# The determinant of block triangular matrix as product of determinants of diagonal blocks

I am given the following partitioned - upper-triangular matrix:

$$\begin{bmatrix} A_1 &* &* &* &* &* \\ 0& A_2 &* &* &* &* \\ .& 0& A_3 &* &* &* \\ .& 0& 0 &... &* &. \\ .& 0& 0& 0& ... &. \\ 0& .& ...& 0&0 & A_m \end{bmatrix}$$ as for $A_i$ are all block-matrices.

I have to prove that the determinant of this general matrix is: $$\prod_{n=1}^m \det A_i$$

we already proved in class that it is true for $m=2$ and we can use it in our proof.

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Use induction on $m$. – Chris Godsil Dec 27 '12 at 19:17

Use a proof by induction on $m$:

$(a)$ It seems to me, then, that you have your base case $m = 2$: $P(2)$,
$\quad$ though it suffices, for a base case, to prove it's (trivially) true for $m = 1$.

$\quad$ But the process used in class for proving it's true for $m = 2$ will be helpful when making the
$\quad$ inductive step of the proof.

$(b)$ Your inductive hypothesis $P(k)$ would be to assume that this is true for $m = k$.

$\quad$ That is, assume the truth of: $$P(K):\quad \det \begin{bmatrix} A_1 &* &* &* &* &* \\ 0& A_2 &* &* &* &* \\ .& 0& A_3 &* &* &* \\ .& 0& 0 &... &* &* \\ .& 0& 0& & ... &* \\ 0& .& ...& 0&0 & A_k \end{bmatrix} = \prod_{i=1}^k \det A_i$$ $(c)$ Then, take the inductive step: you'll need to use the inductive hypothesis to prove that $P(k+1)$ is true: for $m = k + 1$... That is, assuming $P(k)$ is true, prove: $$\det \begin{bmatrix} A_1 &* &* &* &* &| &* \\ 0& A_2 &* &* &* &| &* \\ .& 0& A_3 &* &* &| &* \\ .& 0& 0 &... &* &| &* \\ .& 0& 0& ...& A_k &| &* \\ \hline& & & &\\ 0& 0& ...& 0&0 &| & A_{k+1} \end{bmatrix} = \prod_{i=1}^{k+1} \det A_i$$

Note that to do this, you can partition the matrix into two block matrices on the diagonal,

$(1)$ one of which is triangular (block) matrix with $k$ sub-blocks $A_i$ for $1 < i < k$ along it's diagonal, for which the determinant you know from the inductive hypothesis (having assumed its truth), and

$(2)$ the other with one block on the diagonal which we call $A_{k+1}$.

Here is where you can use the proof used in class for $m = 2$

That is: $$\prod_{i=1}^{k+1} \det A_i = \left(\prod_{i=1}^{k} \det A_i\right)\cdot \det(A_{k+1})$$

Then you will have shown that the determinant of a partitioned triangular matrix is product of the determinants of the block matrices on the diagonal.

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You can use $n$ instead of $m$, and $m$ instead of $k$, if you wish. The process remains the same. – amWhy Dec 27 '12 at 19:19
Thanks, Sami! ;-) – amWhy Apr 10 '14 at 15:54

Sketch of an alternative argument: the determinant of an upper triangular matrix is the product of its diagonal entries. You can do elementary row and column operations to the large matrix that make each $A_i$ upper triangular without changing any of the entries in $A_j$ for $i \neq j$ (taking account of any row swaps you may have to perform). Once you have done so, the whole matrix will be upper triangular, so its determinant is the product of the diagonal entries (and then correct the sign according to the parity of the row swaps you have performed).

I don't think this is what your instructor wanted since it doesn't use the 2 by 2 case, but it's how I think about this fact.

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Tank you! very helpful! – Dor Shalom Dec 28 '12 at 11:43
Good +1. However the fact that the determinant of an upper triangular matrix is the product of its diagonal entries is a special case of what you are trying to prove, so it would be good to reflect on how you got to know that fact in the first place, and see if you can adapt that proof to the more general situation. I think it probably can. – Marc van Leeuwen Dec 28 '12 at 15:06
@Marc van Leeuwen that's a good point. for an upper triangular matrix, you can make it diagonal with row ops unless there's a 0 along the diagonal, and these In the exceptional case that there's a 0 along the diagonal which case the determinant is zero as the first $m$ column vectors (up to and including the zero on the diagonal) can't span more than $m - 1$ dimensions. – user29743 Dec 28 '12 at 17:45