Let $V$ be a finite dimensional inner-product space over $\mathbb{C}$. What does it mean to say that an endomorphism of $V$ is self-adjoint? Prove that a self-adjoint endomorphism has real eigenvalues and may be diagonalised.

An endomorphism $\alpha: V \rightarrow V$ is called positive definite if it is self-adjoint and satisfies $\langle\alpha(x), x\rangle>0$ for all non-zero $x \in V$; it is called negative definite if $-\alpha$ is positive definite. Characterise the property of being positive definite in terms of eigenvalues, and show that the sum of two positive definite endomorphisms is positive definite.

Show that a self-adjoint endomorphism $\alpha: V \rightarrow V$ has all eigenvalues in the interval $[a, b]$ if and only if $\alpha-\lambda I$ is positive definite for all $\lambda<a$ and negative definite for all $\lambda>b$.

Let $\alpha, \beta: V \rightarrow V$ be self-adjoint endomorphisms whose eigenvalues lie in the intervals $[a, b]$ and $[c, d]$ respectively. Show that all of the eigenvalues of $\alpha+\beta$ lie in the interval $[a+c, b+d]$.