MIT 18.06SC Lecture 28: Similar Matrices and Jordan Form

Linear Algebra

MIT 18.06SC

Eigenvalues

Matrix Theory

Author

Chao Ma

Published

November 21, 2025

Overview

This lecture covers: - Properties of $A^{\top}A$ (symmetric, positive semi-definite) - Similar matrices: definition, eigenvalue preservation, and eigenvector transformation - Repeated eigenvalues and the failure of diagonalization - Jordan canonical form for non-diagonalizable matrices - Jordan blocks and their geometric interpretation

1. The Matrix $A^{\top}A$

For any $m \times n$ matrix $A$, the product $A^{\top}A$ has special properties.

Properties

$A^{\top}A$ is:

Square: $n \times n$ (even if $A$ is rectangular)
Symmetric: $(A^{\top}A)^{\top} = A^{\top}(A^{\top})^{\top} = A^{\top}A$
Positive semi-definite: $x^{\top}A^{\top}Ax \geq 0$ for all $x$
Positive definite if $A$ has full column rank

Proof of Positive Semi-definiteness

For any vector $x \in \mathbb{R}^n$:

\[ x^{\top}A^{\top}Ax = (x^{\top}A^{\top})(Ax) = (Ax)^{\top}(Ax) = \|Ax\|^2 \geq 0 \]

Since $\|Ax\|^2$ is the squared length of a vector, it’s always non-negative.

When Is $A^{\top}A$ Positive Definite?

$A^{\top}A$ is positive definite (strictly $> 0$ for $x \neq 0$) if and only if:

\[ x^{\top}A^{\top}Ax = \|Ax\|^2 > 0 \quad \text{for all } x \neq 0 \]

This holds when:

$A$ has no nullspace: $Ax \neq 0$ for all $x \neq 0$
Columns are independent: $A$ has full column rank
Rank of $A$ is $n$: All $n$ columns are linearly independent

Key insight: The Gram matrix $A^{\top}A$ encodes information about the inner products between columns of $A$. If $A = [a_1, a_2, \ldots, a_n]$, then:

\[ A^{\top}A = \begin{bmatrix} a_1^{\top}a_1 & a_1^{\top}a_2 & \cdots & a_1^{\top}a_n \\ a_2^{\top}a_1 & a_2^{\top}a_2 & \cdots & a_2^{\top}a_n \\ \vdots & \vdots & \ddots & \vdots \\ a_n^{\top}a_1 & a_n^{\top}a_2 & \cdots & a_n^{\top}a_n \end{bmatrix} \]

The diagonal entries are $\|a_i\|^2$, and off-diagonal entries measure how aligned the columns are.

2. Similar Matrices

Definition

Matrices $A$ and $B$ are similar if there exists an invertible matrix $M$ such that:

\[ B = M^{-1}AM \]

Notation: $A \sim B$

Interpretation: $A$ and $B$ represent the same linear transformation in different coordinate systems (different bases). The matrix $M$ performs the change of basis.

Example 1: Diagonalization

Every diagonalizable matrix is similar to a diagonal matrix.

If $A$ has eigenvector matrix $S$ and eigenvalue matrix $\Lambda$, then:

\[ S^{-1}AS = \Lambda \]

So $A \sim \Lambda$.

Concrete example:

\[ A = \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix} \]

The eigenvalues are $\lambda_1 = 3, \lambda_2 = 1$, so:

\[ \Lambda = \begin{bmatrix}3 & 0 \\ 0 & 1\end{bmatrix} \]

Thus $A \sim \Lambda$.

Example 2: Different Similarity Transformation

Using a different choice of $M$, we can find other matrices similar to $A$.

Let $M = \begin{bmatrix}1 & 4 \\ 0 & 1\end{bmatrix}$. Then:

\[ B = M^{-1}AM = \begin{bmatrix}1 & -4 \\ 0 & 1\end{bmatrix} \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix} \begin{bmatrix}1 & 4 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}-2 & -15 \\ 1 & 6\end{bmatrix} \]

So $A \sim B$, even though $B$ looks very different from $A$.

Key observation: Similar matrices form equivalence classes. All matrices in the same class represent the same transformation, just in different bases.

3. Properties of Similar Matrices

Main Facts

If $A \sim B$ (i.e., $B = M^{-1}AM$), then:

Same eigenvalues: $A$ and $B$ have identical eigenvalues
Different eigenvectors: The eigenvectors transform as $x_B = M^{-1}x_A$
Same trace: $\text{tr}(A) = \text{tr}(B)$
Same determinant: $\det(A) = \det(B)$
Same rank: $\text{rank}(A) = \text{rank}(B)$

Proof: Eigenvalues Are Preserved

Suppose $Ax = \lambda x$ (so $\lambda$ is an eigenvalue of $A$ with eigenvector $x$).

We want to show that $\lambda$ is also an eigenvalue of $B = M^{-1}AM$.

Step 1: Start with the eigenvalue equation for $A$:

\[ Ax = \lambda x \]

Step 2: Insert $M M^{-1} = I$ to the left of $x$:

\[ A M M^{-1} x = \lambda x \]

Step 3: Multiply both sides by $M^{-1}$ on the left:

\[ M^{-1} A M M^{-1} x = \lambda M^{-1} x \]

Step 4: Recognize that $M^{-1}AM = B$:

\[ B (M^{-1} x) = \lambda (M^{-1} x) \]

Conclusion: $\lambda$ is an eigenvalue of $B$ with eigenvector $M^{-1}x$.

Key insight: The eigenvalues stay unchanged, but the eigenvectors transform according to the change of basis:

\[ x_B = M^{-1} x_A \]

4. Repeated Eigenvalues: The Bad Case

When a matrix has repeated eigenvalues, we must check whether it has enough eigenvectors to be diagonalizable.

Example: Two Contrasting Matrices

Consider two matrices with the same eigenvalues $\lambda_1 = \lambda_2 = 4$:

\[ A = \begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix}, \quad B = \begin{bmatrix}4 & 1 \\ 0 & 4\end{bmatrix} \]

Both have characteristic polynomial $(\lambda - 4)^2 = 0$, so both have double eigenvalue $\lambda = 4$.

Eigenvectors

Matrix $A$:

\[ (A - 4I)x = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}x = 0 \]

Every vector is an eigenvector! The eigenspace is all of $\mathbb{R}^2$.

Matrix $B$:

\[ (B - 4I)x = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}x = 0 \]

This requires $x_2 = 0$, so the eigenvectors are:

\[ x = c\begin{bmatrix}1 \\ 0\end{bmatrix} \]

The eigenspace is only 1-dimensional, even though the eigenvalue has algebraic multiplicity 2.

Classification

Matrix $A$: “Small family” — diagonalizable, full set of eigenvectors
Matrix $B$: “Big family” — non-diagonalizable, deficient in eigenvectors

Important: $A$ and $B$ are not similar because they have different numbers of independent eigenvectors.

5. Jordan Canonical Form

When a matrix is not diagonalizable, the best we can do is put it in Jordan form.

The Family of $2 \times 2$ Matrices with $\lambda_1 = \lambda_2 = 4$

Non-diagonalizable matrices with double eigenvalue 4:

\[ \begin{bmatrix}4 & 1 \\ 0 & 4\end{bmatrix}, \quad \begin{bmatrix}3 & 1 \\ -1 & 5\end{bmatrix}, \quad \ldots \]

These all share:

Trace: $\text{tr}(A) = 8$
Determinant: $\det(A) = 16$
Non-diagonalizable: Only one independent eigenvector

All matrices in this family are similar to the Jordan block:

\[ J = \begin{bmatrix}4 & 1 \\ 0 & 4\end{bmatrix} \]

Jordan Block

A Jordan block of size $n$ for eigenvalue $\lambda$ has the form:

\[ J_i = \begin{bmatrix} \lambda_i & 1 & 0 & \cdots & 0 \\ 0 & \lambda_i & 1 & \cdots & 0 \\ 0 & 0 & \lambda_i & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_i \end{bmatrix} \]

Key property: A Jordan block of size $n$ has exactly one eigenvector (corresponding to the repeated eigenvalue $\lambda_i$).

Structure:

Main diagonal: eigenvalue $\lambda_i$ repeated
Superdiagonal: all 1’s
Everywhere else: 0’s

6. Case Study: Nilpotent Matrices

Consider two $4 \times 4$ nilpotent matrices (all eigenvalues are 0):

Matrix $H_1$

\[ H_1 = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

Jordan form: This has one Jordan block of size 3 and one block of size 1:

\[ J_1 = \begin{bmatrix} 0 & 1 & 0 & | & 0 \\ 0 & 0 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ \hline 0 & 0 & 0 & | & 0 \end{bmatrix} \]

Eigenspace: The eigenvectors (for $\lambda = 0$) are 2-dimensional.

Matrix $H_2$

\[ H_2 = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

Jordan form: This has two Jordan blocks of size 2:

\[ J_2 = \begin{bmatrix} 0 & 1 & | & 0 & 0 \\ 0 & 0 & | & 0 & 0 \\ \hline 0 & 0 & | & 0 & 1 \\ 0 & 0 & | & 0 & 0 \end{bmatrix} \]

Eigenspace: The eigenvectors are also 2-dimensional.

Are $H_1$ and $H_2$ Similar?

No! Even though they have:

Same eigenvalues: $\lambda_1 = \lambda_2 = \lambda_3 = \lambda_4 = 0$
Same dimension of eigenspace: 2

They have different Jordan block structures, so they are not similar.

Key insight: Two matrices are similar if and only if they have the same Jordan form (up to reordering of blocks).

7. General Jordan Canonical Form

Every square matrix $A$ is similar to a unique Jordan form:

\[ J = \begin{bmatrix} J_1 & & & \\ & J_2 & & \\ & & \ddots & \\ & & & J_d \end{bmatrix} \]

where each $J_i$ is a Jordan block.

The Good Case

When $A$ is diagonalizable, the Jordan form is diagonal:

\[ J = \Lambda = \begin{bmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_n \end{bmatrix} \]

This happens when every eigenvalue’s geometric multiplicity (dimension of eigenspace) equals its algebraic multiplicity (number of times it appears as a root of the characteristic polynomial).

The Bad Case

When $A$ is not diagonalizable, some Jordan blocks have size greater than 1. These blocks correspond to eigenvalues with deficient eigenspaces.

Example: If $\lambda = 4$ has algebraic multiplicity 3 but geometric multiplicity 1, the Jordan form might be:

\[ J = \begin{bmatrix} 4 & 1 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{bmatrix} \]

Summary

Concept	Key Idea
$A^{\top}A$	Always symmetric and positive semi-definite; positive definite if $A$ has full column rank
Similar Matrices	$B = M^{-1}AM$ represents the same transformation in a different basis
Eigenvalue Preservation	Similar matrices have identical eigenvalues
Eigenvector Transformation	Eigenvectors transform as $x_B = M^{-1}x_A$
Diagonalizable	$A$ is diagonalizable if it has $n$ independent eigenvectors
Jordan Block	$n \times n$ block with eigenvalue $\lambda$ on diagonal, 1’s on superdiagonal
Jordan Form	Every matrix is similar to a block-diagonal matrix of Jordan blocks
Good Case	Jordan form is diagonal $\Leftrightarrow$ diagonalizable

Main theorem: Two matrices are similar if and only if they have the same Jordan canonical form (up to reordering of blocks).

--- title: "MIT 18.06SC Lecture 28: Similar Matrices and Jordan Form" author: "Chao Ma" date: "2025-11-21" categories: [Linear Algebra, MIT 18.06SC, Eigenvalues, Matrix Theory] --- {{< video https://www.youtube.com/watch?v=TSdXJw83kyA >}} ## Overview This lecture covers: - Properties of $A^{\top}A$ (symmetric, positive semi-definite) - Similar matrices: definition, eigenvalue preservation, and eigenvector transformation - Repeated eigenvalues and the failure of diagonalization - Jordan canonical form for non-diagonalizable matrices - Jordan blocks and their geometric interpretation --- ## 1. The Matrix $A^{\top}A$ For any $m \times n$ matrix $A$, the product $A^{\top}A$ has special properties. ### Properties $A^{\top}A$ is: 1. **Square**: $n \times n$ (even if $A$ is rectangular) 2. **Symmetric**: $(A^{\top}A)^{\top} = A^{\top}(A^{\top})^{\top} = A^{\top}A$ 3. **Positive semi-definite**: $x^{\top}A^{\top}Ax \geq 0$ for all $x$ 4. **Positive definite** if $A$ has full column rank ### Proof of Positive Semi-definiteness For any vector $x \in \mathbb{R}^n$: $$ x^{\top}A^{\top}Ax = (x^{\top}A^{\top})(Ax) = (Ax)^{\top}(Ax) = \|Ax\|^2 \geq 0 $$ Since $\|Ax\|^2$ is the squared length of a vector, it's always non-negative. ### When Is $A^{\top}A$ Positive Definite? $A^{\top}A$ is **positive definite** (strictly $> 0$ for $x \neq 0$) if and only if: $$ x^{\top}A^{\top}Ax = \|Ax\|^2 > 0 \quad \text{for all } x \neq 0 $$ This holds when: - **$A$ has no nullspace**: $Ax \neq 0$ for all $x \neq 0$ - **Columns are independent**: $A$ has full column rank - **Rank of $A$ is $n$**: All $n$ columns are linearly independent **Key insight**: The Gram matrix $A^{\top}A$ encodes information about the inner products between columns of $A$. If $A = [a_1, a_2, \ldots, a_n]$, then: $$ A^{\top}A = \begin{bmatrix} a_1^{\top}a_1 & a_1^{\top}a_2 & \cdots & a_1^{\top}a_n \\ a_2^{\top}a_1 & a_2^{\top}a_2 & \cdots & a_2^{\top}a_n \\ \vdots & \vdots & \ddots & \vdots \\ a_n^{\top}a_1 & a_n^{\top}a_2 & \cdots & a_n^{\top}a_n \end{bmatrix} $$ The diagonal entries are $\|a_i\|^2$, and off-diagonal entries measure how aligned the columns are. --- ## 2. Similar Matrices ### Definition Matrices $A$ and $B$ are **similar** if there exists an invertible matrix $M$ such that: $$ B = M^{-1}AM $$ Notation: $A \sim B$ **Interpretation**: $A$ and $B$ represent the same linear transformation in different coordinate systems (different bases). The matrix $M$ performs the change of basis. ### Example 1: Diagonalization **Every diagonalizable matrix is similar to a diagonal matrix.** If $A$ has eigenvector matrix $S$ and eigenvalue matrix $\Lambda$, then: $$ S^{-1}AS = \Lambda $$ So $A \sim \Lambda$. **Concrete example**: $$ A = \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix} $$ The eigenvalues are $\lambda_1 = 3, \lambda_2 = 1$, so: $$ \Lambda = \begin{bmatrix}3 & 0 \\ 0 & 1\end{bmatrix} $$ Thus $A \sim \Lambda$. ### Example 2: Different Similarity Transformation Using a different choice of $M$, we can find other matrices similar to $A$. Let $M = \begin{bmatrix}1 & 4 \\ 0 & 1\end{bmatrix}$. Then: $$ B = M^{-1}AM = \begin{bmatrix}1 & -4 \\ 0 & 1\end{bmatrix} \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix} \begin{bmatrix}1 & 4 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}-2 & -15 \\ 1 & 6\end{bmatrix} $$ So $A \sim B$, even though $B$ looks very different from $A$. **Key observation**: Similar matrices form **equivalence classes**. All matrices in the same class represent the same transformation, just in different bases. --- ## 3. Properties of Similar Matrices ### Main Facts If $A \sim B$ (i.e., $B = M^{-1}AM$), then: 1. **Same eigenvalues**: $A$ and $B$ have identical eigenvalues 2. **Different eigenvectors**: The eigenvectors transform as $x_B = M^{-1}x_A$ 3. **Same trace**: $\text{tr}(A) = \text{tr}(B)$ 4. **Same determinant**: $\det(A) = \det(B)$ 5. **Same rank**: $\text{rank}(A) = \text{rank}(B)$ ### Proof: Eigenvalues Are Preserved Suppose $Ax = \lambda x$ (so $\lambda$ is an eigenvalue of $A$ with eigenvector $x$). We want to show that $\lambda$ is also an eigenvalue of $B = M^{-1}AM$. **Step 1**: Start with the eigenvalue equation for $A$: $$ Ax = \lambda x $$ **Step 2**: Insert $M M^{-1} = I$ to the left of $x$: $$ A M M^{-1} x = \lambda x $$ **Step 3**: Multiply both sides by $M^{-1}$ on the left: $$ M^{-1} A M M^{-1} x = \lambda M^{-1} x $$ **Step 4**: Recognize that $M^{-1}AM = B$: $$ B (M^{-1} x) = \lambda (M^{-1} x) $$ **Conclusion**: $\lambda$ is an eigenvalue of $B$ with eigenvector $M^{-1}x$. **Key insight**: The eigenvalues stay unchanged, but the eigenvectors transform according to the change of basis: $$ x_B = M^{-1} x_A $$ --- ## 4. Repeated Eigenvalues: The Bad Case When a matrix has repeated eigenvalues, we must check whether it has enough eigenvectors to be diagonalizable. ### Example: Two Contrasting Matrices Consider two matrices with the same eigenvalues $\lambda_1 = \lambda_2 = 4$: $$ A = \begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix}, \quad B = \begin{bmatrix}4 & 1 \\ 0 & 4\end{bmatrix} $$ Both have characteristic polynomial $(\lambda - 4)^2 = 0$, so both have double eigenvalue $\lambda = 4$. ### Eigenvectors **Matrix $A$**: $$ (A - 4I)x = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}x = 0 $$ Every vector is an eigenvector! The eigenspace is all of $\mathbb{R}^2$. **Matrix $B$**: $$ (B - 4I)x = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}x = 0 $$ This requires $x_2 = 0$, so the eigenvectors are: $$ x = c\begin{bmatrix}1 \\ 0\end{bmatrix} $$ The eigenspace is only **1-dimensional**, even though the eigenvalue has algebraic multiplicity 2. ### Classification - **Matrix $A$**: "Small family" — diagonalizable, full set of eigenvectors - **Matrix $B$**: "Big family" — non-diagonalizable, deficient in eigenvectors **Important**: $A$ and $B$ are **not similar** because they have different numbers of independent eigenvectors. --- ## 5. Jordan Canonical Form When a matrix is not diagonalizable, the best we can do is put it in **Jordan form**. ### The Family of $2 \times 2$ Matrices with $\lambda_1 = \lambda_2 = 4$ **Non-diagonalizable matrices** with double eigenvalue 4: $$ \begin{bmatrix}4 & 1 \\ 0 & 4\end{bmatrix}, \quad \begin{bmatrix}3 & 1 \\ -1 & 5\end{bmatrix}, \quad \ldots $$ These all share: - **Trace**: $\text{tr}(A) = 8$ - **Determinant**: $\det(A) = 16$ - **Non-diagonalizable**: Only one independent eigenvector All matrices in this family are similar to the **Jordan block**: $$ J = \begin{bmatrix}4 & 1 \\ 0 & 4\end{bmatrix} $$ ### Jordan Block A **Jordan block** of size $n$ for eigenvalue $\lambda$ has the form: $$ J_i = \begin{bmatrix} \lambda_i & 1 & 0 & \cdots & 0 \\ 0 & \lambda_i & 1 & \cdots & 0 \\ 0 & 0 & \lambda_i & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_i \end{bmatrix} $$ **Key property**: A Jordan block of size $n$ has **exactly one eigenvector** (corresponding to the repeated eigenvalue $\lambda_i$). **Structure**: - Main diagonal: eigenvalue $\lambda_i$ repeated - Superdiagonal: all 1's - Everywhere else: 0's --- ## 6. Case Study: Nilpotent Matrices Consider two $4 \times 4$ nilpotent matrices (all eigenvalues are 0): ### Matrix $H_1$ $$ H_1 = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$ **Jordan form**: This has one Jordan block of size 3 and one block of size 1: $$ J_1 = \begin{bmatrix} 0 & 1 & 0 & | & 0 \\ 0 & 0 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ \hline 0 & 0 & 0 & | & 0 \end{bmatrix} $$ **Eigenspace**: The eigenvectors (for $\lambda = 0$) are 2-dimensional. ### Matrix $H_2$ $$ H_2 = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$ **Jordan form**: This has two Jordan blocks of size 2: $$ J_2 = \begin{bmatrix} 0 & 1 & | & 0 & 0 \\ 0 & 0 & | & 0 & 0 \\ \hline 0 & 0 & | & 0 & 1 \\ 0 & 0 & | & 0 & 0 \end{bmatrix} $$ **Eigenspace**: The eigenvectors are also 2-dimensional. ### Are $H_1$ and $H_2$ Similar? **No!** Even though they have: - Same eigenvalues: $\lambda_1 = \lambda_2 = \lambda_3 = \lambda_4 = 0$ - Same dimension of eigenspace: 2 They have **different Jordan block structures**, so they are **not similar**. **Key insight**: Two matrices are similar if and only if they have the **same Jordan form** (up to reordering of blocks). --- ## 7. General Jordan Canonical Form Every square matrix $A$ is similar to a unique Jordan form: $$ J = \begin{bmatrix} J_1 & & & \\ & J_2 & & \\ & & \ddots & \\ & & & J_d \end{bmatrix} $$ where each $J_i$ is a Jordan block. ### The Good Case **When $A$ is diagonalizable**, the Jordan form is diagonal: $$ J = \Lambda = \begin{bmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_n \end{bmatrix} $$ This happens when every eigenvalue's **geometric multiplicity** (dimension of eigenspace) equals its **algebraic multiplicity** (number of times it appears as a root of the characteristic polynomial). ### The Bad Case **When $A$ is not diagonalizable**, some Jordan blocks have size greater than 1. These blocks correspond to eigenvalues with deficient eigenspaces. **Example**: If $\lambda = 4$ has algebraic multiplicity 3 but geometric multiplicity 1, the Jordan form might be: $$ J = \begin{bmatrix} 4 & 1 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{bmatrix} $$ --- ## Summary | **Concept** | **Key Idea** | |------------|-------------| | **$A^{\top}A$** | Always symmetric and positive semi-definite; positive definite if $A$ has full column rank | | **Similar Matrices** | $B = M^{-1}AM$ represents the same transformation in a different basis | | **Eigenvalue Preservation** | Similar matrices have identical eigenvalues | | **Eigenvector Transformation** | Eigenvectors transform as $x_B = M^{-1}x_A$ | | **Diagonalizable** | $A$ is diagonalizable if it has $n$ independent eigenvectors | | **Jordan Block** | $n \times n$ block with eigenvalue $\lambda$ on diagonal, 1's on superdiagonal | | **Jordan Form** | Every matrix is similar to a block-diagonal matrix of Jordan blocks | | **Good Case** | Jordan form is diagonal $\Leftrightarrow$ diagonalizable | **Main theorem**: Two matrices are similar if and only if they have the same Jordan canonical form (up to reordering of blocks).