Proof regarding determinant of block matrices

[Adgorn]

Homework Statement

Let A,B,C,D be commuting n-square matrices. Consider the 2n-square block matrix $M= \begin{bmatrix} A & B \\ C & D \\ \end{bmatrix}$. Prove that $\left | M \right |=\left | A \right |\left | D \right |-\left | B \right |\left | C \right |$. Show that the result may not be true if the matrices do not commute.

Homework Equations

$det(M)= det(A_1)det(A_2)...det(A_n)$ (Where M is an upper (lower) triangular block matrix with $A_1,A_2,...,A_n$ diagonal blocks.)

The Attempt at a Solution

At first I tried using the theorems of determinants in combinations with the properties of (commuting) matrices to try and get the desired expression, but had no success. I then tried expressing the determinant using the elements of the square matrices in M and dividing the expression to 4 separate determinants, but could not figure out how.
The closest I managed to get to a solution is prove the above equation where C=0.
Any help would be appreciated.

 

[Adgorn]

Thank you Mr. AI bot thingy but I said all I really have to say, It's pretty straight forward.

 

[fresh_42]

Thank you Mr. AI bot thingy but I said all I really have to say, It's pretty straight forward.

Yes, but your attempts could have been a bit more elaborated. Unfortunately, the only solution I can think of are either an induction or the explicit formula via diagonals. Doesn't sound funny though.

 

[Adgorn]

Yes, but your attempts could have been a bit more elaborated. Unfortunately, the only solution I can think of are either an induction or the explicit formula via diagonals. Doesn't sound funny though.

Unfortunately I cannot explain any further simply because I don't even know how to approach this, I have never worked with commutative matrices before (Learning from Shaum's outlines), let alone under the context of determinants. So I don't really know what to exploit with the given information that the matrices are commutative.
I am certain that the proof uses induction, I managed to prove something similar with C=0:
$M=\begin{bmatrix} A & C & \\ 0 & B & \end{bmatrix}$
Prove $det(M)=det(A)det(B)$

If A is r-square and B is s-square, then n=r+s and
$\sum_{σ \in S_n}$ (sgn $σ$) $m_{1σ(1)} m_{2σ(2)}...m_{nσ(n)}$.
If $i\gt r$ and $j\leq r$. $m_{ij}=0$. Therefor we only consider the permutations $σ(r+1,r+2,...r+s)=(r+1,r+2,...,r+s)$ and $σ(1,2,...,r)=(1,2,...,r)$.
Let $σ_1(k)=σ(k)$ for $k\leq r$ and let $σ_2(k)=σ(k+r)-r$ for $k\leq s$.
Thus, (sgn $σ$) $m_{1σ(1)} m_{2σ(2)}...m_{nσ(n)}$$=$(sgn $σ_1$)$a_{1σ_1(1)} a_{2σ_1(2)}...a_{rσ_1(r)}$(sgn $σ_1$)$b_{1σ_2(1)} b_{2σ_2(2)}...b_{sσ_2(s)}$
Which implies $det(M)=det(A)det(B)$.

So I proved this theorem, which is the closest I got to proving the question of this post. I tried using the raw formula to no sucess, perhaps there is some hidden property of commutative matrices that will allow me to solve this.

 

[fresh_42]

The commutativity requirement should automatically show up during the proof. I wouldn't start with it. But it may be used early in a proof. So without any brute force methods to apply, I think there could be a tricky decomposition of $M$. However, it should be a multiplicative decomposition or a basis transformation $TMT^{-1}$, since we don't really want to calculate something like $\det (M+N)$. If you already have a proof for triangular matrices, maybe you can find a decomposition into factors like this.

One can always write any matrix $M = \frac{1}{2}(M+M^t) + \frac{1}{2}(M-M^t)$ as a sum of a symmetric and a skew-symmetric matrix. Don't know whether this helps. Or apply some normal forms you know about.

 

[StoneTemplePython]

Homework Statement

Let A,B,C,D be commuting n-square matrices. Consider the 2n-square block matrix $M= \begin{bmatrix} A & B \\ C & D \\ \end{bmatrix}$. Prove that $\left | M \right |=\left | A \right |\left | D \right |-\left | B \right |\left | C \right |$.

This is an old thread, but it has been left open and I thought about this earlier today for some reason.

The assertion in the problem statement is wrong.

A simple and general counter example is: consider the case where $n$ is some even natural number, and $\mathbf A = \mathbf D = \mathbf 0_{nxn}$ and $\mathbf B = \mathbf C = \mathbf I_{nxn}$

The zero matrix and the identity matrix commute with all matrices. Yet the above formula indicates that the determinant is $-1$ when in fact it is $+1$. (I.e. in this case we have a permutation matrix that becomes the identity matrix after an even number of pairwise column swaps and hence has determinant of 1.)

real simple example: consider $n = 2$

$1 =det\Big(\begin{bmatrix} 0 & 0 & 1 &0 \\ 0 & 0 & 0 & 1\\ 1 & 0 & 0&0 \\ 0 & 1 & 0& 0 \end{bmatrix}\Big) = det\Big( \begin{bmatrix} \mathbf 0 & \mathbf I \\ \mathbf I & \mathbf 0 \\ \end{bmatrix}\Big) \neq det\big(\mathbf 0\big)det\big(\mathbf 0\big) - det\big(\mathbf I\big)det\big(\mathbf I\big) = 0*0 - 1*1 = -1$

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note: the specific example I am giving is problem 6.2.5 in Meyer's Matrix Analysis.