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$D$ is a diagonal matrix of eigenvalues and $X$ is a matrix of eigenvectors. Does $AA = A$ make $DD = D$? If so, how can you show that this is true?

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Hint: $A(XD) = A(AX) = AX$, so letting $B = AX$ we have $BD = B$ with $D$ being diagonal. Now think about what relationship between the $i$th column of the left-hand side and the $i$th column on the right-hand side must be. (And, be careful not to arrive at *too* strong of a conclusion about $D$ either.) – cardinal Oct 29 '11 at 19:49
up vote 3 down vote accepted

I am assuming this may be homework, so I'm starting with a (strong) hint.

Hint: $\newcommand{\m}{\mathbf} \m A(\m X \m D) = \m A (\m A\m X) = \m A \m X$, so letting $\m B = \m A \m X$ we have $\m B \m D= \m B$ with $\m D$ being diagonal. Now think about what the relationship between the $i$th column of the left-hand side and the $i$th column of the right-hand side must be. (And, be careful not to arrive at too strong of a conclusion about $\m D$ either.)

Alternatively, and maybe simpler, break down your original matrix relation columnwise. That is, if $\m x_i$ is the $i$th column of $\m X$ and $\lambda_i$ is the corresponding eigenvalue of $\m A$, then

$$ \lambda_i \m x_i = \m A \m x_i = \m A^2 \m x_i = \m A (\m A \m x_i) = \m A (\lambda_i \m x_i) = \lambda_i (\m A \m x_i) \> \ldots $$

Now, hopefully you can finish the rest.

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