# Prop: If $T : V \rightarrow V$ and if $v_1,…,v_n$ is a basis such that $T$ is diagonal, then the $v_i$ are eigenvectors.

Prop: If $$T : V \rightarrow V$$ and if $$v_1,...,v_n$$ is a basis such that $$T$$ is diagonal, then the $$v_i$$ are eigenvectors.

Pf: The eigenvalues of a diagonal matrix are its diagonal entries.

Let $$T = \begin{pmatrix} a_1 & 0 & \dots & 0 \\ 0 & a_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots &a_n \end{pmatrix}$$, where the columns of $$[T]$$ are basis vectors.

$$Tv_i = \begin{pmatrix} a_1 & 0 & \dots & 0 \\ 0 & a_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots &a_n \end{pmatrix}\begin{pmatrix} a_1 \\ 0 \\ \vdots \\ 0\end{pmatrix}=\begin{pmatrix} a^2_1 \\ 0 \\ \vdots \\ 0\end{pmatrix}=\lambda\begin{pmatrix} a_1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} =\lambda v_i \square$$

Does my proof look right?

• What exactly is your question? – Severin Schraven Dec 30 '19 at 18:33
• Then you should formulate this in your post and also put the proof-verification tag :) – Severin Schraven Dec 30 '19 at 18:35
• Your second equality is false. – amd Dec 30 '19 at 19:04
• It is interesting to note that this means that, up to the order the eigenvalues occur in, there is only one diagonal matrix that a matrix can be similar to. That is, the diagonalization, if it exists, is essentially unique. See the Jordan-Chevalley normal form. This is a more general form. – Chris Custer Dec 30 '19 at 20:47
• Sorry, it's Jordan-Chevalley decomposition. – Chris Custer Dec 30 '19 at 21:03

Anyhow, we want to show that $$Tv_i= \lambda_i v_i$$ for some $$\lambda_i$$ a scalar, for every $$i$$.
Now $$Tv_i= \begin{pmatrix} a_1 & 0 & \dots & 0 \\ 0 & a_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots &a_n \end{pmatrix} \begin{pmatrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ \vdots \\ a_i \\ \vdots \\ 0 \end{pmatrix}=a_i\begin{pmatrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0 \end{pmatrix}=a_iv_i$$
So every $$v_i$$ is an eigenvector with eigenvalue $$a_i$$