We know that elementary row operations do not change the determinant of a matrix but may change the associated eigenvalues.
Consider an example, say two $5 \times 5$ matrix are given:
$$A = \begin{pmatrix}
0 & 1 & 0 & 0 & 0\\
a & b & 0 & 0 & 0\\
0 & 0 & p & q & r\\
0 & 0 & s & t & u\\
0 & 0 & v & w & x\\
\end{pmatrix}, \hspace{2cm} B = \begin{pmatrix}
0 & 1 & 0 & 0 & 0\\
a & b & 0 & 0 & 0\\
0 & 0 & p & q & r\\
0 & 0 & s & t & u\\
ka & kb & v & w & x\\
\end{pmatrix} $$
Now $B$ can easily be reduced to $A$ by using the following operation on $B$
$$R_5 – kR_2$$
Now these two have the same eigenvalues. It is cumbersome to try to symbolically calculate the eigenvalues to show they are indeed same (I have tried tons of matrices with random numbers in Mathematica). $A$ is a block diagonal matrix and $B$ is reduceable to one. If you talked about systems, $A$ shows two decoupled spaces (of dimensions $2$ and $3$) within the $5-D$ vector space of $A$. Can anyone prove that such a pair of matrices always have same eigenvalues? Is there any property that says so? Does $A$ being a block diagonal matrix have to do anything with the eigenvalues being the same? Any insight or discussion is welcome!
Please correct me if I used any term loosely or wrongly.
Best Answer
You may already know that $$\det\pmatrix{A&0\\B&C}=\det\pmatrix{A&0\\0&C}=\det A\cdot\det C$$ which can be shown using the fact that the determinant doesn't change by elementary row operations.
Also note that the eigenvalues of $M$ are the roots of $\det(\lambda I-M)=0$. Now let $M=\pmatrix{A&0\\B&C}$ then $$\begin{align}\det(\lambda I-M)&=\det\pmatrix{\lambda I-A&0\\B&\lambda I-C}\\&=\det\pmatrix{A_1&0\\B&C_1}\\&=\det\pmatrix{A_1&0\\0&C_1}=\det\pmatrix{\lambda I-A&0\\0&\lambda I-C} \end{align}$$