# 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.

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

Does my proof look right?

• What exactly is your question? Commented Dec 30, 2019 at 18:33
• Then you should formulate this in your post and also put the proof-verification tag :) Commented Dec 30, 2019 at 18:35
• Your second equality is false.
– amd
Commented Dec 30, 2019 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.
– user403337
Commented Dec 30, 2019 at 20:47
• Sorry, it's Jordan-Chevalley decomposition.
– user403337
Commented Dec 30, 2019 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 $$a_i$$
I don't think so: It looks like you have assumed that $$i=1$$ in your proof, so the proof is not complete. In addition, we have that $$[v_i]=e_i$$, not $$[v_i]=a_ie_i$$.
I noticed that the other answer is incomplete as well because what it really proves is that $$[T][v_i]=a_i[v_i]$$. Let me complete the proof: Since $$[T][v_i]=[Tv_i]$$ and $$a_i[v_i]=[a_iv_i]$$, we have the following equivalence: $$$$\forall i:[T][v_i]=a_i[v_i]\Leftrightarrow[Tv_i]=[a_iv_i]$$$$ In addition, since the function $$$$V\ni v\mapsto[v]\in F^n$$$$ is bijective (here $$F$$ is the field associated to $$V$$), we get the following equivalence: $$$$\forall i:[T][v_i]=a_i[v_i]\Leftrightarrow[Tv_i]=[a_iv_i]\Leftrightarrow Tv_i=a_iv_i$$$$