# Showing that $\|\alpha(v) \|\leq \| v\|$ for all $v \in V$, if $\alpha\in End(V )$ is self-adjoint.

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 $v \in V$.

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 $$\langle v,v\rangle=\sum^{n}_{i=1}|a_i|^2.$$
• 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. – ChocolateAndCheese Jan 4 '17 at 5:29
• 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? – Marc van Leeuwen Jan 4 '17 at 7:15