# Determinant of Matrix of Matrices

My question concerns a situation where you are looking for a determinant of a matrix which is in itself composed of other matrices (in my example, all the inner matrices are square and of equal dimensions).

Say we have matrix $A_{cl}$: $$A_{cl}= \left[\begin{matrix} 0 & I\\ -kL_e & -kL_e \end{matrix}\right]$$ where $L$ is a laplacian matrix of a graph (meaning it is symmetric and positive definite in this example because the graph is a spanning tree).

I presume the following: $L$ is $n$ x $n$, therefore $A_{cl}$ is $2n$ x $2n$.

I see the following development, which I don't understand:

$$det(\lambda I-A_{cl}) = det(\lambda^2I + (\lambda+1)kL_e)) = 0$$ Since $\lambda = -1$ does not satisfy this equation, it is not an eigenvalue of $A_{cl}$. The eigenvalues of $A_{cl}$ thus satisfy $$det(\lambda^2/(\lambda+1)I + kL_e) = 0$$ Denoting the eigenvalues of $-kL_e$ by $\mu$, one has that, for each $i$, $$\mu_i = \lambda^2/(\lambda+1)$$ and hence $$\lambda_i = \frac12(\mu_i+\sqrt{\mu_i^2+4\mu_i})$$

My beef with this development is mostly in the first sentence of it, where they say: $$det(\lambda I-A_{cl}) = det(\lambda^2I + (\lambda+1)kL_e)) = 0$$ This is a determinant of a matrix of matrices, and they treat it like it is a 2x2 matrix determinant (and keep the det() operation after, which is even more confusing). If anybody could explain the mechanics behind this first part of the development I would be very grateful.

Thank you

• Also, forgot this but would like to ask how was the transformation from the determinant with $L_e$ in it to $\mu_i$ done? – grap Jun 6 '13 at 11:30
• We need to go deeper – gukoff Jun 6 '13 at 11:39

• NO. Read the Wikipedia article again. It says "if $C$ and $D$ commute (i.e., $CD = DC$), then the following formula comparable to the determinant of a 2×2 matrix holds". So, there are premises to satisfy. Your matrix just happens to satisfy these premises, but in general, you cannot calculate the determinant of a block matrix using the formula for an ordinary $2\times2$ matrix. – user1551 Jun 6 '13 at 13:10
• In fact, $\det(AD-BC)$ is not necessarily equal to, say, $\det(DA-CB)$. So, even if the determinant formula for $2\times2$ matrices is applicable, the order of multiplication may matter. For more details, see formula (2) in my answer to another question. – user1551 Jun 6 '13 at 13:11