Videos for Section 4.2


A matrix $A$ is said to be diagonalizable if there is a basis of ${\bf R}^n$ consisting of eigenvectors of $A$. Finding the eigenvalues and eigenvectors of $A$ is called diagonalizing the matrix.

If $A$ is an $n \times n$ matrix with eigenvectors $\mathcal{B}=\{ {\bf b}_1, {\bf b}_2, \ldots , {\bf b}_n \}$, and if $L({\bf x}) = A{\bf x}$, then $[L]_\mathcal{E} = A$ and $[L]_\mathcal{B} = D$, where $$D = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \cr 0 & \lambda_2 & \cdots & 0 \cr \vdots & \vdots & \ddots & \vdots \cr 0&0&\cdots & \lambda_n \end{pmatrix}$$ is a diagonal matrix whose diagonal entries are the eigenvalues. To get $A$ we just apply a change-of-basis: $$A = [L]_\mathcal{E} = P_{\mathcal{EB}} [L]_\mathcal{B} P_{\mathcal{EB}}^{-1} = P D P^{-1},$$ where $$P = P_{\mathcal{EB}} = \begin{pmatrix} \phantom{A}&&& \cr {\bf b}_1 & {\bf b}_2 & \cdots & {\bf b}_n \cr \phantom{A}&&& \end{pmatrix}. $$

If $L: V \to V$ is a linear operator, and if $\mathcal{B}$ is an arbitrary basis (not necessarily a basis of eigenvectors), then the eigenvalues of $L$ are the same as the eigenvalues of $[L]_\mathcal{B}$, and the eigenvectors of $L$ are the vectors whose coordinates (in the $\mathcal{B}$ basis) are eigenvectors of $[L]_\mathcal{B}$. To see this, note that the following statements are equivalent: \begin{eqnarray*} L({\bf x}) &=& \lambda {\bf x} \cr [L({\bf x})]_\mathcal{B} &=& [\lambda {\bf x}]_\mathcal{B} \cr [L]_\mathcal{B} [{\bf x}]_\mathcal{B} &=& \lambda [{\bf x}]_\mathcal{B}, \end{eqnarray*} so ${\bf x}$ is an eigenvector of $L$ with eigenvalue $\lambda$ if and only if $[{\bf x}]_\mathcal{B}$ is an eigenvector of $[L]_\mathcal{B}$, also with eigenvalue $\lambda$.