The Diagonalization Theorem

Definition 1.26 — why \(AP = PD\) means “columns of \(P\) are eigenvectors”. 5-minute lesson.

Definition 1.26 (Diagonalization Theorem). A matrix \(A \in \mathbb{R}^{n \times n}\) is non-defective iff it is diagonalizable, that is, \(\exists\) an invertible \(P \in \mathbb{R}^{n \times n}\) such that \[ D = P^{-1} A P. \] \(D\) is diagonal. We say \(A\) is diagonalized by \(P\).

1. The one idea: diagonal = eigenvector basis

Multiplying by a diagonal matrix only scales coordinates. So \(D = P^{-1}AP\) says:

\(A\) in standard coordinates = scale in the \(P\)-coordinates.

\(P^{-1}\) converts into the new basis, \(D\) scales, \(P\) converts back. That new basis must be one where \(A\) just stretches vectors — i.e. an eigenbasis.

Takeaway: Diagonalizable \(\iff\) there are \(n\) linearly independent eigenvectors to build \(P\) from. \(D\) holds the eigenvalues, \(P\) holds the eigenvectors.

2. The proof trick: use \(AP = PD\)

\(D = P^{-1}AP\) is awkward. Multiply on the left by \(P\):

\[ AP = PD \tag{6} \]

Define, exactly as in your image:

\[ P := [p_1, \dots, p_n], \qquad D = \mathrm{diag}(\lambda_1,\dots,\lambda_n). \]

Claim: (6) holds iff each \(\lambda_i\) is an eigenvalue of \(A\) with corresponding eigenvector \(p_i\). Why? Compare columns.

Left side: \(A\) acts on each column

\[ AP = A[p_1,\dots,p_n] = [Ap_1,\dots,Ap_n] \tag{7} \]

Matrix multiplication acts column-by-column. No assumption about \(p_i\) yet.

Right side: \(D\) scales each column

\[ PD = [p_1,\dots,p_n]\begin{bmatrix}\lambda_1 & 0 & \cdots & 0\\0 & \lambda_2 & \cdots & 0\\\vdots & \vdots & \ddots & \vdots\\0 & 0 & \cdots & \lambda_n\end{bmatrix} = [\lambda_1p_1,\dots,\lambda_np_n] \tag{8} \]

Right-multiplying by a diagonal matrix scales column \(i\) by \(\lambda_i\).

So \(AP = PD\) is just \(n\) column equations glued together:

\[ Ap_1 = \lambda_1 p_1, \tag{9} \] \[ \vdots \] \[ Ap_n = \lambda_n p_n. \tag{10} \]

That is literally the eigenvector equation. Therefore the columns of \(P\) must be eigenvectors, and the diagonal entries of \(D\) must be the matching eigenvalues.

3. Why “\(n\) independent eigenvectors”?

Definition 1.26 demands \(P\) be invertible (full rank, Theorem 4.3 in your book). An \(n \times n\) matrix is invertible iff its columns are linearly independent. Hence:

\(A\) diagonalizable \(\iff\) its eigenvectors \(p_1,\dots,p_n\) form a basis of \(\mathbb{R}^n\). That is what “non-defective” means. A “defective” matrix is missing an eigenvector direction.

4. Concrete 2×2 check

Let \(A = \begin{bmatrix}4 & 1\\2 & 3\end{bmatrix}\). Trace \(=7\), det \(=10\), so eigenvalues \(\lambda_1=5,\ \lambda_2=2\).

Eigenvectors: \(p_1=\begin{bmatrix}1\\1\end{bmatrix}\) since \(Ap_1=5p_1\); \(p_2=\begin{bmatrix}1\\-2\end{bmatrix}\) since \(Ap_2=2p_2\).

Build \(P=[p_1\ p_2]=\begin{bmatrix}1&1\\1&-2\end{bmatrix}\), \(D=\begin{bmatrix}5&0\\0&2\end{bmatrix}\). Then:

\[ AP = [5p_1,\ 2p_2] = PD. \]

Since \(p_1,p_2\) are independent, \(P\) is invertible (\(\det P=-3\)), and \(P^{-1}AP=D\). If we had only one eigenvector direction, \(P\) would be singular and diagonalization would fail.

Object	What it holds	Condition
\(D\)	eigenvalues \(\lambda_i\)	diagonal
\(P\)	eigenvectors \(p_i\)	invertible (independent columns)
Pairing	column \(i\) ↔ diagonal \(i\)	\(Ap_i=\lambda_i p_i\)

5. If the professor says: “Prove it” — follow this flow

Memorize 5 verbs: Claim → Define → Rewrite → Expand → Conclude. One sentence per box. Write the gray line, say the white line.

Memory hook: “Cats Drink Really Expensive Coffee” = Claim, Define, Rewrite, Expand, Conclude. The Expand step splits into ⇒ and ⇐.

1CLAIM — state both directions

Non-defective ⟺ diagonalizable, i.e. ∃ invertible P: P⁻¹AP = D

Say ► “A is non-defective if and only if A is diagonalizable, that is, there exists an invertible P such that P-inverse A P equals D, where D is diagonal. I will prove both directions.”

↓

2DEFINE — name P and D

P = [p₁ … pₙ], D = diag(λ₁ … λₙ)

Say ► “Let P be the matrix with columns p-one through p-n, and let D be the diagonal matrix with lambda-one through lambda-n on the diagonal.”

↓

3REWRITE — kill the inverse

D = P⁻¹AP ⟺ AP = PD (6)

Say ► “D equals P-inverse A P holds if and only if A P equals P D. I multiply on the left by P, so I never have to work with the inverse directly.”

↓

4EXPAND — columns both sides

AP = [Ap₁ … Apₙ] (7) vs PD = [λ₁p₁ … λₙpₙ] (8)

Say ► “A times P equals A times the block p-one through p-n, which is A p-one through A p-n. P times D equals lambda-one p-one through lambda-n p-n, because right-multiplying by a diagonal matrix scales column i by lambda-i. So A P equals P D means A p-i equals lambda-i p-i for each i.”

↓ ⇐ split ⇒

5aCONCLUDE ⇒ — diagonalizable → eigenbasis

AP = PD ⇒ Apᵢ = λᵢpᵢ, pᵢ ≠ 0 independent (P invertible) ⇒ non-defective

Say ► “Assume A is diagonalizable, so there is an invertible P with P-inverse A P equals D. Then A P equals P D, so A p-i equals lambda-i p-i. The columns of an invertible P are nonzero and linearly independent, so each p-i is a genuine eigenvector, and I have n of them. Hence A is non-defective.”

5bCONCLUDE ⇐ — eigenbasis → diagonalizable

Given n indep. Apᵢ = λᵢpᵢ → build P,D → AP = PD, P invertible → P⁻¹AP = D

Say ► “Conversely, assume A is non-defective, so there exist n linearly independent eigenvectors p-one through p-n with A p-i equals lambda-i p-i. Define P as the block p-one through p-n and D as diag lambda-one through lambda-n. Then A P equals P D by the same column calculation. Since the p-i are independent, P is invertible by Theorem 4.3. Multiply A P equals P D on the left by P-inverse to get P-inverse A P equals D. So A is diagonalizable.”

↓

∎QED — one closing line

Say ► “Therefore A is non-defective if and only if A is diagonalizable. Q-E-D.”

If grilled: “Why pᵢ ≠ 0?” → columns of invertible P can’t be zero. “Why independent?” → Theorem 4.3, P invertible ⟺ full rank. “Swap columns?” → swap the λ’s too; pairing i↔i must hold.

6. Check yourself (retrieval practice)

Pick an answer — feedback is immediate. Answers in each question are matched in length so formatting gives no hints.

Q1. Which equation unlocks the whole proof? AP equals PD AP equals DP PA equals PD

Q2. What is entry D[i,i]? lambda_i vector_i matrix_A

Q3. Why must P be invertible? rank n basis rank n image rank n nulls

7. Go to the source

Primary source (best high-trust exposition): Gilbert Strang, Introduction to Linear Algebra, §6.2 “Diagonalizing a Matrix” — or MIT 18.06 Lecture 22 (Diagonalization and Powers of A), freely available. It proves the same \(AP=PD\) → eigenvector argument and shows why \(A^k = PD^kP^{-1}\) is the payoff.

Next: 0002 — singular value decomposition · Reference: diagonalization cheatsheet (upcoming).