01 / 20

Matrix
Diagonalization

Learn when and how a matrix can be expressed as \(A = PDP^{-1}\), unlocking efficient computation of powers, exponentials, and more.

Linear Algebra

What Is Diagonalization?

An \(n \times n\) matrix \(A\) is diagonalizable if there exists an invertible matrix \(P\) and a diagonal matrix \(D\) such that:

Equivalently, \(D = P^{-1}AP\). The columns of \(P\) are eigenvectors of \(A\), and the diagonal entries of \(D\) are the corresponding eigenvalues.

Eigenvalues and Eigenvectors

A scalar \(\lambda\) is an eigenvalue of \(A\) if there exists a nonzero vector \(\mathbf{v}\) such that \(A\mathbf{v} = \lambda\mathbf{v}\). That vector \(\mathbf{v}\) is the corresponding eigenvector.

Characteristic Equation
\(\det(A - \lambda I) = 0\) gives us all eigenvalues.
Eigenspace
\(\text{Nul}(A - \lambda I)\) is the eigenspace: all eigenvectors for \(\lambda\), plus \(\mathbf{0}\).

Why Diagonalize?

Diagonal matrices are extremely easy to work with. If \(A = PDP^{-1}\), then:

Matrix Powers
\(A^k = PD^kP^{-1}\), and \(D^k\) simply raises each diagonal entry to the \(k\)-th power.
Systems of ODEs
Decouples \(\mathbf{x}' = A\mathbf{x}\) into independent equations, each solvable instantly.
Matrix Exponential
\(e^{At} = Pe^{Dt}P^{-1}\), where \(e^{Dt}\) is diagonal with entries \(e^{\lambda_i t}\).

The Diagonalization Theorem

Equivalently, \(A\) is diagonalizable if and only if the sum of the dimensions of all eigenspaces equals \(n\):

A Handy Sufficient Condition

Distinct eigenvalues guarantee linearly independent eigenvectors. However, the converse is not true: a matrix can be diagonalizable even with repeated eigenvalues, as long as each eigenspace has sufficient dimension.

Algebraic vs Geometric Multiplicity

Algebraic Multiplicity
The number of times \(\lambda\) appears as a root of \(\det(A - \lambda I) = 0\). Denoted \(\text{AM}(\lambda)\).
Geometric Multiplicity
The dimension of the eigenspace \(\text{Nul}(A - \lambda I)\). Denoted \(\text{GM}(\lambda)\).

The Big Picture

\(A\)
\(P\)\(D\)\(P^{-1}\)
Diagonal Matrix \(D\)
\(\lambda_1\)
\(0\)
\(0\)
\(0\)
\(\lambda_2\)
\(0\)
Matrix \(P\)
\(\mathbf{v}_1\) \(\mathbf{v}_2\) \(\cdots\)
 eigenvectors change of basis → diagonal → change back

How to Diagonalize a Matrix

1
Find eigenvalues. Solve the characteristic equation \(\det(A - \lambda I) = 0\).
2
Find eigenvectors. For each eigenvalue \(\lambda_i\), solve \((A - \lambda_i I)\mathbf{x} = \mathbf{0}\) to find a basis for the eigenspace.
3
Check diagonalizability. Verify that the total number of linearly independent eigenvectors equals \(n\).
4
Build \(P\) and \(D\). Let \(P = [\mathbf{v}_1 \;\; \mathbf{v}_2 \;\; \cdots \;\; \mathbf{v}_n]\) and \(D = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)\), matching column order.

Example 1: Diagonalizing a \(2 \times 2\) Matrix

Diagonalize \(A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}\).

Example 1: Result

We have 2 linearly independent eigenvectors for a \(2 \times 2\) matrix, so \(A\) is diagonalizable:

Example 2: Not Diagonalizable

Consider \(A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}\).

This matrix is a defective matrix: it lacks enough eigenvectors. Jordan normal form would be needed instead.

Example 3: Repeated \(\lambda\), Still Diagonalizable

Consider \(A = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}\). Here \(\lambda = 3\) with \(\text{AM} = 2\).

Common Mistakes

✕ Confusing AM and GM
Assuming every eigenvalue automatically has enough eigenvectors. Always compute the eigenspace dimension explicitly.
✕ Wrong column order
Placing eigenvectors in \(P\) in a different order than the corresponding eigenvalues in \(D\). Column \(j\) of \(P\) must match entry \(D_{jj}\).
✕ Forgetting invertibility of \(P\)
If the eigenvectors are not linearly independent, \(P\) is singular and the factorization fails. This is exactly the non-diagonalizable case.

Diagonalizable vs Not Diagonalizable

Diagonalizable Not Diagonalizable
Eigenvectors \(n\) linearly independent Fewer than \(n\)
For every \(\lambda\) \(\text{GM} = \text{AM}\) \(\text{GM} < \text{AM}\) for some \(\lambda\)
Factorization \(A = PDP^{-1}\) Use Jordan form instead
Powers of \(A\) \(A^k = PD^kP^{-1}\) More complex computation
Distinct eigenvalues? Guaranteed if \(n\) distinct Repeated roots require checking

Question 1 of 5

Which equation correctly expresses the diagonalization of a matrix \(A\)?

Question 2 of 5

A \(3 \times 3\) matrix has eigenvalues \(\lambda_1 = 2\), \(\lambda_2 = 5\), and \(\lambda_3 = -1\). Is it diagonalizable?

Question 3 of 5

The matrix \(\begin{pmatrix} 5 & 1 \\ 0 & 5 \end{pmatrix}\) has eigenvalue \(\lambda = 5\) with algebraic multiplicity 2. What is its geometric multiplicity?

Question 4 of 5

If \(A = PDP^{-1}\), what is \(A^{50}\)?

Question 5 of 5

Which of the following matrices is always diagonalizable (over \(\mathbb{R}\))?