Question regarding $A = B^{-1}DB$ and determinants

Question

Consider $A = B^{-1}DB$, where $A$ is a normal matrix represented by unitary matrices $B, B^{-1}$ and the diagonal matrix $D$.

Although $B^{-1}B = BB^{-1} = I_B$ why doesn't $B^{-1}DB$ give you $D$? What special algebraic properties are revealed about two matrices being similar? (In other words, what is the significance of this equation other than a change-of-base)?

What does it mean for two matrices to have the same determinant? I can do all these "cool" calculations yet have no access to insight of them. (For example, two similar matrices have the same determinant...

More specifically, this is what I'm after: Are $A$ and $D$ algebraically related?

In general, matrix multiplication isn't commutative: you can't change DB to BD, which you'd need to do before you could cancel the $B^{-1}B$. — John Hughes, Nov 27 '13 at 18:24
Have a look at the properties listed at http://en.wikipedia.org/wiki/Matrix_similarity — Ivan Loh, Nov 27 '13 at 18:28
@John: More specifically, what does it give you? A multiple of $A$? — Don Larynx, Nov 27 '13 at 18:40

score 1 · Accepted Answer · answered Nov 27 '13 at 18:45

1

Matrix multiplication is not commutative. That means that we cannot assume that $MN = NM$ for any pair of matrices where $MN$ and $NM$ make sense, i.e. square matrices. There are some matrices which satisfy the equation $MN=NM$, but not all.

One direct consequence is that $B^{-1}DB$ is not necessarily the same as $D$. You will be able to find specific matrices for which $B^{-1}DB=D$ but, in general, this is impossible.

If we consider non-singular matrices then we have a group. Given a fixed matrix $D$, there set of matrices $B^{-1}DB$, where $B$ is any other non-singular matrix is called the conjugacy class of $D$.

If you want $B^{-1}DB=D$ for all $D$ then you require the conjugacy class of $D$ to comprise $D$ alone. For that to happen, you need $D$ to lie in the centre of the matrix group.

answered Nov 27 '13 at 18:45

Fly by Night

32,272

Are the nice matrices the normal matrices and the unitary matrices, that is, those that satisfy $MN = NM$? (I'm asking if they are nice) – Don Larynx Nov 27 '13 at 18:49
$B^{-1}DB = D$ iff it's unitary? – Don Larynx Nov 27 '13 at 18:51
What do you mean by unitary? The Unitary matrices $U(n)$ are $n$-by-$n$ complex matrices which preserve the Hermitian inner product, i.e. $X\overline{X}^{\top} = E$, where $E$ is the $n$-by-$n$ identity matrix. – Fly by Night Nov 27 '13 at 19:16
isn't it unitary if the product $MN = NM = I$? – Don Larynx Nov 27 '13 at 20:12
Are they nice? – Don Larynx Nov 27 '13 at 20:12
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https://en.wikipedia.org/wiki/Unitary_matrix – Fly by Night Nov 27 '13 at 20:13

Question regarding $A = B^{-1}DB$ and determinants

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