1
$\begingroup$

Can $-3$ and $2$ be eigenvalues of and nxn matrix B such that $A = B^{2}+B-6I$ and A's determinant is $0$?

So this is what I concluded:

At first glance, it can be seen that the matrix $A$ can be factored into two different terms. \begin{align*} A = B^{2}+B-6I = (B+3I)(B-2I) \end{align*} \begin{align*}(B+3I)(B-2I) = B^{2}-2IB+3IB-6I^{2} \end{align*} \begin{align*} 2IB=2B, 3IB=3B, I^{2}=I.\end{align*} \begin{align*} (B+3I)(B-2I) = B^{2}-2IB+3ID-6I^{2} = B^{2}+B-6I = A. \end{align*}

Given that the $det(A) = 0$, let the matrix $C = B+3I$, the matrix $D = B-2I$, and \begin{align*} det(A) = det(CD)= \begin{vmatrix}CD\end{vmatrix} = \begin{vmatrix} C\end{vmatrix} \begin{vmatrix} D\end{vmatrix} = 0. \end{align*}

In turn, this suggests that either \begin{align*} \begin{vmatrix} C \end{vmatrix}, \begin{vmatrix} D \end{vmatrix}, or ( \begin{vmatrix}C\end{vmatrix} and \begin{vmatrix}D\end{vmatrix}) = 0.\end{align*}

So my question is... is this the correct way of proving this so far?

$\endgroup$
  • $\begingroup$ Take $B=\begin{pmatrix}-3&0\\0&2\end{pmatrix}$, for example. How exactly do you determine that $\mathrm{det}(C+3I)=0$? That's not correct, $-6$ is not necessarily an eigenvalue of $B$. $\endgroup$ – Kirill Nov 24 '14 at 19:04
  • $\begingroup$ How did you get $-6$? Also, I said "From these cases I can see that this is not the case." Meaning the eigenvalues $-3$ does not satisfy the fact that the determinant is always going to be $0$. $\endgroup$ – Telo Springs Nov 24 '14 at 19:09
  • $\begingroup$ If $C=B+3I$, then $C+3I=B+6I$, so if $\mathrm{det}(C+3I) = \mathrm{det}(B+6I)=0$, then $-6$ is an eigenvalue of $B$. $\endgroup$ – Kirill Nov 24 '14 at 19:11
  • $\begingroup$ @Kirill Thank you. Where in this can you make a suggestion to change? $\endgroup$ – Telo Springs Nov 24 '14 at 19:14
1
$\begingroup$

Your reasoning starts out fine, up to "In turn, this suggests that either

$$ |C|,|D|, \textrm{ or } |C| \textrm{ and } |D| = 0.'' $$

After that, I'm not sure what you're trying to do, but it is something circular/unnecessary.

To start over from your last correct assertion, you now know that

$$ \det[(B + 3I)(B-2I)] = 0 $$

which tells you that either

$$ \det(B+3I) = 0 $$ or $$ \det(B-2I) = 0 $$ (or both). If the first statement is true, then $-3$ is an eigenvalue of $B$. This is (equivalent to) the definition of eigenvalue. Similarly, if the second statement is true, then $2$ is an eigenvalue. So, at least one of those two things is true, and thus both could be eigenvalues. (But it isn't required that $B$ have both as eigenvalues, consider $B = \begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}$; so the correct statement is "at least one of $2$ or $-3$ is an eigenvalue of $B$")

$\endgroup$
  • 1
    $\begingroup$ Small nit: $\mathrm{det}(A-\lambda)=0$ isn't quite the definition of an eigenvalue; usually the definition is the equivalent statement that $Ax=\lambda x$ has a solution in $x$. $\endgroup$ – Kirill Nov 24 '14 at 19:13

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.