Necessary and sufficient conditions of eigen values of a real symmetric matrix to be in the interval $[a ,b]$ Prove that all the eigenvalues of a real symmetric matrix A  lie on the interval  $[a,b]$ if and only if the quadratic form  with matrix $A-\lambda I$  is positive definite for any $\lambda <a$ and is negative definite for any $\lambda > b$. I know that for a positive definite matrix eigenvalues are strictly positive and for negative definite matrix the eigenvalues are strictly negative.
 A: The spectral theorem asserts that any real symmetric matrix is diagonalizable, i.e. there exist an orthogonal matrix $P$ such that $$P^{-1}AP = D$$ where $D$ is diagonal matrix containing the eigenvalues of $A$. Observe, $$P^{-1}(A-\lambda I)P = P^{-1}AP-\lambda I = D-\lambda I$$ Recall that the process of conjugation by an orthogonal matrix does not change the set of eigenvalues of the matrix. Therefore, $D-\lambda I$ has the same eigenvalues as $A-\lambda I$. Next we must relate the eigenvalues of $D$, (and therefore the eigenvalues of $A$), to $D-\lambda I$. But this is simple since the addition or subtraction of a scalar multiple of the identity matrix shifts each of the entries (and therefore eigenvalues) by $\lambda$.
If $D-\lambda I$ is positive definite for any $\lambda <a$, then for any eigenvalue $c$ of $D$, $c-a \geq 0$. Similarly, $D-\lambda I$ is negative definite for any $\lambda > b$, then for any eigenvalue $c$ of $D$, $c-b \leq 0$. Hence, $a \le c \le b$ for any eigenvalue $c$.
