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I want to prove a property on the eigenvalues of a positive linear combination of p.d. matrices.

I have the following:

$$ z \in \mathbb R^m_{++} $$

$$ A(z) = \Sigma z_i A_i $$

$$A_i \in S^n_{++} , \forall i \in [1,m]$$

I want to prove that the eigenvalues of $A(z)$ are linear in $z$

my take:

since $z_i > 0 $ We get $A(z) \succ 0$

for any $d \in \mathbb R^n $ such that $\|d\| = 1$ we have: $$ \sum z_i\lambda_{min} (A_i) \leq d^TA(z)d \leq \sum \lambda_{max}(A_i)$$

since we can take $d =$ $q\over \|q\| $ and $q$ is and eigenvector of $A(z)$

I know that any $\lambda(A(z)) $ is bounded above and below by a linear function of $z$ and hence must be linear.

Is that a reasonable proof? I'm I missing anything?

Many Thanks.

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  • $\begingroup$ I have 1) a little remark : $z \in \mathbb R^m_{++}$ should be $z \in \mathbb R^m_{+}$ 2) a question: what do you mean by "the eigenvalues are linear in z" ? That the spectrum of $A(z)$ is a linear combination of the spectra of the $A_i$ (e.g., in decreasing order) with the $z_i$ as coefficients ? $\endgroup$ – Jean Marie Feb 14 '16 at 14:32
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I assume here that we deal with symmetric p.d. (s.p.d) matrices.

I am sorry but I am unable to see what is demontrated, because bounding the spectrum is one thing, establishing that it is linear is another.

About the linearity, I fear, unless I have not well understood what is meant by "linearity", that the result is not exact.

In fact, the spectrum of a linear combination of S.P.D. matrices is not the linear combination (with the same coefficients) of the spectra of these matrices. A counter-example among others:

Take $A=\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ and $B=\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$.

The spectrum of $\dfrac{1}{2}(A+B)$ (it is a s.p.d. matrix because their set is a convex cone) is $\approx \{2.20711, 0.792893\}$ whereas $\dfrac{1}{2}(S_A+S_B) \approx \{2.30902, 0.690983\}$ where $S_M$ stands for the vector of eigenvalues of $M$ sorted by descending order.

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