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?

  • $\begingroup$ 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

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 $B$")

  • 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

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