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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?

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    $\begingroup$ What exactly is your question? $\endgroup$ Commented Dec 30, 2019 at 18:33
  • $\begingroup$ Then you should formulate this in your post and also put the proof-verification tag :) $\endgroup$ Commented Dec 30, 2019 at 18:35
  • $\begingroup$ Your second equality is false. $\endgroup$
    – amd
    Commented Dec 30, 2019 at 19:04
  • $\begingroup$ 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. $\endgroup$
    – user403337
    Commented Dec 30, 2019 at 20:47
  • $\begingroup$ Sorry, it's Jordan-Chevalley decomposition. $\endgroup$
    – user403337
    Commented Dec 30, 2019 at 21:03

2 Answers 2

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I can't understand your proof.

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 $a_i$

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Does my proof look right?

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: \begin{equation} \forall i:[T][v_i]=a_i[v_i]\Leftrightarrow[Tv_i]=[a_iv_i] \end{equation} In addition, since the function \begin{equation} V\ni v\mapsto[v]\in F^n \end{equation} is bijective (here $F$ is the field associated to $V$), we get the following equivalence: \begin{equation} \forall i:[T][v_i]=a_i[v_i]\Leftrightarrow[Tv_i]=[a_iv_i]\Leftrightarrow Tv_i=a_iv_i \end{equation}

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