LINEAR-ALGEBRA · Interactive Practice | Unit 21 · Video 1

The Matrix Factorization That Unlocks Eigenvalues: A = SΛS⁻¹

IKey Formulas

Formula	Name	What It Says
$A x_i = \lambda_i x_i$	Eigenvalue equation	$A$ stretches eigenvector $x_i$ by $\lambda_i$
$AS = S\Lambda$	Master equation	Stack eigenvectors; $\Lambda$ is diagonal of eigenvalues
$A = S\Lambda S^{-1}$	Diagonalization	Valid when $S$ has $n$ independent columns
$A^k = S\Lambda^k S^{-1}$	Power formula	Raise each diagonal entry to the $k$

IIVisualization 1 — Eigenvectors Only Stretch

In its eigenvector basis $A$ only stretches: eigenvectors keep their direction while a generic vector rotates.

λ₁ 2 λ₂ 1.5

💡 Those two directions are the only ones $A$ leaves unrotated — $\Lambda$ is $A$ written in exactly that basis.

IIIVisualization 2 — Why Powers Are Trivial

Powers of a diagonalized matrix are trivial — $\Lambda^k$ just exponentiates each eigenvalue, and the largest soon dominates.

λ₁ 1.5 λ₂ 0.75 power k 5

💡 The largest $|\lambda|$ sets the long-run growth rate of $A^k x$ — the engine behind difference equations, ODEs, and matrix exponentials.

IVVisualization 3 — Diagonalizable vs. Defective

Diagonalization needs two independent eigenvectors — edit $A$ and watch when the second direction vanishes.

a 2 b 1 c 0 d 3

💡 Distinct eigenvalues guarantee diagonalizability; symmetric matrices go further, with orthogonal eigenvectors.

VQuiz Questions

Question 1 · The Master Equation

When you stack the eigenvectors of $A$ as columns of $S$ and apply $A$ , you get the master equation $AS = S\Lambda$ . Given

$A = \begin{pmatrix} 4 & 0 \\ 0 & 5 \end{pmatrix}, \quad S = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},$

what is the diagonal eigenvalue matrix $\Lambda$ ?

✅ Correct! The $i$ -th diagonal entry of $\Lambda$ is the eigenvalue of the $i$ -th column of $S$ .

❌ Not quite. Remember: the order of eigenvalues in $\Lambda$ must match the order of eigenvectors in $S$ .

Show solution

Solution:

Since $A$ is already diagonal with entries $4$ and $5$ , its eigenvalues are $\lambda_1 = 4$ (with eigenvector $e_1$ ) and $\lambda_2 = 5$ (with eigenvector $e_2$ ). With $S = I$ , we have

$AS = AI = A = \begin{pmatrix} 4 & 0 \\ 0 & 5 \end{pmatrix}, \qquad S\Lambda = I\Lambda = \Lambda.$

For $AS = S\Lambda$ to hold, $\Lambda = \text{diag}(4, 5)$ .

The order of the diagonal entries of $\Lambda$ must match the order of eigenvectors in $S$ .

Question 2 · When Diagonalization Fails

True or False: Every square matrix $A$ can be written as $A = S\Lambda S^{-1}$ .

✅ Correct! Defective matrices (like $\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ ) cannot be diagonalized.

❌ Not quite. Having eigenvalues is not enough — you need $n$ independent eigenvectors so that $S^{-1}$ exists.

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Solution:

The statement is False.

Diagonalization requires $S$ to be invertible, which means its columns — the eigenvectors — must be linearly independent. When $A$ has a repeated eigenvalue that doesn't supply enough independent eigenvectors, $A$ is called defective and cannot be diagonalized.

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

Characteristic polynomial: $(\lambda - 2)^2 = 0$ , so $\lambda = 2$ is a double eigenvalue. But solving $(A - 2I)x = 0$ gives only the single eigenvector direction $(1, 0)^T$ . Since we need 2 independent eigenvectors but have only 1, $S$ is singular and $A = S\Lambda S^{-1}$ fails.

❌ Not diagonalizable: defective matrices
Diagonalizable: matrices with $n$ independent eigenvectors (e.g., distinct eigenvalues, or symmetric matrices)

Question 3 · Powers Made Easy

Suppose $A = S\Lambda S^{-1}$ with $\Lambda = \text{diag}(2, 3)$ . What is $\Lambda^{10}$ , the tenth power of the eigenvalue matrix?

✅ Correct! For a diagonal matrix, $(\Lambda^k)_{ii} = \lambda_i^k$ .

❌ Not quite. You multiplied by 10 instead of raising to the 10th power. $\Lambda^{10}$ means ten matrix products, which for diagonals means $\lambda_i^{10}$ .

❌ Not quite. Raise each diagonal entry to the 10th power.

Show solution

Solution:

Powers of a diagonal matrix are trivial: raise each diagonal entry to the power.

$\Lambda^k = \begin{pmatrix} \lambda_1^k & 0 \\ 0 & \lambda_2^k \end{pmatrix}$

For $\Lambda = \text{diag}(2, 3)$ and $k = 10$ :

$2^{10} = 1024, \quad 3^{10} = 59049$

So $\Lambda^{10} = \text{diag}(1024, 59049)$ .

Why this matters: Using $A^k = S\Lambda^k S^{-1}$ , computing $A^{10}$ reduces to two scalar exponentiations plus one matrix multiply on each side — far cheaper than multiplying $A$ by itself 10 times.

Question 4 · What Does Λ Really Represent?

In the factorization $A = S\Lambda S^{-1}$ , which statement best describes the role of $\Lambda$ ?

✅ Correct! $\Lambda$ is $A$ viewed from the eigenvector basis — the world where $A$ is pure scaling.

❌ Not quite. Think of $S^{-1}$ as changing to the eigenvector basis, $\Lambda$ as stretching in that basis, and $S$ as changing back.

Show solution

Solution:

$\Lambda$ is $A$ expressed in the eigenvector basis. Reading $A = S\Lambda S^{-1}$ right-to-left:

$S^{-1}$ converts a vector from standard coordinates into eigenvector coordinates.
$\Lambda$ applies $A$ 's action in that basis — but in this basis it's just stretching (no rotation, no mixing).
$S$ converts the result back to standard coordinates.

Important subtleties:

$\Lambda \neq A$ in general. They represent the same linear map, but in different bases.
$S$ is not unique. You can reorder the eigenvectors (the columns of $S$ ), and $\Lambda$ 's diagonal reorders to match. You can also rescale any eigenvector by a nonzero constant.
$\Lambda$ is not the inverse of $A$ . (The inverse of $A$ , when it exists, would be $S\Lambda^{-1}S^{-1}$ .)

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