Diagonalize a Matrix

1. Find eigenvalues
2. $\mathcal{B}=\{\vec{b}_1, \dots, \vec{b}_n\}$ are eigenvectors.
3. $[T]_\mathcal{B}=\begin{bmatrix} \lambda_1 & \dots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \dots & \lambda_n \end{bmatrix}$

Diagonalization Theorems

• $T:\R^n\to\R^n$ can be represented by a diagonal matrix $\Leftrightarrow \exist$ basis for $\R^n$ consisting of eigenvectors of $T$. If $B$ is such a basis, then $[T]_B$ is a diagonal matrix.
• A matrix is diagonalizable if it's similar to a diagonal matrix.
• Eigenvectors are always linearly independent

Spectral Decomposition

Given $A$, find $A=Q\Lambda Q^{-1}$

1. Find eigenvalues $\lambda_i$ by solving a characteristic polynomial of $A$
2. Let $\Lambda=\begin{bmatrix} \lambda_1 & \dots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \dots & \lambda_n \end{bmatrix}$
3. Find eigenvectors $\vec{v}_i$ by solving $\text{null}(A-\lambda_iI)$ for each $\lambda_i$
4. Let base $\mathcal{L}=\{\vec{v}_1, \dots, \vec{v}_n\}$
5. Let $Q=[\mathcal{E}\leftarrow\mathcal{L}]=\begin{bmatrix}\vec{v}_1 & \dots & \vec{v}_n\end{bmatrix}$
6. Find $Q^{-1}=[\mathcal{L}\leftarrow\mathcal{E}]$