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This was a question in my linear algebra course. Can Someone please check If what I've done is correct?

Prove that If $A = PDP^{t}$, where $D$ is the diagonal matrix, and $P$ the matrix with eigenvectors of $A$ as columns, that $A$ is symmetric.

If $A$ by hypothesis is symmetric, it's correct to say that: $$ = $$

Se want to prove that the equality above is correct for our $A$, so we get:

$$ = < v, (pdp^t)^t w> $$

$$ = < v, dpp^t w> $$

But $dp = pd$, hence

$$ = < v, pdp^t w>$$

So $A$ is indeed symmetric.

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    How did you get $(pdp^t)^t = d p p^t$? Generally $(abc)^t = c^t b ^t a ^t$. – 2017-01-27
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    It's just because DP = PD that I wrote that way... – 2017-01-27
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    Whether a **matrix** is symmetric is an intrinsic property of the matrix itself, but whether a **linear operator** is self-adjoint [depends on the inner product](http://math.stackexchange.com/q/2110967). The question asks you to prove that $A$ is symmetric, without referring to any inner product. You should state what inner product (such as the Euclidean inner product if the matrices are real) you are using. – 2017-01-27
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    Why is $dp=pd$? – 2017-01-27
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    If $A=P D P^T$ then $A^T = P D^T P^T = P D P^T = A$. – 2017-01-27
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    @user1551 Yeah, I'm using euclidean inner product – 2017-01-27
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    @copper.hat That's what I've concluded, but using inner product... So my answer is correct right? – 2017-01-27
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    If it is the standard inner product then yes. – 2017-01-27
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    Yes it is. Thank you for answering my friend. I was afraid. It was a question in my test haha – 2017-01-27

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