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Martin Sleziak
Martin Sleziak
10:41 AM
Just a reminder - to read MathJax/LaTeX in chat you can use the bookmarklet mentioned in this post on meta or go directly to robjohn's website: math.ucla.edu/~robjohn/math/mathjax.html
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A: How to find eigenvalues of the matrix

farruhotaFor block matrices, you can use Schur complement: $$0=\det(M-I\lambda)=\begin{vmatrix}A&B\\ C&D\end{vmatrix}=\det(D)\cdot \det(A-B\cdot D^{-1}\cdot C)= \\ \begin{vmatrix}4-\lambda&0\\ 0&4-\lambda\end{vmatrix}\cdot \det\left(A-B\cdot \begin{pmatrix}\frac1{4-\lambda}&0\\ 0&\frac1{4-\lambda}\end{pma...

Schur complement: $$\begin{vmatrix}A&B\\ C&D\end{vmatrix}=\det(D)\cdot \det(A-B\cdot D^{-1}\cdot C)$$
@MartinSleziak; my question is can we use Schur complement formula even when $D$ is not invertible — Math_Freak 40 mins ago
I don't think so - if $D$ is not invertible then it is not clear what $D^{-1}$ is.
Also it is not true that if $D$ is not invertible then the whole determinant $\begin{vmatrix}A&B\\ C&D\end{vmatrix}$ has to be zero.
Still, in this specific case I have already mentioned how to avoid this problem:
@Math_Freak You get $\det(D)=0$ only for $\lambda=4$. It's not difficult to check the case $\lambda=4$ separately and to see that the equality $\det(M-\lambda I)=(4-\lambda)^4(\lambda^2-12\lambda+24)$ is true in this case, too. — Martin Sleziak 58 mins ago
Another possible workaround - if you look at $\det(M-\lambda I)=(4-\lambda)^4(\lambda^2-12\lambda+24)$, both sides of the equation are continuous as function of $\lambda$.
So if we know that they are equal for all real $\lambda$'s with the exception of the finitely many roots, by continuity we can extend the equality to all real $\lambda$'s. (In fact, if we have continuity, it suffices to have equality on a dense set.)
I see that the answerer also commented on this: math.stackexchange.com/questions/3348089/…
It depends. $\begin{vmatrix}\color{red}A&B\\C&D\end{vmatrix}=\begin{vmatrix}\color{red}1&2\\3&0\end{vmatrix}=-6\ne 0$, but $\begin{vmatrix}\color{red}A&B\\C&D\end{vmatrix}=\begin{vmatrix}\color{red}2&\color{red}3&1&1\\\color{red}4&\color{red}5&1&1\\ 1&1&2&3\\ 1&1&4&6\end{vmatrix}=\det(D)\cdot \det(A-B\cdot D^{-1}\cdot C)=0$, where $D^{-1}$ must be treated as generalized inverse. In this problem we did not actually calculate the determinant of $D$ numerically, we transformed it algebraically to get final answer, which is free of singularity. — farruhota 4 mins ago
@Math_Freak I left some comments about this in chat - so that I do not create too long comment thread here. (It is rather long already.) — Martin Sleziak 8 secs ago
Though, in the first case, by the generalized inverse it will work too (looks strange): $$\begin{vmatrix}\color{red}A&B\\C&D\end{vmatrix}=\begin{vmatrix}\color{red}1&2\\3&0\end{vmatrix}=0\cdot \det(\color{red}1-2\cdot \frac10 \cdot 3)=0\cdot \det(\frac{0-6}{0})=0\cdot \frac{-6}{0}=-6.$$ — farruhota 6 mins ago
BTW, I did not notice until now that there is a separate tag schur-complement.
 

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