= Solution
A <self-adjoint operator> $T$ satisfies $\langle Tx,y\rangle=\langle x,Ty\rangle$. If $Tx=\lambda x$, then $\lambda\langle x,x\rangle=\langle Tx,x\rangle=\langle x,Tx\rangle=\overline\lambda\langle x,x\rangle$, so $\lambda\in\mathbb R$. An eigenvector exists over $\mathbb C$; its <orthogonal complement> is $T$-invariant. Induction on dimension produces an <orthonormal basis> of eigenvectors, proving the <finite-dimensional spectral theorem>.
A self-adjoint $T$ is positive definite exactly when every eigenvalue is positive, since $\langle Tx,x\rangle=\sum_i\lambda_i|x_i|^2$ in an eigenbasis. The quadratic forms of two positive-definite operators add, so their sum is positive definite.
The same eigenbasis shows that $\operatorname{spec}T\subset[a,b]$ exactly when $T-\lambda I$ is positive definite for every $\lambda<a$ and negative definite for every $\lambda>b$. Finally, if $v$ is a unit eigenvector of $\alpha+\beta$ with eigenvalue $\gamma$, the <Rayleigh quotient> bounds give
$$
\gamma=\langle\alpha v,v\rangle+\langle\beta v,v\rangle\in[a+c,b+d].
$$
Thus \b[every eigenvalue of $\alpha+\beta$ lies in $[a+c,b+d]$].
Back to article page
