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EDIT:

In one direction, assuming that $A$ has $n$ linearly independent eigenvectors, then I can compute the action of $A$ on the n eigenvectors, and expand each image as a linear combination of the n eigenvectors:

$$Av_1 = \lambda_1v_1 = \lambda_1v_1 + 0v_2 + ... +0v_n$$ $$Av_2 = \lambda_2v_2 = 0v_1 + \lambda_2 v_2 + ... +0v_n$$

...

$$Av_n = \lambda_nv_n = 0v_1 + 0v_2 + ... +\lambda_n v_n$$

And now taking the transpose of the coefficients gives the matrix of the linear operator, with respect to the basis of eigenvectors, which is diagonal.

Is my proof in this direction ok? It is purely...computational, so I hope it suffices, and that I don't need something a little more algebraic?

Problem statement:

I want to prove that an $nxn$ matrix $A$ has $n$ linearly independent eigenvectors if and only if it is similar to a diagonal matrix.

I'm not sure how to get started on a proof - of course, the concrete computations is no problem at all.

Any hints and suggestions are welcome.

Thanks,

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    $\begingroup$ construct a matrix $P$ whose columns are these eigen vectors. Now think about $PAP^{-1}$. $\endgroup$ – Anurag A Nov 21 '15 at 0:22
  • $\begingroup$ Any guesses on what the diagonal matrix might be? You can make an educated guess based on the context of the problem $\endgroup$ – user217285 Nov 21 '15 at 0:22
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    $\begingroup$ In basis of eigenvectors, the matrix will be diagonal by definition. $\endgroup$ – Bernard Nov 21 '15 at 0:47
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This is fairly standard stuff.

If $A v_k = \lambda_k v_k$ then let $V = \begin{bmatrix} v_1 & \cdots & v_n\end{bmatrix}$ and $\Lambda = \operatorname{diag}(\lambda_1, \cdots , \lambda_n )$. Note that $V$ is invertible. Then the above equations can be written as $AV = V \Lambda$ which gives $V^{-1} A V = \Lambda$.

If $A$ is similar to a diagonal matrix $\Lambda\operatorname{diag}(\lambda_1, \cdots , \lambda_n )$, then there is some $V$ such that $V^{-1}A V = \Lambda$. Hence $A V = V \Lambda$. Let $e_k$ be the $k$th unit vector, and let $\lambda_k$ be the $k$th diagonal entry of $\Lambda$ then $AV e_k = V \Lambda e_k = \lambda_k V e_k$. If we let $v_k = V e_k$ then we have $A v_k = \lambda v_k$. Since $V$ is invertible, it follows that the $v_k$ are linearly independent.

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  • $\begingroup$ Hi @copper.hat -- thanks for your quick response and answer. I have not read the second part of your answer but I will take a peak at it soon, if I do not come up with something. However, is my proof (just edited my question) in one direction ok? It is basically what you wrote up, but yours is purely algebraic. Is mine...incomplete? Thanks, $\endgroup$ – User001 Nov 21 '15 at 0:29
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    $\begingroup$ I don't follow your statement 'And now taking the transpose...'. Once you have $Av_k = \lambda_k v_k$ you are essentially finished, since in the basis $v_1,...,v_n$ the matrix corresponding to $A$ is just the diagonal with entries $\lambda_k$. I just wanted to given an explicit $V$ above. $\endgroup$ – copper.hat Nov 21 '15 at 0:34
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    $\begingroup$ I'm not sure I follow. This is just ordinary matrix multiplication. Since $S=\begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix}$ we have $AS = \begin{bmatrix} A v_1 & \cdots & A v_n \end{bmatrix}$. $\endgroup$ – copper.hat Nov 21 '15 at 3:36
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    $\begingroup$ Well, in general two matrices don't commute. The result follows just by computation, there's no magic. $\endgroup$ – copper.hat Nov 21 '15 at 3:58
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    $\begingroup$ Ok got it - thanks again @copper.hat :-) $\endgroup$ – User001 Nov 21 '15 at 4:17

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