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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.

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  • $\begingroup$ Instead of doing elementary row operations on $B$, you can test such operations on $\lambda I-B$. I don't think you would need any special property in this way $\endgroup$
    – polfosol
    Nov 30, 2016 at 21:51
  • $\begingroup$ How is that going to prove anything? Can you please elaborate? $\endgroup$
    – Zero
    Nov 30, 2016 at 22:02

1 Answer 1

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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 $$\begin{align}\det(\lambda I-M)&=\det\pmatrix{\lambda I-A&0\\B&\lambda I-C}\\&=\det\pmatrix{A_1&0\\B&C_1}\\&=\det\pmatrix{A_1&0\\0&C_1}=\det\pmatrix{\lambda I-A&0\\0&\lambda I-C} \end{align}$$

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  • $\begingroup$ Thanks for the answer...I knew what you meant by trying with $\lambda I - B$...but my doubt was the first identity itself...how are those determinants equal? For scalars it is pretty straight forward but I didn't know about block matrices $\endgroup$
    – Zero
    Dec 1, 2016 at 4:08
  • $\begingroup$ In general I know that you can use blocks in block matrices like you would scalars in a normal matrix. By that logic I know that the first identity is easy to prove but I am not sure to the extent this statement is true. $\endgroup$
    – Zero
    Dec 1, 2016 at 4:16
  • $\begingroup$ @Zero I added a link to the proof $\endgroup$
    – polfosol
    Dec 1, 2016 at 7:54

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