## Determinant of a block triangular matrix

I am sure that the fact I am going to discuss here is well known but I do not know a good source and so I decided to prove it for myself and record the answer here. Suppose we have an $n\times n$ matrix $M$ with complex entries $m_{ij}$ and that it is partitioned into four blocks by considering the index ranges $1\le i\le k$ and $k+1\le i\le n$ for some positive integer $k. Suppose that the lower left block is zero so that the matrix is block upper triangular. Denote by $A$, $B$ and $C$ the upper left, upper right and lower right blocks. Then $\det M=\det A\det C$. This is already interesting in the block diagonal case $B=0$. To begin the proof of this I use the fact that over the complex numbers diagonalizable matrices with distinct diagonal elements are dense in all matrices. Approximate $A$ and $C$ by diagonalizable matrices with distinct diagonal elements $A_n$ and $C_n$, respectively. Replacing $A$ and $C$ by $A_n$ and $C_n$ gives an approximating sequence $M_n$ for $M$. If we can show that $\det M_n=\det A_n\det C_n$ for each $n$ then the desired result follows by continuity. The reason I like this approach is that the density statement may be complicated to prove but it is very easy to remember and can be applied over and over again. The conclusion of this is that it suffices to treat the case where $A$ and $C$ are diagonalizable with distinct eigenvalues. Since the determinant of a matrix of this kind is the product of its eigenvalues it is enough to show that every eigenvalue of $A$ or $C$ is an eigenvalue of $M$. In general the determinant of a matrix is equal to the determinant of its transpose. Thus the matrix and its transpose have the same eigenvalues. Putting it another way, left eigenvectors define the same set of eigenvalues as right eigenvectors. Let $\lambda$ be an eigenvalue of $A$ and $x$ a corresponding eigenvector. Then $[x\ 0]^T$ is an eigenvector of $M$ corresponding to that same eigenvalue. Hence any eigenvalue of $A$ is an eigenvalue of $M$. Next let $\lambda$ be an eigenvalue of $C$ and $y$ a corresponding left eigenvector. Then $[0\ y]$ is a left eigenvector of $M$ with eigenvalue $\lambda$. We see that all eigenvalues of $A$ and $C$ are eigenvalues of $M$. To see that we get all eigenvalues of $M$ in this way it suffices to do the approximation of $A$ and $C$ in such a way that $A_n$ and $C_n$ have no common eigenvalues for any $n$. Then we just need to compare the numbers of eigenvalues of the matrices $A$,$C$ and $M$. A consequence of this result is that the characteristic polynomial of $M$ is the product of those of $A$ and $C$. An application which interests me is the following. Suppose we have a partially decoupled system of ODE $\dot x=f(x,y)$, $\dot y=g(y)$, where $x$ and $y$ are vectors. Then the derivative of the right hand side of this system at any point has the block triangular form. This is in particular true for the linearization of the system about a steady state and so the above result can be helpful in stability analyses.

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