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,
Best Answer
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.