Question

Determinant question.

Answer 1

If \[A \in K^{n \times n}, B \in K^{m \times m}\] and \[C \in K^{n \times m}\]
Let be \[M \in K^{(n+m) \times (n+m)}\]
Defined as:
\[M = \begin{matrix} A &C \\ 0 &B \end{matrix}\]
Prove that det(M)=det(A).det(B)

Answer 2

That's the last problem I have to solve :D, I solved it a month ago but I didn't write it down and I don't remember how I did the induction neither :(

Answer 3

Sorry for my bad English :P

A and B are both squared matrix.

If n=1
det(A)=a_11
And clearly:

\[\det \begin{bmatrix} a_{11} & C\\ 0 & B \end{bmatrix} =a_{11}.B\]
Because when you calculate column that's not 1 to calculate the determinant, you'll have a column full of 0 in the new submatrix, so the determinant is 0.


Now we assume it works for n'
Then the matrix

\[M'=\begin{bmatrix} A' & C'\\ 0 & B \end{bmatrix}\]
\[\det(M')=\det(A').\det(B)\]
\[A=\begin{bmatrix} a_{11} &ar \\ ac & A' \end{bmatrix}, C=\begin{bmatrix} \ cr \\ C' \end{bmatrix}\]
\[M= \begin{bmatrix} a_{11} &ar & c.r \\ ac &A' &C' \\ 0 &0 &B \end{bmatrix}\]
\[\det(M)=\sum_{i=1}^{n+1}m_{1i}.M(1 | i)=\sum_{i=2}^{n+1}m_{1i}.M(1 | i)+a_{11}.\det(M')\]

I don't know why, but I had to put a dot in c.r because it didn't show.
I'll continue later.

Answer 4

Works for n=n' *

Answer 5

When you are calculating the determinant, you have to chose n numbers, each time you chose one, you have a matrix of one row and one column minus, if you chose one of the numbers in the matrix C, you'll have to chose a zero at some point because you eliminated a row of A and a column of B, so the determinant of A,C,0,B matrix is the same than the one of A,0,0,B matrix, thus ... damn I got an easier answer.

Answer 6

If the Matrix A is singular, then M have a dimension missing on the columns, so M is singular, then 0.det(B)=0 and it's true.
If the Matrix B is singular, then M have a dimension missing on the rows, so M is singular, then det(A).0=0 and it's true.
If neither A or B are singular, you can put it in a triangular form, using only permutations, row Ai -> row Ai+ a*row Aj, and because the determinant is an alternating multilinear form, it don't change the determinant (only the permutation, but it multiply it by -1 and I can be solved multiplying that row for -1 so you'll have a matrix with the same determinant).
Then you can put A, and B in a triangular form, without changing the determinants.
Then, M is in a triangular form.
And the triangular form matrixes determinant can be computed as:
Let's be D the triangular form matrix, then it's determinant is:
\[\prod _{i=1}^{n}d_{ii}\]

Which in the case of M it's equal to:
\[\prod _{i=1}^{n+m}m_{ii}=\prod _{i=1}^{n}a_{ii}.\prod _{i=n+1}^{n+m}b_{ii}=\det(A).\det(B)\]

Answer 7

@hartnn I solved it, I didn't put it all online because my English is quite bad, plus having to put it in LaTex only makes it worse, but I think it's very clear.

Answer 8

I corrected a little mistake:
\[\prod _{i=1}^{n+m}m_{ii}=\prod _{i=1}^{n}m_{ii}\prod _{i=n+1}^{n+m}m_{ii}=\prod _{i=1}^{n}a_{ii}\prod _{i=1}^{m}b_{ii}\]

Answer 9

When you put A in triangular form, C can change but it doesn't matter.