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A linear transformation $T:\mathbb{V}$ $\rightarrow$ $\mathbb{V}$ on an inner product space is called self-adjoint if: $\langle Tx,y \rangle = \langle x,Ty \rangle$ for all $x,y \in V$. Prove that $T:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is self-adjoint if and only if its associated matrix (in the standard basis) is symmetric.

I'm not too sure how to tackle this, I'm guessing using the properties of a matrix being symmetric in inner product terms is a starting point. i.e. $\langle$$Ax,y$$\rangle$ = $\langle$$x,Ay$$\rangle$ where $A\in{M}_n(\mathbb{R})$.

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2 Answers 2

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Hint: the standard inner product on $\mathbb{R}^n$ is just matrix multiplication, by $$\langle u,v\rangle=u^Tv.$$

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  • $\begingroup$ This helpped me to prove $<Au,v> = <A^Tv,u>$ for any matrix $A$: $<Au,v> = (Au)^Tv = u^TA^Tv = <u, A^Tv> = <A^Tv,u>$ $\endgroup$ Feb 20, 2019 at 17:19
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Hint: Compute $\langle x,Ay \rangle$ and $\langle Ax,y\rangle$ in terms of the entries of $A$ when $x$ and $y$ are both standard basis vectors.

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