13.1: Diagonalization

Suppose we are lucky, and we have $L \colon V\to V$, and the ordered basis $B=(v_{1}, \ldots, v_{n} )$ is a set of linearly independent eigenvectors for $L$, with eigenvalues $\lambda_{1}, \ldots, \lambda_{n}$. Then:

$$\begin{eqnarray*} L(v_{1})&=&\lambda_{1} v_{1} \\ L(v_{2})&=&\lambda_{2} v_{2} \\ &\vdots & \\ L(v_{n})&=&\lambda_{n} v_{n} \\ \end{eqnarray*}$$

As a result, the matrix of $L$ in the basis of eigenvectors $B$ is diagonal:

$$L\begin{pmatrix} x^{1}\\ x^{2}\\ \vdots\\ x^{n} \end{pmatrix}_{B} = \left( \begin{pmatrix} \lambda_{1} \\ & \lambda_{2} & & \\ & & \ddots & \\ & & & \lambda_{n} \end{pmatrix} \begin{pmatrix} x^{1}\\ x^{2}\\ \vdots\\ x^{n} \end{pmatrix} \right)_{B} ,$$

where all entries off of the diagonal are zero.

Suppose that $V$ is any $n$-dimensional vector space. We call a linear transformation $L \colon V\mapsto V$ $\textit{diagonalizable}$ if there exists a collection of $n$ linearly independent eigenvectors for $L$. In other words, $L$ is diagonalizable if there exists a basis for $V$ of eigenvectors for $L$.

In a basis of eigenvectors, the matrix of a linear transformation is diagonal. On the other hand, if an $n \times n$ matrix is diagonal, then the standard basis vectors $e_{i}$ must already be a set of $n$ linearly independent eigenvectors. We have shown:

Theorem $\PageIndex{1}$:

Given an ordered basis $B$ for a vector space $V$ and a linear transformation $L \colon V\rightarrow V$, then the matrix for $L$ in the basis $B$ is diagonal if and only if $B$ consists of eigenvectors for $L$.

Typically, however, we do not begin a problem with a basis of eigenvectors, but rather have to compute these. Hence we need to know how to change from one basis to another.

Contributor

• David Cherney, Tom Denton, and Andrew Waldron (UC Davis)