TITLE: The determinant of block triangular matrix as product of determinants of diagonal blocks
QUESTION [16 upvotes]: I am given the following partitioned, upper-triangular matrix:
$$
\begin{bmatrix}
  A_1 &*  &*  &*  &*  &*  \\
   0& A_2 &*  &*  &*  &*  \\
   .&  0& A_3 &*  &*  &*  \\
   .&  0&  0 &...  &*  &. \\
   .&  0&  0&  0& ... &.  \\
   0&  .&  ...&  0&0 & A_m
\end{bmatrix}
$$ 
where each of the $A_i$ are 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.
Thanks for your answers!

REPLY [5 votes]: 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.