Let \(A,B\) be Hermitian \(n\times n\) matrices. Suppose every eigenvalue of \(A\) lies in \([a,a']\) and every eigenvalue of \(B\) lies in \([b,b']\). Prove that every eigenvalue of \(A+B\) lies in \[ [a+b,a'+b']. \]
Solution
1. For every \(v\in\CC^n\),\[ a\norm{v}^2 \leq v^*Av \leq a'\norm{v}^2. \]
By the spectral theorem, there is an orthonormal eigenbasis \(e_1,\ldots,e_n\) for \(A\), with corresponding eigenvalues \(\lambda_j\in[a,a']\). Write \[ v=\sum_{j=1}^n c_je_j. \] Then \[ v^*Av = \sum_{j=1}^n\lambda_j\abs{c_j}^2, \qquad \norm{v}^2 = \sum_{j=1}^n\abs{c_j}^2. \] Bounding every \(\lambda_j\) between \(a\) and \(a'\) gives the claim.
2. For every \(v\in\CC^n\),\[ b\norm{v}^2 \leq v^*Bv \leq b'\norm{v}^2. \]
Apply the argument of step 1 to the Hermitian matrix \(B\), whose eigenvalues lie in \([b,b']\).
3. For every \(v\in\CC^n\),\[ (a+b)\norm{v}^2 \leq v^*(A+B)v \leq (a'+b')\norm{v}^2. \]
Add the inequalities from steps 1 and 2 and use \[ v^*(A+B)v=v^*Av+v^*Bv. \]
4. Every eigenvalue \(\lambda\) of \(A+B\) satisfies\[ \boxed{a+b\leq\lambda\leq a'+b'}. \]
The matrix \(A+B\) is Hermitian, so its eigenvalues are real. Let \(v\neq0\) satisfy \[ (A+B)v=\lambda v. \] Then \[ v^*(A+B)v = \lambda\norm{v}^2. \] Apply step 3 and divide by the positive number \(\norm{v}^2\).
5. Q.E.D.
Step 4 is the required spectral interval bound.