# 13.1: Diagonalization

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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

This page titled 13.1: Diagonalization is shared under a not declared license and was authored, remixed, and/or curated by David Cherney, Tom Denton, & Andrew Waldron.