# Easy condition for positive definite endomorphism

This problem is taken from Golan's linear algebra book.

Problem: Let $V$ be an inner product space over $\mathbb{R}$ and let $\alpha$ be an endomorphism of $V$. Show that $\alpha$ is positive definite if and only if $\alpha+\alpha^*$ is positive definite.

Definition: An endomorphism $\alpha$ is positive definite if and only if it is selfadjoint and satisfies the condition that $\langle \alpha(v), v\rangle$ is a positive real number for all nonzero $v\in V$.

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First, assume $\alpha$ is positive definite. We have $\langle \alpha(v), v\rangle\in \mathbb{R}^+.$, and because $\alpha=\alpha^*$, their sum is self-adjoint.

Now suppose $\alpha+\alpha^*$ is positive definite. We have $\langle (\alpha+\alpha^*)(v), v\rangle \in \mathbb{R}^+$. Also,

$$\langle \alpha(v), v\rangle =\langle v, \alpha^*(v)\rangle =\langle \alpha^*(v), v\rangle.$$

This shows that

$$\langle (\alpha+\alpha^*)(v), v\rangle =\langle \alpha(v), v\rangle+ \langle \alpha^*(v), v\rangle = \langle \alpha(v), v\rangle+\langle \alpha(v), v\rangle$$

is always nonnegative, which shows that $\langle \alpha(v), v\rangle$ is always nonnegative, and it is easy to check that equality only holds when $v=0$.

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