Re: linear operator, determinant

On Thu, 16 Feb 2006 00:37:05 +0100, Kiuhnm wrote:

Hi, I am having trouble with the following exercise.

Let V be the vector space of all nxn matrices over the field of complex
numbers, and let B be a fixed nxn matrix over C. Define a linear
operator M_B on V by M_B(A) = BAB*. Show that det M_B = |det B|^{2n}.

The determinant of M_B is equal to that of the matrix P representing M_B
in an ordered base S for V. Let E_{xy} be a matrix in V such that
[E_{xy}]_{ij} = 1 if (i,j)=(x,y), 0 otherwise. Let S = {E_{11}, E_{12},
..., E_{1n}, E_{21}, ...., E_{2n}, .........., E_{n1}, ..., E_{nn}} be
an ordered base of V.
First of all I start to determine the generic element ij of BE_{xy}B*:
[BE_{xy}B*]_{ij}
= Sum_k [B E_{xy}]_{ik}{B*}_{kj} =
= Sum_k ~B_{jk}[B E_{xy}]_{ik} =
= Sum_k ~B_{jk}Sum_h B_{ih}[E_{xy}]_{hk} =
= ~B_{jy}Sum_h B_{ih}[E_{xy}]_{hy} =
= ~B_{jy}B_{ix}[E_{xy}]_{xy} =
= ~B_{jy}B_{ix}
Now I am in big trouble. How can I write P?
I could write
P_{st} = ~B_{jy}B_{ix}
where
s = (i-1)n + j and
t = (x-1)n + y
but what should I do now?
Should I write P explicitly hoping for a miracle or should I use the
general determinant formula (the sum extended over the permutation)
expecting some simplification?

Kiuhnm

Not sure if this helps but could you prove the result for the matrices for
elementary row operations? If so, then since any B can be written as a
product of these, and M_(B1*B2) = M_B1 composed M_B2 etc, and det product
= product det, you're done.
Duncan


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