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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.