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$A,B$ are $n\times n$ Hermitian matrices. If the eigenvalues of $A$ and $B$ are all in an interval $I$, then the eigenvalues of any convex combination of $A,B$ are also in $I$.

In the book Bhatia, Matrix Analysis, the author says "This is an easy consequence of results in Chapter III.

From the answers (lemmas, in fact) to the question Eigenvalues of Matrix Sums (mathoverflow) the statement follows easily.

However I would like to obtain a direct (possibly "simple") proof. If it is possible, for the infinite dimensional Hilbert spaces also.

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We can use – Davide Giraudo Sep 9 '12 at 19:19
up vote 2 down vote accepted

If $A_j$ are self-adjoint bounded operators on a Hilbert space with $\sigma(A_j) \subseteq [a,b]$, then $b I - A_j \ge 0$ and $A_j - a I \ge 0$, i.e. $\langle (bI - A_j) v, v \rangle \ge 0$ and $\langle ((A_j - a I) v, v \rangle \ge 0$ for all vectors $v$. These properties are obviously preserved by convex combinations.

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this is I wanted. Thanks. – vesszabo Sep 10 '12 at 9:01

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