# Finding the determinant of a block diagonal matrix

We have the following square block matrix $X= \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}$, where $A$ and $B$ are square. I want to get the determinant of this matrix (I know it equals $\det A \times \det B$).

I've looked through the question catalog and found a lot of questions similar to this one, but the answers they received use techniques I am not familiar with - Leibniz' formula, techniques beyond the scope of an introductory linear algebra course, etc.

So I want to know how to do this with Laplace expansion. I know how Laplace expansion works for regular matrices but I don't know how to use it here. Let's say we want to use Laplace expansion along the first row of $A$, then we get that $\det A = \displaystyle \sum_{j=1}^n (-1)^{j+1}a_{1j}\det A_{1j}$, but how do we use this to obtain the result for the entire block matrix?

Can we say that because the elements of $A$ are always in different rows and columns as compared to the elements of $B$, the entire matrix $B$ is always present in the minors of $A$? If so, the result seems a bit more intuitive to me, but I still don't know how to computationally obtain it.

## 1 Answer

You should already know that $\det(MN) = \det(M)\det(N)$.

With that in mind, we can write $$\pmatrix{A&0\\0&B} = \pmatrix{A&0\\0&I} \pmatrix{I&0\\0&B}$$ where $I$ denotes an identity matrix. So now, it's sufficient to show that $$\det\pmatrix{A&0\\0&I} = \det \pmatrix{I&0\\0&A} = \det(A)$$ you'll find that it's fairly easy to show that this one is true using Laplace expansion along the top "$I$-row".

Computation: let's take that last matrix. Suppose that $$X = \pmatrix{I_{k \times k} \\&A} = \\ \pmatrix{1\\&1\\&&\ddots\\&&&1\\ &&&&a_{11} & \cdots & a_{1n}\\ &&&& \vdots & \ddots & \vdots\\ &&&& a_{n1} & \cdots & a_{nn}}$$ Taking the Laplace expansion along the first row gives us $$\det(X) = 1 \det\pmatrix{I_{(k-1)\times (k-1)} \\&A}$$ and this will work whenever $k \geq 1$. When $k = 1$, we have $$\det(X) = \pmatrix{1&0 & \cdots & 0\\ 0&a_{11} & \cdots & a_{1n}\\ 0& \vdots & \ddots & \vdots\\ 0& a_{n1} & \cdots & a_{nn}} = 1 \det(A)$$ So, by induction, we will have $\det(X) = \det(A)$ for any integer $k \geq 1$.

• I don't know how to apply the Laplace expansion to block matrices, that is my issue. For the last matrix you give, I guess we could say it's a diagonal matrix so we know that the determinant must be $1 \times 1 \times \dots \times 1 \times A = A$, but here I didn't use Laplace expansion, I used a result for diagonal matrices. – YakSal Tafri Nov 22 '16 at 18:50
• It's not a diagonal matrix though since $A$ is a matrix, not a number. – Ben Grossmann Nov 22 '16 at 18:53
• See my latest edit – Ben Grossmann Nov 22 '16 at 18:58
• Ah thanks, the edit helps, all other minors are zero so we effectively keep shaving off rows/columns. – YakSal Tafri Nov 22 '16 at 19:01
• Yes. In particular, switching the "blocked" rows from the top to the bottom may switch the sign of the determinant, but in this case, switching the "blocked" columns from the left to the right will undo this effect. Of course, we could always repeat the original proof by taking Laplace expansions along the bottom row. – Ben Grossmann Nov 22 '16 at 19:37