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Let $V$ be a finitely-generated inner product space and let $\alpha\in End(V )$ be self-adjoint. Show that $\|\alpha(v) \|\leq \| v\| $ for all $v \in V$.

I'm stuck on this problem. Here is what I have in my mind about this.

Since $V$ is finitely generated inner product space and $\alpha$ is a self-adjoint then $\alpha$ is orthogonally diagonalizable.

Let $\lambda_1,\dots,\lambda_n$ be the distinct eigenvalues of $\alpha$. Let $v\in V$, there exist scalars $a_1,\dots,a_n$ such that $v=\sum_{i=1}^{n}a_iv_i$, where $v_1,\dots,v_n$ are eigenvectors of $\alpha$ associated with eigenvalues $\lambda_1,\dots,\lambda_n$.

If I am correct then we can get $$\langle \alpha(v),\alpha(v)\rangle=\sum^{n}_{i=1}|a_i|^2\lambda_i^2,$$ and $$\langle v,v\rangle=\sum^{n}_{i=1}|a_i|^2.$$

However, I can't go further, any help would be appreciated.

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    $\begingroup$ I don't think that this is true in general. For example, if $V$ is a one dimensional real vector space, then we could take $\alpha$ to be multiplication by $2$. In this case, $\alpha$ is self-adjoint but clearly fails the condition you are trying to prove. $\endgroup$ – ChocolateAndCheese Jan 4 '17 at 5:29
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    $\begingroup$ This obviously fails to be true for $\alpha$ given on an orthogonal basis by a diagonal matrix with real entries (such endomorphisms are always self-adjoint), when some diagonal entry has absolute value${}>1$. It beats me why somebody would think of stating such a claim. Where did you find this statement? $\endgroup$ – Marc van Leeuwen Jan 4 '17 at 7:15
  • $\begingroup$ @MarcvanLeeuwen This book, The linear Algebra a beginning graduate students ought to know. third Edition Chapter 17, Exercise 1094 $\endgroup$ – Parisina Jan 4 '17 at 17:21
  • $\begingroup$ It's one of the reference books for my school! But this question is obviously wrong. you're right. Thank you! $\endgroup$ – Parisina Jan 4 '17 at 17:25

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