Is the determinant a linear operation?

1. Jul 26, 2011

EngWiPy

Is the determinant a linear operation? I mean can we say that:

$$\mathbb{E}[\text{det}]=\text{det}[\mathbb{E}]$$

where E is the expectation operator?

2. Jul 26, 2011

Char. Limit

First impression says no, but I'm testing it out now, I might be wrong.

3. Jul 26, 2011

pmsrw3

No, it's not. For instance, det(aM) = andet(M). And det(M+N) doesn't have any simple relationship to det(M) and det(N). (See, for instance, http://en.wikipedia.org/wiki/Matrix_determinant_lemma" [Broken].)

Last edited by a moderator: May 5, 2017
4. Jul 26, 2011

EngWiPy

Ok, now we have this inequality:

$$\mathbb{E}\left[\log(X)\right]\leq\log\left(\mathbb{E}[X]\right)$$

Can we say in the same manner that:

$$\mathbb{E}\left[\log\left(\text{det}\left\{H\right\}\right)\right]\leq\log\left(\text{det}\left\{\mathbb{E}\left[H\right]\right\}\right)$$

5. Jul 26, 2011

pmsrw3

Nope. Suppose H=((1 0), (0 1)) with 50% probability and ((-1 0), (0 -1)) with 50% probability. Both of those have determinant 1, so E[log det H] = 0. But E[H] = ((0 0), (0, 0)), with determinant 0, so log det E[H] = -infinity.

6. Jul 26, 2011

EngWiPy

Ok, I see. Thanks a lot