7.3: Diagonal matrices

Note that if $T$ has $n=\dim(V)$ distinct eigenvalues, then there exists a basis $(v_1,\ldots,v_n)$ of $V$such that
$$\begin{equation*} Tv_j = \lambda_j v_j, \quad \text{for all \(j=1,2,\ldots,n\).} \end{equation*}$$

Then any $v\in V$ can be written as a linear combination $v=a_1v_1+\cdots+a_nv_n$of $v_1,\ldots,v_n$. Applying $T$ to this, we obtain $$\begin{equation*} Tv = \lambda_1 a_1 v_1 + \cdots + \lambda_n a_n v_n. \end{equation*}$$

Hence the vector $$\begin{equation*} M(v) = \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix} \end{equation*}$$

is mapped to $$\begin{equation*} M(Tv) = \begin{bmatrix} \lambda_1 a_1 \\ \vdots \\ \lambda_n a_n \end{bmatrix}. \end{equation*}$$

This means that the matrix $M(T)$for $T$ with respect to the basis of eigenvectors $(v_1,\ldots,v_n)$ is diagonal, and so we call $T$ diagonalizable: $$\begin{equation*} M(T) = \begin{bmatrix} \lambda_1 & & 0 \\ & \ddots & \\ 0& & \lambda_n \end{bmatrix}. \end{equation*}$$

We summarize the results of the above discussion in the following Proposition.

Proposition 7.3.1. If $T\in \mathcal{L}(V,V)$ has $\dim(V)$ distinct eigenvalues, then $M(T)$ is diagonal with respect to some basis of $V$. Moreover, $V$ has a basis consisting of eigenvectors of $T$. ​

Contributors

• Isaiah Lankham, Mathematics Department at UC Davis
• Bruno Nachtergaele, Mathematics Department at UC Davis
• Anne Schilling, Mathematics Department at UC Davis

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