Construct example - $A$ and $B$ have strictly positive eigenvalues, but $A+B$ and $AB$ have strictly negative eigenvalues.

Question

As a follow-up to this question I asked, I wondered what would happen if I imposed the weaker condition of having positive eigenvalues, rather than being positive definite.

How do I construct an example of two matrices $A$ and $B$ such that:

1) $A$ and $B$ have strictly positive eigenvalues.
2) $A + B$ has strictly negative eigenvalues (is this even possible?).
3) $AB$ has strictly negative eigenvalues.

Generally, I'm unsure how to begin going about constructing an example of a matrix that satisfies these properties.

Okay, I see that, but I'm not sure how to use it for this purpose. — Thev, Mar 13 '18 at 18:24
Does it have anything to do with using that trace($AB$) = trace($BA$)? — Thev, Mar 13 '18 at 18:25
Similarly, using $\det(AB) = \det(A) \det(B)$, you could conclude 1 and 3 are incompatible in the case of odd dimension. — Daniel Schepler, Mar 13 '18 at 18:33

score 6 · Accepted Answer · edited Mar 13 '18 at 21:02

6

The trace of a matrix is the sum of its eigenvalues. Also, $$\mathrm{tr}(A+B) = \mathrm{tr}(A)+\mathrm{tr}(B).$$ If $A$ and $B$ have strictly positive eigenvalues, then $$\mathrm{tr}(A),\mathrm{tr}(B)>0\implies \mathrm{tr}(A+B)>0,$$ so it is not the case that $A+B$ has negative eigenvalues.

For a case where $A,B$ have only positive eigenvalues and $AB$ has only negative eigenvalues, take $$A=\pmatrix{1&0\\0&4}, B=\pmatrix{4&-3\\3&-2}$$ The eigenvalues of $A$ are obviously $1$ and $4$, both eigenvalues of $B$ are $1$, and both eigenvalues of $AB$ are $-2.$ I constructed this just by messing around with the characteristic polynomials. I don't have any insight into the problem at all.

edited Mar 13 '18 at 21:02

mechanodroid

46,490

answered Mar 13 '18 at 18:34

saulspatz

53,131

Sum, not product. – Robert Israel Mar 13 '18 at 18:37
*sum of eigenvalues, and in your first line of working, I think you meant tr(A+B). – Thev Mar 13 '18 at 18:37
@RobertIsrael Yes, that's what I meant. Thanks. – saulspatz Mar 13 '18 at 18:41
@TheveshTheva Thanks. I'm not sure if I'm worse at typing, or at proofreading. – saulspatz Mar 13 '18 at 18:42
Well, it's no issue, since you pointed me in the right direction to begin with! Do you have any pointers on case (3), comparing A and B to AB? – Thev Mar 13 '18 at 18:44
As Daniel Schelper pointed out in his comment, since the determinant is the product of the eigenvalues, you can rule out odd dimension. I've been thinking about even dimension, but I don't have any insight yet. – saulspatz Mar 13 '18 at 18:50

score 1 · Answer 2 · answered Mar 13 '18 at 18:21

1

if $A$ and $B$ are symmetric and strictly positive definit, you can not have neither $A+B$ nor $AB$ to be negative definite

answered Mar 13 '18 at 18:21

im hs

49

Yep, I agree with that. Does that still hold true if $A$ and $B$ have positive eigenvalues, but are not necessarily symmetric or positive definite? – Thev Mar 13 '18 at 18:21

score 1 · Answer 3 · answered Mar 13 '18 at 19:03

1

For an example where $A$ and $B$ have all positive eigenvalues while $AB$ has all negative eigenvalues, consider $$ A = \pmatrix{10 & 8\cr -9 & -7\cr},\ B = \pmatrix{1 & 0\cr 0 & 2\cr} $$ $A$ and $B$ both have eigenvalues $1$ and $2$, while $AB$ has a double eigenvalue $-2$.

answered Mar 13 '18 at 19:03

Robert Israel

448,999

Brilliant - could I ask how you constructed this example? – Thev Mar 13 '18 at 19:08
1

A little optimization and experimentation. – Robert Israel Mar 13 '18 at 22:26

Construct example - $A$ and $B$ have strictly positive eigenvalues, but $A+B$ and $AB$ have strictly negative eigenvalues.

3 Answers3