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

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

Never mind, I googled determinant of Block Matrix, which gave me:


So it turns out when you have a block matrix, assuming the dimensions agree for matrix multiplication rules, you can actually treat it as a regular matrix. In my case it looked like a 2 by 2 matrix, therefore that development was perfectly legal.

  • 4
    $\begingroup$ 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. $\endgroup$ – user1551 Jun 6 '13 at 13:10
  • 1
    $\begingroup$ 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. $\endgroup$ – user1551 Jun 6 '13 at 13:11
  • $\begingroup$ By the way, since you have found out an answer by yourself, please do the community a favor and accept your own answer (by clicking the "tick" symbol on the left of your answer), so that this question will not be rendered as "UNANSWERED" in the question queue. $\endgroup$ – user1551 Jun 6 '13 at 13:15
  • $\begingroup$ Ok, thanks for the clarification. $\endgroup$ – grap Jun 7 '13 at 5:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.