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I was asked this question:

There's $A,B \in M_3 (\mathbb{R})$ a square matrices from order $3\times3$ from the real numbers, that satisifes: $BA^2+A^2=I$.

  1. Prove that $A$ is invertible and that the matrices $A$ and $B$ have the commutative property. $(AB=BA=I)$
  2. Given that $B+A^2B=O$. Prove that $B$ isn't invertible and calculate $|A|$, write the two possibilities.

My first attempt (But then I saw that determinant does not preserve the addition, then this attempt is completely wrong): Suppose $A$ isn't invertible, meaning $|A|=0$.

$BA^2+A^2=I \rightarrow A^2=I-BA^2$

And since we assumed that A isn't invertible: $|A^2|=|I|-B|A|^2 \rightarrow 0=1-B*0 \rightarrow 0=1$ Contradiction, therefore $A$ is invertible.

My second attempt: $BA^2+A^2=I \rightarrow (B+I)A^2=I$

Now, we need to show that $A^2(B+I)=(B+I)A^2=I$, which will prove that $A$ is invertible and found an inverse for it.

But I wasn't sure how to continue from here...

And since $A$ is invertible then: $BA^2+A^2=AB^2+A^2=I \rightarrow A^2(B+I)=(B+I)A^2=I$

Which proved that the commutative property exists.

My attempt on the second question: (But it also relies on adding determinants, then it must be incorrect too) We'll prove it by contradiction, $B$ is inverse, meaning $|B|\neq 0$: $B+A^2B=O \rightarrow B=O-A^2B \rightarrow |B|=-|A^2||B|$

And from $1$. we know that $|A|\neq 0$, and since we assumed that $|B| \neq 0$ then the determinant of $B$ isn't necessarily always $0$, therefore it is a contradiction.

I'd love for guidance on how to continue. Thanks for the help!

Christian E. Ramirez
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therealcain
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1 Answers1

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$\bullet$ You have shown that $BA^2+A^2 = I$. This can be written $$ (BA+A)A=id. $$ This shows that $BA+A$ is the inverse of $A$. Since $A$ commutes with its inverse, we know that $A$ commutes with $BA+A$. So $$ A(BA+A) = (BA+A)A, \quad ABA = BAA, $$ and multiplying on the right by $A^{-1}$, we end with $BA=AB$.

$\bullet$ For the second question, $B = -A^2B$ together with $BA^2+A^2=I$ and the commutativity of $A$ and $B$, $$ A^2 = B+I, \quad B+A^2B = B+(I+B)B = 0, \quad 2B+B^2 = 0. $$ If $B$ is invertible, multiplying the last equality by its inverse would show that $B = -2I$ and $-A^2 = I$. As pointed out by Ben Grossmann, this is impossible for 3x3 matrices, as can be checked using the determinant. $$\det(-A^2) = -\det(A)^2 \neq \det(I).$$

$\bullet$ We now need to compute the determinant of $A$. From $A^2 = B+I$ and $2B+B^2 = 0$, we obtain $$A^4 = (B+I)^2 = B^2+2B+I = I.$$ Taking the determinant, this gives $\det(A)^4 = 1$. We have seen previously that $\det(A)^2 \neq -1$, so $\det(A)^2 = 1$ and $\det(A)$ is equal either to 1 or $-1$.

coudy
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