Talk:Determinant of Block Diagonal Matrix

Not sure, but wasn't the original question given as (paraphrased):


 * $\displaystyle \mathbf A = \begin{bmatrix}

\mathbf{A}_{11} & \mathbf{A}_{12} & \cdots & \mathbf{A}_{1k} \\ 0 & \mathbf{A}_{22} & \cdots & \mathbf{A}_{2k} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf{A}_{kk} \end{bmatrix}$ ... in other words upper triangular?

It comes to the same thing in the end, from what I understand (don't you just multiply the determinants on the main diagonal?) but it's probably worth doing like this. --prime mover 16:03, 24 October 2011 (CDT)
 * You are entirely correct. However, I was planning to do some more work on block matrices in general. The upper triangular block matrices are also included in that planning. --Lord_Farin 16:07, 24 October 2011 (CDT)