Diagonalization

1. What is Diagonalization?

Diagonalization expresses a matrix in terms of its eigenvalues and eigenvectors, revealing its essential structure.

Definition 6.8: Diagonalizable Matrix

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

A = PDP^{-1}

Equivalently: $D = P^{-1}AP$ (A is similar to D).

Definition 6.9: Diagonalizable Linear Operator

A linear operator $T: V \to V$ is diagonalizable if there exists a basis $\mathcal{B}$ of $V$ such that $[T]_{\mathcal{B}}$ is diagonal.

Remark 6.3: The P and D Matrices

$P$ : Columns are eigenvectors of $A$
$D$ : Diagonal entries are eigenvalues (in corresponding order)
Column $i$ of $P$ is an eigenvector for eigenvalue $d_{ii}$

Theorem 6.20: Equivalent Conditions

For an $n \times n$ matrix $A$ , the following are equivalent:

$A$ is diagonalizable
$A$ has $n$ linearly independent eigenvectors
$\mathbb{R}^n$ (or $\mathbb{C}^n$ ) has a basis of eigenvectors of $A$
The sum of eigenspace dimensions equals $n$

Proof:

(1) ⟺ (2): If $A = PDP^{-1}$ , then the columns of $P$ are eigenvectors and $P$ invertible means they're independent. Conversely, if we have $n$ independent eigenvectors, form $P$ from them.

(2) ⟺ (3): $n$ independent vectors in an $n$ -dimensional space form a basis.

(2) ⟺ (4): Eigenspaces for distinct eigenvalues are in direct sum, so total dimension equals number of independent eigenvectors.

∎

Example 6.25: Verifying A = PDP^{-1}

For $A = \begin{pmatrix} 5 & 4 \\ 2 & 3 \end{pmatrix}$ :

Step 1: Find eigenvalues: $\chi_A = \lambda^2 - 8\lambda + 7 = (\lambda-1)(\lambda-7)$

Eigenvalues: $\lambda_1 = 1, \lambda_2 = 7$

Step 2: Find eigenvectors:

For $\lambda = 1$ : $\ker(A - I) = \text{span}\{(-1, 1)^T\}$

For $\lambda = 7$ : $\ker(A - 7I) = \text{span}\{(2, 1)^T\}$

Step 3: Form P and D:

P = \begin{pmatrix} -1 & 2 \\ 1 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 1 & 0 \\ 0 & 7 \end{pmatrix}

Verify: $PDP^{-1} = A$ ✓

2. The Diagonalizability Criterion

The key question: when is a matrix diagonalizable? The answer involves comparing algebraic and geometric multiplicities.

Theorem 6.21: Complete Diagonalizability Criterion

An $n \times n$ matrix $A$ is diagonalizable if and only if:

The characteristic polynomial splits completely (factors into linear terms)
For each eigenvalue $\lambda$ : $g(\lambda) = a(\lambda)$

Equivalently: $\sum_{i} g(\lambda_i) = n$

Proof:

(⟹): If $A$ is diagonalizable, then $A = PDP^{-1}$ where $D$ has eigenvalues on diagonal. The char poly of $A$ equals that of $D$ , which splits completely. Each eigenspace has dimension equal to the multiplicity of that eigenvalue in $D$ .

(⟸): If the char poly splits and $g = a$ for all eigenvalues, then the sum of eigenspace dimensions is $\sum g_i = \sum a_i = n$ . Taking a basis from each eigenspace gives $n$ independent eigenvectors.

∎

Theorem 6.22: Distinct Eigenvalues Suffice

If an $n \times n$ matrix $A$ has $n$ distinct eigenvalues, then $A$ is diagonalizable.

Proof:

Each distinct eigenvalue has $a = 1$ , and always $g \geq 1$ . Combined with $g \leq a$ , we get $g = 1 = a$ for each. Thus the criterion is satisfied.

∎

Corollary 6.8: Generic Matrices

"Most" matrices are diagonalizable: the set of diagonalizable matrices is dense in the space of all matrices.

Example 6.26: Checking Diagonalizability

Is $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}$ diagonalizable?

Eigenvalues: $\lambda = 2$ with $a = 3$ (from triangular form)

Geometric multiplicity: $g = \dim\ker(A - 2I) = \dim\ker\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} = 1$

Conclusion: $g = 1 \neq 3 = a$ , so NOT diagonalizable.

Example 6.27: Diagonalizable Despite Repeated Eigenvalue

Is $A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ diagonalizable?

Eigenvalues: $\lambda_1 = 2$ with $a = 2$ , $\lambda_2 = 3$ with $a = 1$

For $\lambda = 2$ : $g = 2$ (two free variables in ker)

For $\lambda = 3$ : $g = 1$

Conclusion: $g = a$ for both, so diagonalizable (it's already diagonal!).

Remark 6.4: Defective Matrices

A matrix that is NOT diagonalizable is called defective. For defective matrices, we use Jordan normal form instead.

3. The Diagonalization Algorithm

Step-by-Step Algorithm

Find eigenvalues: Solve $\det(A - \lambda I) = 0$
Check multiplicities: Verify $g(\lambda) = a(\lambda)$ for each
Find eigenvectors: For each $\lambda$ , find basis of $\ker(A - \lambda I)$
Form P: Columns are eigenvectors (any order, but remember it!)
Form D: Diagonal entries are eigenvalues (matching order with P)
Verify: Check $AP = PD$ or $A = PDP^{-1}$

Example 6.28: Complete 3×3 Diagonalization

Diagonalize $A = \begin{pmatrix} 1 & 2 & 2 \\ 0 & 2 & 1 \\ -1 & 2 & 2 \end{pmatrix}$ .

Step 1: Characteristic polynomial: $\chi_A = (\lambda - 1)(\lambda - 2)(\lambda - 2) = (\lambda - 1)(\lambda - 2)^2$

Eigenvalues: $\lambda_1 = 1$ (a=1), $\lambda_2 = 2$ (a=2)

Step 2: Check multiplicities...

For $\lambda = 1$ : g = 1 = a ✓

For $\lambda = 2$ : need g = 2. Find $\ker(A - 2I)$ ...

If g = 2: diagonalizable. If g = 1: NOT diagonalizable.

Remark 6.5: Verification Shortcut

Instead of computing $P^{-1}$ , you can verify $AP = PD$ (column by column: $Av_i = \lambda_i v_i$ ).

4. Computing Matrix Powers

One of the most important applications of diagonalization: computing $A^k$ efficiently.

Theorem 6.23: Matrix Powers via Diagonalization

If $A = PDP^{-1}$ , then for any integer $k \geq 0$ :

A^k = PD^kP^{-1}

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

Proof:

By induction or direct computation:

A^2 = (PDP^{-1})(PDP^{-1}) = PD(P^{-1}P)DP^{-1} = PD^2P^{-1}

In general: $A^k = PD^kP^{-1}$ .

∎

Example 6.29: Computing A^{100}

For $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$ , compute $A^{100}$ .

Step 1: Eigenvalues: 3, 2 (from diagonal of triangular matrix)

Step 2: Eigenvectors: $v_1 = (1,0)^T$ , $v_2 = (1,-1)^T$

Step 3: $P = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$ , $D = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$

Step 4: $D^{100} = \begin{pmatrix} 3^{100} & 0 \\ 0 & 2^{100} \end{pmatrix}$

Step 5: $A^{100} = PD^{100}P^{-1}$

Corollary 6.9: Matrix Functions

If $A = PDP^{-1}$ and $f$ is a function defined on eigenvalues:

f(A) = P \cdot \text{diag}(f(\lambda_1), \ldots, f(\lambda_n)) \cdot P^{-1}

This defines matrix exponential, logarithm, square root, etc.

Example 6.30: Matrix Exponential Preview

For diagonal $D = \text{diag}(\lambda_1, \ldots, \lambda_n)$ :

e^D = \text{diag}(e^{\lambda_1}, \ldots, e^{\lambda_n})

So $e^A = Pe^DP^{-1}$ when A is diagonalizable.

5. Special Classes of Matrices

Theorem 6.24: Symmetric Matrices

Every real symmetric matrix is diagonalizable. Moreover, it is orthogonally diagonalizable:

A = A^T \Rightarrow A = QDQ^T

where $Q$ is orthogonal ( $Q^{-1} = Q^T$ ).

Theorem 6.25: Normal Matrices

A complex matrix is unitarily diagonalizable iff it is normal ( $AA^* = A^*A$ ).

Remark 6.6: Classes Always Diagonalizable

Symmetric: $A = A^T$ ⟹ diagonalizable (real eigenvalues)
Skew-symmetric: $A = -A^T$ ⟹ diagonalizable (imaginary eigenvalues)
Orthogonal: $A^TA = I$ ⟹ diagonalizable (eigenvalues on unit circle)
Normal: $AA^* = A^*A$ ⟹ unitarily diagonalizable
Distinct eigenvalues: Always diagonalizable

Example 6.31: Orthogonal Diagonalization

For symmetric $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ :

Eigenvalues: 1, 3

Orthonormal eigenvectors: $\frac{1}{\sqrt{2}}(1,-1)^T, \frac{1}{\sqrt{2}}(1,1)^T$

Q = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, \quad A = QDQ^T

6. Common Mistakes

Order of eigenvectors matters

Column $i$ of $P$ must correspond to diagonal entry $i$ of $D$ ! Mismatching gives wrong result.

Not all matrices are diagonalizable

Always check $g = a$ for each eigenvalue before proceeding. Defective matrices need Jordan form.

Forgetting the inverse

It's $A = PDP^{-1}$ , not $A = PD$ . The inverse is essential for the similarity transform.

Real vs Complex diagonalization

A matrix might not be diagonalizable over $\mathbb{R}$ but is over $\mathbb{C}$ . Specify the field!

7. Applications

Differential Equations

System $\mathbf{x}' = A\mathbf{x}$ has solution $\mathbf{x}(t) = e^{At}\mathbf{x}_0$ . With diagonalization: $e^{At} = Pe^{Dt}P^{-1}$ .

Markov Chains

Long-term behavior: $P^n$ as $n \to \infty$ . Diagonalization reveals stationary distribution.

Recurrence Relations

Fibonacci: $F_n = \frac{\phi^n - \hat{\phi}^n}{\sqrt{5}}$ comes from diagonalizing the companion matrix.

Quadratic Forms

Diagonalizing symmetric $A$ simplifies $\mathbf{x}^TA\mathbf{x}$ to sum of squares.

Example 6.32: System of Differential Equations

Solve $\begin{cases} x' = 3x + y \\ y' = x + 3y \end{cases}$ .

Matrix form: $\mathbf{x}' = A\mathbf{x}$ where $A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$

Diagonalize: $A = PDP^{-1}$ with eigenvalues 2, 4

Solution: $\mathbf{x}(t) = c_1 e^{2t} v_1 + c_2 e^{4t} v_2$

Example 6.33: Fibonacci via Diagonalization

The Fibonacci matrix $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ has eigenvalues $\phi, \hat{\phi}$ (golden ratio).

Since $\begin{pmatrix} F_{n+1} \\ F_n \end{pmatrix} = A^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ , we get:

F_n = \frac{\phi^n - \hat{\phi}^n}{\sqrt{5}}

8. More Worked Examples

Example 6.34: Non-Diagonalizable Matrix

Show $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ is not diagonalizable.

Eigenvalues: $\chi_A = (\lambda - 1)^2$ , so $\lambda = 1$ with $a = 2$

Geometric: $\ker(A - I) = \ker\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \text{span}\{(1,0)^T\}$

So $g = 1 \neq 2 = a$ . NOT diagonalizable.

Example 6.35: Complex Eigenvalues

Diagonalize $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ over $\mathbb{C}$ .

Eigenvalues: $\lambda^2 + 1 = 0 \Rightarrow \lambda = \pm i$

Eigenvectors:

For $\lambda = i$ : $v_1 = (1, -i)^T$

For $\lambda = -i$ : $v_2 = (1, i)^T$

P = \begin{pmatrix} 1 & 1 \\ -i & i \end{pmatrix}, \quad D = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}

Example 6.36: Projection Matrix

Diagonalize projection $P = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix}$ .

Eigenvalues: 0, 1 (projections always have eigenvalues 0 and 1)

Eigenvectors:

$\lambda = 1$ : $(1, 1)^T$ (image of P)

$\lambda = 0$ : $(1, -1)^T$ (kernel of P)

P_{mat} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \quad D = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}

9. Conceptual Understanding

What does diagonalization mean geometrically?

In the eigenvector basis, the linear transformation is just scaling along each axis. No rotation, no shearing—pure stretching.

Why do repeated eigenvalues cause problems?

Repeated eigenvalues mean the eigenspace might not have enough dimensions. We need dim = multiplicity for each eigenvalue.

Is the zero matrix diagonalizable?

Yes! $0 = I \cdot 0 \cdot I^{-1}$ . It's already diagonal (all zeros).

Can I always find P using any eigenvectors?

Yes, any basis of eigenvectors works. Different choices give different P but all valid diagonalizations.

Key Takeaways

The Formula

$A = PDP^{-1}$

P = eigenvectors, D = eigenvalues

Matrix Powers

$A^k = PD^kP^{-1}$

Powers become trivial!

The Criterion

$g(\lambda) = a(\lambda)$ for all $\lambda$

Geometric = algebraic multiplicity

Sufficient Condition

$n$ distinct eigenvalues

⟹ Always diagonalizable

Quick Reference

Algorithm Steps

Find eigenvalues
Check g = a for each
Find eigenvector bases
Form P and D
Verify AP = PD

Always Diagonalizable

• Distinct eigenvalues
• Symmetric matrices
• Normal matrices
• Diagonal matrices

10. Additional Practice

Problem 1

Diagonalize $A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$ .

Answer: Eigenvalues 2, 3. P and D can be formed from eigenvectors.

Problem 2

Is $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ diagonalizable?

Answer: Check if g(2) = 2. If not, NOT diagonalizable.

Problem 3

If $A = PDP^{-1}$ with $D = \text{diag}(2, 3)$ , find $A^{10}$ in terms of P.

Answer: $A^{10} = P \cdot \text{diag}(2^{10}, 3^{10}) \cdot P^{-1}$

Problem 4

Prove: If A is diagonalizable and all eigenvalues are 1, then A = I.

Hint: $A = PDP^{-1} = PIP^{-1} = I$

Problem 5 (Challenge)

Show that if A is diagonalizable and B commutes with A (AB = BA), then A and B are simultaneously diagonalizable.

11. Challenge Problems

Challenge 1

Prove that if $A$ is diagonalizable and invertible, then $A^{-1}$ is also diagonalizable.

Challenge 2

If A and B are both diagonalizable and AB = BA, prove they share a common eigenvector.

Challenge 3

Find all 2×2 matrices that are diagonalizable and satisfy $A^2 = I$ .

Hint: Eigenvalues must satisfy $\lambda^2 = 1$ .

Challenge 4

Prove: A matrix is diagonalizable iff its minimal polynomial has no repeated roots.

Challenge 5

For $A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}$ , diagonalize A over $\mathbb{C}$ .

Hint: This is a cyclic permutation; eigenvalues are cube roots of unity.

12. Exam Preparation

What You Should Know

• Definition of diagonalizability
• The complete diagonalizability criterion
• Step-by-step diagonalization algorithm
• Computing matrix powers via $A^k = PD^kP^{-1}$
• Recognizing always-diagonalizable classes
• Verifying diagonalization with AP = PD

Common Exam Questions

• Diagonalize a given 2×2 or 3×3 matrix
• Determine if a matrix is diagonalizable
• Compute $A^n$ using diagonalization
• True/False on diagonalizability properties
• Apply to differential equations
• Work with complex eigenvalues

Verification Checklist

After diagonalizing, verify:

P is invertible (det ≠ 0)

D is diagonal

Columns of P are eigenvectors

AP = PD (column by column)

Order matches: column i ↔ entry (i,i)

Eigenvalues match char poly

Diagonalizability Summary

Matrix Type	Diagonalizable?	Notes
n distinct eigenvalues	Always ✓	Most common case
Symmetric ( $A = A^T$ )	Always ✓	Orthogonally diagonalizable
Normal ( $AA^* = A^*A$ )	Always ✓	Unitarily diagonalizable
Diagonal	Always ✓	Already diagonal!
Identity multiple (cI)	Always ✓	Already diagonal
Repeated eigenvalues	Check g = a	May or may not be
Nilpotent (non-zero)	Never ✗	Only eigenvalue is 0 with g < a
Jordan block (size > 1)	Never ✗	g = 1 but a = block size

Study Tips

Start with eigenvalues: Can't diagonalize without them!
Check multiplicities early: If any g < a, stop—not diagonalizable.
Use AP = PD: Easier to verify than computing $P^{-1}$ .
Remember order matters: Column i of P ↔ entry (i,i) of D.
Know the shortcuts: Symmetric, distinct eigenvalues → always works.
Practice 2×2 first: Master small cases before 3×3 and larger.

Module Summary

Diagonalization $A = PDP^{-1}$ reveals a matrix's structure by expressing it in terms of eigenvalues and eigenvectors. The key criterion is $g(\lambda) = a(\lambda)$ for all eigenvalues. When satisfied, matrix powers become trivial, differential equations simplify, and the geometry becomes pure scaling along eigenvector directions.

25+

Theorems

36+

Examples

12

Quiz Questions

8

FAQs

Historical Notes

Cauchy (1829): First systematic study of eigenvalue problems for symmetric matrices in the context of quadratic forms.

Jacobi (1846): Developed iterative methods for finding eigenvalues of symmetric matrices.

Sylvester (1852): Developed matrix algebra and studied invariants under similarity transformations.

Weierstrass (1858): Complete theory of canonical forms, including Jordan form for non-diagonalizable cases.

Spectral Theorem: The culmination—every symmetric/normal matrix is diagonalizable in a particularly nice way with orthonormal eigenvectors.

13. Geometric Interpretation

Diagonalization has a beautiful geometric meaning: it finds directions where the linear transformation acts simply.

The Key Insight

In the standard basis, a matrix transformation can stretch, rotate, and shear. But in the eigenvector basis:

Each eigenvector direction is simply scaled by its eigenvalue
No rotation between eigenvector directions
No shearing—pure stretching/compression

Remark 6.7: Change of Basis View

$A = PDP^{-1}$ says: "To apply A, first convert to eigenvector coordinates ( $P^{-1}$ ), then scale ( $D$ ), then convert back ( $P$ )."

Example 6.37: Geometric Visualization

Consider $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ with eigenvectors $(1,0)^T$ (for $\lambda=2$ ) and $(1,1)^T$ (for $\lambda=3$ ).

In standard coordinates: A shears and stretches.

In eigenvector basis: A just stretches by 2 along $(1,0)$ and by 3 along $(1,1)$ .

14. Simultaneous Diagonalization

Theorem 6.26: Commuting Diagonalizable Matrices

If $A$ and $B$ are both diagonalizable and commute ( $AB = BA$ ), then they are simultaneously diagonalizable:

\exists P: A = PD_AP^{-1}, \quad B = PD_BP^{-1}

The same $P$ works for both!

Corollary 6.10: Functions of Commuting Matrices

If $AB = BA$ and both are diagonalizable, then $f(A)g(B) = g(B)f(A)$ for any functions $f, g$ .

Example 6.38: Simultaneous Diagonalization

If $A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 0 \\ 0 & 4 \end{pmatrix}$ :

Both are already diagonal in the same basis (standard basis).

They commute: $AB = BA = \begin{pmatrix} 3 & 0 \\ 0 & 8 \end{pmatrix}$

15. Spectral Decomposition

Theorem 6.27: Spectral Decomposition

If $A$ is diagonalizable with distinct eigenvalues $\lambda_1, \ldots, \lambda_k$ and projection matrices $P_1, \ldots, P_k$ onto eigenspaces:

A = \lambda_1 P_1 + \lambda_2 P_2 + \cdots + \lambda_k P_k

where $P_i P_j = 0$ for $i \neq j$ and $\sum P_i = I$ .

Remark 6.8: Why "Spectral"

The "spectrum" of a matrix is its set of eigenvalues. Spectral decomposition expresses A as a sum over its spectrum.

Example 6.39: Spectral Form

For $A = \begin{pmatrix} 3 & 0 \\ 0 & 5 \end{pmatrix}$ :

A = 3 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + 5 \cdot \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}

16. Connection to Minimal Polynomial

Theorem 6.28: Diagonalizability via Minimal Polynomial

A matrix $A$ is diagonalizable if and only if its minimal polynomial has no repeated roots:

m_A(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_k)

where $\lambda_1, \ldots, \lambda_k$ are distinct.

Proof:

(Sketch): If the minimal polynomial has no repeated roots, it splits into distinct linear factors. By the primary decomposition theorem, the space decomposes into eigenspaces. Each factor contributes dimension equal to its multiplicity in the characteristic polynomial.

∎

Example 6.40: Comparing Char and Min Poly

For $A = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ :

Characteristic: $\chi_A = (\lambda - 2)^2$ (repeated root)

Minimal: $m_A = \lambda - 2$ (NO repeated root)

⟹ Diagonalizable (it's 2I, already diagonal!)

Example 6.41: Minimal Polynomial with Repeated Root

For $A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ :

Characteristic: $\chi_A = (\lambda - 2)^2$

Minimal: $m_A = (\lambda - 2)^2$ (HAS repeated root)

⟹ NOT diagonalizable

17. More Theoretical Results

Theorem 6.29: Diagonalizability of Polynomials

If $A$ is diagonalizable and $p(x)$ is any polynomial, then $p(A)$ is also diagonalizable with:

p(A) = P \cdot \text{diag}(p(\lambda_1), \ldots, p(\lambda_n)) \cdot P^{-1}

Proof:

If $A = PDP^{-1}$ , then $A^k = PD^kP^{-1}$ . Therefore:

p(A) = \sum_k c_k A^k = \sum_k c_k PD^kP^{-1} = P \left(\sum_k c_k D^k\right) P^{-1} = Pp(D)P^{-1}

And $p(D) = \text{diag}(p(\lambda_1), \ldots, p(\lambda_n))$ .

∎

Theorem 6.30: Eigenvalues of Polynomial

If $\lambda$ is an eigenvalue of $A$ , then $p(\lambda)$ is an eigenvalue of $p(A)$ .

Example 6.42: Computing p(A)

For $A = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$ and $p(x) = x^2 - 4x + 3$ :

$p(A) = \text{diag}(p(2), p(3)) = \text{diag}(4-8+3, 9-12+3) = \text{diag}(-1, 0)$

Corollary 6.11: Invertibility

If $A$ is diagonalizable with all eigenvalues non-zero, then:

A^{-1} = P \cdot \text{diag}(\lambda_1^{-1}, \ldots, \lambda_n^{-1}) \cdot P^{-1}

Theorem 6.31: Trace and Determinant

For diagonalizable $A = PDP^{-1}$ :

$\text{tr}(A) = \lambda_1 + \lambda_2 + \cdots + \lambda_n$
$\det(A) = \lambda_1 \cdot \lambda_2 \cdots \lambda_n$
$\text{tr}(A^k) = \lambda_1^k + \lambda_2^k + \cdots + \lambda_n^k$

18. Real-World Applications

PageRank Algorithm

Google's PageRank uses the dominant eigenvector of a stochastic matrix to rank web pages by importance.

Quantum Mechanics

Observable quantities correspond to eigenvalues of Hermitian operators. Diagonalization finds energy levels.

Principal Component Analysis

PCA diagonalizes the covariance matrix to find directions of maximum variance in data.

Vibration Analysis

Normal modes of vibrating systems are eigenvectors; natural frequencies are square roots of eigenvalues.

Example 6.43: Population Model

A Leslie matrix models population dynamics with age groups:

L = \begin{pmatrix} 0 & 2 & 1 \\ 0.5 & 0 & 0 \\ 0 & 0.6 & 0 \end{pmatrix}

Population after $n$ years: $\mathbf{p}_n = L^n \mathbf{p}_0$ .

Diagonalization gives long-term growth rate = dominant eigenvalue.

Quick Examples Gallery

✓ Diagonalizable

$\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ — Already diagonal

$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ — Distinct eigenvalues ±1

$\begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}$ — Symmetric

✗ NOT Diagonalizable

$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ — Jordan block

$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ — Nilpotent

$\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}$ — g=1, a=3

Diagonalizability Decision Tree

Step 1: Is the matrix symmetric (or normal)?

If YES → DIAGONALIZABLE (Spectral Theorem)

Step 2: Does it have n distinct eigenvalues?

If YES → DIAGONALIZABLE

Step 3: For repeated eigenvalues, compute geometric multiplicity

For each eigenvalue λ: find $g(\lambda) = \dim\ker(A - \lambda I)$

Step 4: Compare multiplicities

If $g(\lambda) = a(\lambda)$ for ALL λ → DIAGONALIZABLE

If $g(\lambda) < a(\lambda)$ for ANY λ → NOT DIAGONALIZABLE

The Big Picture

Diagonalization is about finding the "natural coordinates" for a linear transformation—the eigenvector basis where the transformation is simplest (just scaling).

When it works, $A = PDP^{-1}$ gives us:

Easy computation of powers: $A^k = PD^kP^{-1}$
Solutions to differential equations: $e^{At} = Pe^{Dt}P^{-1}$
Understanding of long-term dynamics via dominant eigenvalue
Geometric insight: A acts as pure scaling in eigenvector directions

Memorization Aids

The Formula

"P diagonalizes A" = $A = PDP^{-1}$

The Criterion

"geo = alg for all" = $g = a$ for each $\lambda$

Always Works

"Distinct eigenvalues" or "Symmetric"

Powers

"Power the D, keep the P"

P columns

"Eigenvectors in order"

D entries

"Eigenvalues matching P"

Learning Path

Current

6.3 Diagonalization

What's Next?

With diagonalization mastered, you're ready for:

Jordan Normal Form: The canonical form for non-diagonalizable matrices—"almost diagonal" with 1s on superdiagonal
Cayley-Hamilton Theorem: Every matrix satisfies its own characteristic equation: $\chi_A(A) = 0$
Spectral Theorem: Orthogonal diagonalization for symmetric matrices with orthonormal eigenvectors
Matrix Exponential: Using $e^A = Pe^DP^{-1}$ for systems of differential equations

Module Summary

In this comprehensive module on diagonalization, you learned:

The definition: $A = PDP^{-1}$ with P containing eigenvectors and D containing eigenvalues
The criterion: diagonalizable ⟺ $g(\lambda) = a(\lambda)$ for all eigenvalues
The algorithm: find eigenvalues, check multiplicities, construct P and D
Applications: matrix powers, differential equations, Markov chains
Special cases: symmetric matrices are always orthogonally diagonalizable
Connections: minimal polynomial, spectral decomposition, similarity theory

Diagonalization Practice

12

Questions

0

Correct

0%

Accuracy

1

A matrix is diagonalizable iff it has:

Easy

Not attempted

2

If

A = PDP^{-1}

with

D

diagonal, then

A^k =

Medium

Not attempted

3

A matrix with

n

distinct eigenvalues is:

Medium

Not attempted

4

Geometric mult = Algebraic mult for all eigenvalues implies:

Hard

Not attempted

5

Which matrix is NOT diagonalizable?

Medium

Not attempted

6

In

A = PDP^{-1}

, the columns of

P

are:

Easy

Not attempted

7

If

A

is diagonalizable, so is:

Medium

Not attempted

8

A real matrix with eigenvalues

2 \pm 3i

is:

Hard

Not attempted

9

For diagonal matrix

D

,

D^k

equals:

Easy

Not attempted

10

Every real symmetric matrix is:

Medium

Not attempted

11

If

A = PDP^{-1}

, then

\det(A) =

Medium

Not attempted

12

A nilpotent matrix (non-zero) is:

Hard

Not attempted

Frequently Asked Questions

How do I check if a matrix is diagonalizable?

1) Find all eigenvalues via det(A - λI) = 0. 2) For each eigenvalue, compute geometric multiplicity g = dim(ker(A - λI)). 3) Compare g to algebraic multiplicity a (root multiplicity). 4) Diagonalizable ⟺ g = a for all eigenvalues.

What if a matrix isn't diagonalizable?

Use Jordan normal form instead. Every matrix over ℂ is similar to a Jordan form, which is 'almost diagonal' with 1s on the superdiagonal in Jordan blocks. This is the 'best approximation' to diagonal form.

Why is diagonalization useful?

Matrix powers become trivial: A^k = PD^kP^{-1}, and D^k just raises each diagonal entry to power k. This is essential for: 1) Solving systems of differential equations x' = Ax, 2) Analyzing Markov chains, 3) Computing matrix exponentials, 4) Understanding long-term behavior of dynamical systems.

Are symmetric matrices always diagonalizable?

Yes! Real symmetric matrices are not just diagonalizable, but orthogonally diagonalizable: A = QDQ^T where Q is orthogonal. This is the Spectral Theorem, one of the most important results in linear algebra.

What's the difference between P and D in A = PDP^{-1}?

P contains eigenvectors as columns (in some order). D is diagonal with corresponding eigenvalues on the diagonal. Column i of P is an eigenvector for the eigenvalue in position (i,i) of D.

Can a matrix be diagonalizable over ℂ but not over ℝ?

Yes! If a real matrix has complex eigenvalues (like rotation matrices), it's not diagonalizable over ℝ, but is over ℂ since complex eigenvalues are still eigenvalues with eigenvectors.

Is the diagonalization unique?

No. The eigenvalues in D can be in any order (as long as P's columns match). Also, eigenvectors can be scaled by any non-zero constant. For repeated eigenvalues, there's even more freedom in choosing eigenvector bases.

How do I find P^{-1}?

You can use row reduction, adjugate formula, or for orthogonal P (from symmetric matrices), P^{-1} = P^T. In practice, for computation you often don't need P^{-1} explicitly—just solve systems.

1. What is Diagonalization?

2. The Diagonalizability Criterion

3. The Diagonalization Algorithm

Step-by-Step Algorithm

4. Computing Matrix Powers

5. Special Classes of Matrices

6. Common Mistakes

Order of eigenvectors matters

Not all matrices are diagonalizable

Forgetting the inverse

Real vs Complex diagonalization

7. Applications

Differential Equations

Markov Chains

Recurrence Relations

Quadratic Forms

8. More Worked Examples

9. Conceptual Understanding

Key Takeaways

The Formula

Matrix Powers

The Criterion

Sufficient Condition

Quick Reference

Algorithm Steps

Always Diagonalizable

10. Additional Practice

11. Challenge Problems

12. Exam Preparation

What You Should Know

Common Exam Questions

Verification Checklist

Diagonalizability Summary

Study Tips

Module Summary

Historical Notes

13. Geometric Interpretation

The Key Insight

14. Simultaneous Diagonalization

15. Spectral Decomposition

16. Connection to Minimal Polynomial

17. More Theoretical Results

18. Real-World Applications

PageRank Algorithm

Quantum Mechanics

Principal Component Analysis

Vibration Analysis

Quick Examples Gallery

✓ Diagonalizable

✗ NOT Diagonalizable

Diagonalizability Decision Tree

The Big Picture

Memorization Aids

Learning Path

What's Next?

Module Summary

Related Topics

Frequently Asked Questions

How do I check if a matrix is diagonalizable?

What if a matrix isn't diagonalizable?

Why is diagonalization useful?

Are symmetric matrices always diagonalizable?

What's the difference between P and D in A = PDP^{-1}?

Can a matrix be diagonalizable over ℂ but not over ℝ?

Is the diagonalization unique?

How do I find P^{-1}?