11: Diagonalization and Similarity

Diagonalization converts a complicated matrix into the simplest possible form: a diagonal matrix. When you can diagonalize a matrix $A$, you’ve found the “eigenvector coordinate system” where $A$ just stretches along axes. This makes everything easier,computing powers, solving differential equations, understanding long-term behavior. The key is finding enough eigenvectors to build a new basis.

What is Diagonalization?

(The Goal)

A square $n \times n$ matrix $A$ is diagonalizable if it can be written as:

where:

• $D$ is a diagonal matrix (eigenvalues on the diagonal)
• $P$ is an invertible matrix (eigenvectors as columns)

Geometric meaning: In the eigenvector basis, $A$ is just a scaling transformation. The columns of $P$ define this special coordinate system, $D$ tells you the scaling factors, and $P^{-1}$ converts back.

(Why This Matters)

Once you have $A = PDP^{-1}$, everything becomes easier:

Powers:

Computing $D^k$ is trivial,just raise each diagonal entry to the $k$-th power.

Matrix exponential:

where $e^{Dt}$ has $e^{\lambda_i t}$ on the diagonal.

Long-term behavior: The dominant eigenvalue (largest $|\lambda|$) determines if $A^k \to 0$, explodes, or oscillates.

Eigenvalues and Eigenvectors

(Definition)

A scalar $\lambda$ is an eigenvalue of $A$ if there exists a nonzero vector $\mathbf{v}$ such that:

The vector $\mathbf{v}$ is an eigenvector corresponding to $\lambda$.

Interpretation: Eigenvectors are the special directions where $A$ acts like pure scaling,no rotation, just stretch or compression by factor $\lambda$.

(Finding Eigenvalues)

Rewrite $A\mathbf{v} = \lambda \mathbf{v}$ as:

For a nontrivial solution to exist, $A - \lambda I$ must be singular:

This is the characteristic equation. Expanding it gives a polynomial of degree $n$ in $\lambda$,the characteristic polynomial.

(Example: Finding Eigenvalues)

Compute $\det(A - \lambda I)$:

Eigenvalues: $\lambda_1 = 5$, $\lambda_2 = 2$.

(Finding Eigenvectors)

For each eigenvalue $\lambda$, solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$ to find the corresponding eigenvector(s).

For $\lambda_1 = 5$:

Row reduce: Both rows give $-v_1 + v_2 = 0$, so $v_2 = v_1$.

Eigenvector: $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ (or any nonzero scalar multiple)

For $\lambda_2 = 2$:

Row reduce: $2v_1 + v_2 = 0$, so $v_2 = -2v_1$.

Eigenvector: $\mathbf{v}_2 = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$

The Diagonalization Process

(Diagonalization Theorem)

An $n \times n$ matrix $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors.

When this holds:

1. Let $P = [\mathbf{v}_1 \mid \mathbf{v}_2 \mid \cdots \mid \mathbf{v}_n]$ (eigenvectors as columns)
2. Let $D = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)$ (corresponding eigenvalues)

Then:

(Why This Works)

The equation $AP = PD$ says:

Column by column, this is exactly $A\mathbf{v}_i = \lambda_i\mathbf{v}_i$,the defining property of eigenvectors.

Multiplying on the right by $P^{-1}$ gives $A = PDP^{-1}$.

(Example: Diagonalizing a Matrix)

From before: $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$ with:

• $\lambda_1 = 5$, $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$
• $\lambda_2 = 2$, $\mathbf{v}_2 = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$

Build $P$ and $D$:

Compute $P^{-1}$:

Verify:

Similar Matrices

(Definition)

Two matrices $A$ and $B$ are similar if there exists an invertible matrix $P$ such that:

Interpretation: Similar matrices represent the same linear transformation in different bases. They have the same intrinsic properties but different coordinate representations.

(Properties Preserved by Similarity)

If $A$ and $B$ are similar, they share:

1. Determinant: $\det(B) = \det(A)$
2. Trace: $\text{tr}(B) = \text{tr}(A)$ (sum of diagonal entries)
3. Eigenvalues: Same characteristic polynomial, same eigenvalues (with multiplicity)
4. Rank: $\text{rank}(B) = \text{rank}(A)$
5. Invertibility: $A$ invertible $\iff$ $B$ invertible

Why? All these properties are basis-independent,they depend only on the transformation itself, not the coordinate system.

(Diagonalization as Similarity)

When $A = PDP^{-1}$, we’re saying $A$ is similar to a diagonal matrix $D$.

Key insight: Diagonalizable matrices are exactly those that are similar to diagonal matrices. The “nicest” matrices are those you can represent diagonally in some basis.

When Is a Matrix Diagonalizable?

(Sufficient Condition: Distinct Eigenvalues)

Theorem: If an $n \times n$ matrix has $n$ distinct eigenvalues, it is diagonalizable.

Why? Eigenvectors from different eigenvalues are automatically linearly independent. So $n$ distinct eigenvalues gives you $n$ independent eigenvectors.

Example: $A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$ has eigenvalues $3$ and $-1$ (distinct), so it’s diagonalizable.

(Repeated Eigenvalues: It Depends)

When eigenvalues repeat, diagonalizability depends on whether there are enough eigenvectors.

Algebraic vs Geometric Multiplicity:

For an eigenvalue $\lambda$:

• Algebraic multiplicity: How many times $\lambda$ appears as a root of the characteristic polynomial
• Geometric multiplicity: $\dim(\ker(A - \lambda I))$,the number of linearly independent eigenvectors for $\lambda$

Key fact: Always: geometric multiplicity $\leq$ algebraic multiplicity

Diagonalizability condition: $A$ is diagonalizable if and only if for every eigenvalue, geometric multiplicity equals algebraic multiplicity.

(Example: Diagonalizable with Repeated Eigenvalue)

Eigenvalue $\lambda = 2$ has algebraic multiplicity 2.

Check geometric multiplicity:

$\dim(\ker(A - 2I)) = 2$ (the first two basis vectors are eigenvectors).

Since geometric = algebraic for all eigenvalues, $A$ is diagonalizable (it’s already diagonal!).

(Example: Not Diagonalizable)

Eigenvalue: $\lambda = 2$ with algebraic multiplicity 2.

$\dim(\ker(A - 2I)) = 1$,only one independent eigenvector $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$.

Since geometric (1) < algebraic (2), $A$ is not diagonalizable.

Computing Powers of Matrices

(The Power Formula)

If $A = PDP^{-1}$, then:

where $D^k = \text{diag}(\lambda_1^k, \lambda_2^k, \ldots, \lambda_n^k)$.

Why this works:

By induction, $A^k = PD^kP^{-1}$.

(Example)

Compute $A^{10}$ for $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$.

We found:

Computing this is far easier than multiplying $A$ ten times!

(Long-Term Behavior)

As $k \to \infty$, $A^k$ is dominated by the largest eigenvalue (in absolute value).

• If $|\lambda_{\max}| < 1$: $A^k \to 0$ (everything decays)
• If $|\lambda_{\max}| = 1$: Bounded behavior (might oscillate)
• If $|\lambda_{\max}| > 1$: $A^k$ explodes (growth along dominant eigenvector)

Application: In Markov chains, Leslie models, discrete dynamical systems,eigenvalues control the long-term fate.

Change of Basis Perspective

(What $P$ Does)

The matrix $P$ performs a change of basis from the standard basis to the eigenvector basis.

In the eigenvector basis:

• Coordinates: $[\mathbf{x}]_{\mathcal{B}} = P^{-1}\mathbf{x}$
• Transformation: $[A]_{\mathcal{B}} = D$ (diagonal!)
• The transformation is just scaling along each eigenvector direction

In the standard basis:

• Coordinates: $\mathbf{x}$
• Transformation: $A$
• The transformation looks complicated because we’re using the “wrong” coordinates

The diagonalization formula:

can be read as:

1. $P^{-1}$: Convert from standard to eigenvector basis
2. $D$: Apply the simple diagonal transformation
3. $P$: Convert back to standard basis

Applications

(Differential Equations)

The system $\frac{d\mathbf{x}}{dt} = A\mathbf{x}$ has solution:

If $A = PDP^{-1}$:

The eigenvectors give you the “modes” of the system, and eigenvalues tell you whether each mode grows or decays.

(Fibonacci Numbers)

The Fibonacci recurrence $F_{n+1} = F_n + F_{n-1}$ can be written as:

Diagonalizing $A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ gives eigenvalues $\phi = \frac{1+\sqrt{5}}{2}$ (golden ratio) and $\hat{\phi} = \frac{1-\sqrt{5}}{2}$.

This leads to Binet’s formula:

Summary: The Diagonalization Playbook

To diagonalize an $n \times n$ matrix $A$:

1. Find eigenvalues: Solve $\det(A - \lambda I) = 0$
2. Find eigenvectors: For each $\lambda$, solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$
3. Check linear independence: You need $n$ independent eigenvectors
4. Build $P$ and $D$:
   • $P = [\mathbf{v}_1 \mid \cdots \mid \mathbf{v}_n]$ (eigenvectors as columns)
   • $D = \text{diag}(\lambda_1, \ldots, \lambda_n)$ (eigenvalues in matching order)
5. Verify: $AP = PD$

If you can’t find $n$ independent eigenvectors, the matrix is not diagonalizable,but you might still use the Jordan normal form (a nearly-diagonal form with 1’s above some diagonal entries).

Diagonalization reveals the coordinate system where a transformation is simplest,just scaling, no mixing. It’s the key to understanding matrix powers, exponentials, and the long-term behavior of linear systems.