# Block Matrix Zero Determinant Implication?

Recently I've been working with a number of square (order of 2n) matrices whose determinants are zero. That is,

$$\det\begin{bmatrix}A&B\\C &D \end{bmatrix} = 0$$ where each of A,B,C, and D are also matrices.

I couldn't help but notice that in all of my cases, det(A) = det(D) = 0 as well. This is fairly surprising to me, and I'm curious: what is the limit of validity of this special case? Alternatively, under what conditions is this true?

If it matters, I'm working with passive electrical transmission networks, which typically never have zeroes on the main diagonal.

Thanks!

• For what it's worth consider the case where $A=B=C=D=1$. Commented Jul 2, 2015 at 20:33

If $A$ and $D$ are of the same size, and $\det A = \det B = \det C = \det D = 0$ then of course $\det \begin{bmatrix}A&B\\C &D \end{bmatrix} = 0$.

But the converse does not hold: It is easy to produce square matrices of determinant zero none of whose blocks have determinant zero. An easy example is $$A= \begin{bmatrix}4I&2I\\2I &I \end{bmatrix}$$ where $A$ is $2n \times 2n$ and $I$ is an $n \times n$ unit matrix.

For instance, look at $$A= \begin{bmatrix}4&0&2&0\\0&4&0&2\\2&0&1&0\\0&2&0&1 \end{bmatrix}$$

• If all submatrices have determinant zero, then the determinant of the full matrix can still be non-zero: $\pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 }$. Hence your first statement is incorrect.
– daw
Commented Jul 3, 2015 at 17:59
• Ooops, you are right. Commented Jul 6, 2015 at 22:21

The determinant of your matrix (or any such matrix in block diagonal form) is equal to the product of the determinants of the diagonal blocks, i.e. $\operatorname{det}(A)\operatorname{det}(D)$. If the determinant of the block diagonal matrix is $0$, this means $\operatorname{det}(A)\operatorname{det}(D)=0$ so that either $\operatorname{det}(A)=0$ or $\operatorname{det}(D)=0$, but not necessarily both.

• Your answer is not right - it is only in the case where $C \equiv 0$ or $D \equiv 0$ Commented Jul 2, 2015 at 20:48