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Given a non-scalar matrix $A$ in size $n\times n$ over $\mathbb{R}$ that maintains the following equation $$A^2 + 2A = 3I$$ given matrix $B$ in size $n\times n$ too $$B = A^2 + A- 6I$$ Is $B$ an invertible matrix?

Idan
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3 Answers3

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An alternative version to the other excellent answers by Quark and user1551.

Note first, as the others did, that $(A + 3 I) (A - I) = 0$, and $B = (A - 2 I)(A + 3 I)$.

Then $$ B (A - I) = 0. $$ So if $A \ne I$, then $B$ is definitely not invertible.

If $A = I$, as noted by Quark, then $B = -4 I$ is invertible.

But Tharsis (thanks!) made me notice that $A$ is non-scalar by assumption, so the case $A = I$ is excluded (although easy to deal with).

Andreas Caranti
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It depends, If you factorize the first equation you get, $$(A-I)*(A+3I)=0$$

This implies$\; \det(A-I)\;\; or\;\; \det(A-3I)\;$ must be zero.

Now factorizing B gives you (A-2I)*(A+3I)

So $\det(B) = \det(A-2I)*\det(A+3I)$

If $\det(A-3I)$ is zero from the first condition, then $\det(B)$ is also zero, so B is not invertible.

But, if $\det(A-I)$ is zero and not $\det(A-3I)$, I don't think anything can be said

(not very sure, but I can give a example: If you take A=I, the first equation is satisfied, and B comes out to be -4I which is invertible...)

Quark
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Hint: The first equation implies that $A^2+2A-3I=(A-I)(A+3I)=0$. Hence the minimal polynomial $m_A(x)$ of $A$ divides $(x-1)(x+3)$. What happens if $x+3$ is not a factor of $m_A(x)$?

user1551
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  • hello.thanks for your response . could you explain your answer. – Idan May 10 '13 at 06:53
  • @idannik If $x+3$ is not a factor of the minimal polynomial of $A$ (I hope you have learnt what a minimal polynomial is), then the minimal polynomial must be equal to $x-1$, which means $A=I$, which contradicts our assumption. Therefore $x+3$ must be a factor of the minimal polynomial, i.e. $A+3I$ is singular. But then $B$ is singular too. – user1551 May 10 '13 at 08:20
  • oh, I got it . thanks – Idan May 11 '13 at 07:24