Is the determinant a linear operation?

[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?

 

[Char. Limit]

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

 

[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" .)

 

[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)

 

[pmsrw3]

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)

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.

 

[EngWiPy]

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.

Ok, I see. Thanks a lot