatrix} 2&1\\ 0&2 \end{bmatrix} diagonalizable?