Diagonalization

$A$ is diagonalizable if there exists $S$ such that:

$$A = S\Lambda S^{-1}$$

Where:

• $S$ — matrix of eigenvectors (columns)
• $\Lambda$ — diagonal matrix of eigenvalues

When Is It Possible?

$A$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors.

• Distinct eigenvalues → always diagonalizable
• Repeated eigenvalues → may or may not be diagonalizable

Why It’s Useful

$$A^k = S\Lambda^k S^{-1}$$

Raising a diagonal matrix to a power is trivial! Just raise each eigenvalue.

Similarity

$A$ and $B$ are similar if $B = M^{-1}AM$ for some invertible $M$. Similar matrices have:

• Same eigenvalues
• Same trace
• Same determinant
• Same rank

Diagonalization is a special case of similarity where $B = \Lambda$.