# On the sum of definite matrices

Let $$A,B$$ be two $$n\times n$$ symmetric positive semi-definite matrices. I have the intuition that:

$$A-B$$

has, at most, $$\operatorname{rank}(A)$$ positive eigenvalues, and at most $$\operatorname{rank}(B)$$ negative eigenvalues. But I have not been able to prove or disprove this.

Is this statement true? If so, can we prove it? Otherwise, find a counterexample?

Yes, this statement is true. One way to prove this is using Weyl's matrix inequality. That is:

Theorem: For a symmetric matrix $$M$$, let $$\lambda_i(M)$$ denote the $$i$$th eigenvalues of $$M$$, where $$\lambda_1(M) \geq \lambda_2(M) \geq \cdots \geq \lambda_n(M)$$. For symmetric matrices $$P,Q$$ and indices $$1 \leq i,j \leq n$$, $$\lambda_{i + j - 1}(P + Q) \leq \lambda_i(P) + \lambda_j(Q) \leq \lambda_{i + j - n}(P + Q).$$

The cases where $$j = 1$$ and $$j = n$$ tend to be particularly useful. These values of $$j$$ yield the inequalities $$\lambda_{i}(P + Q) \leq \lambda_i(P) + \lambda_1(Q), \\ \lambda_i(P + Q) \geq \lambda_i(P) + \lambda_n(Q).$$

Now, let $$r = \text{rank}(A)$$. To see that $$A - B$$ has at most $$r$$ positive eigenvalues, we note that the eigenvalues of $$A$$ are such that $$\lambda_{r}(A) > \lambda_{r+1}(A) = 0$$. Thus, \begin{align} \lambda_{r+1}(A - B) &\leq \lambda_{r+1}(A) + \lambda_1(-B) = 0 - \lambda_n(B) \\ &\leq 0 - 0 = 0. \end{align} To see that $$A - B$$ has at most $$\text{rank}(B)$$ negative eigenvalues, it suffices to apply the above reasoning to the matrix $$-(A-B) = B-A$$.

You might find it interesting that, with a similar approach, we can also show that $$A - B$$ has at least $$\text{rank}(A) - \text{rank}(B)$$ positive eigenvalues (and similarly at least $$\text{rank}(B) - \text{rank}(A)$$ negative eigenvalues).

• Very nice answer, thanks. I was just reading about Weyl's inequality, and suspected it could be useful.
– a06e
Commented Apr 23 at 19:05