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Let $A^3=0$; check if $A^2-A+I$ is invertible and, if it is, find its inverse.

Is it possible to solve it this way?

$A^2-A+I=A(A-I)+I=B$

Multiply both sides by $A^2$:

$A^3(A-I)+ A^2= A^2B$

$A^2= A^2B$

$I=B$

$I$ is an invertible matrix!

egreg
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Nemanja Strkic
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4 Answers4

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$$(A+I)(A^2-A+I)=A^3+I$$

Hence,

$$(A+I)(A^2-A+I)=I$$

Ahmed S. Attaalla
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    Three answers form a whole here: This answer shows why $B$ is invertible. Wirius' answer explains what OP did wrong in his solution, and egreg showed how the mistake OP did can actually be proven wrong with a counterexample. – 5xum Aug 22 '17 at 10:21
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No, you can't remove $A^2$ in the last step, because $A$ is not invertible.

For instance, $A=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$ satisfies $A^3=0$, but $$ A^2-A+I=\begin{pmatrix}1 & -1 \\ 0 & 1\end{pmatrix}\ne I $$ contradicting your conclusion.

egreg
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3

The problem with your proof is that you consider by saying $A^2 = A^2 B \implies I = B$ that $\mathcal{M}_n(\mathbb{K})$ is an integral domain. An integral domain is a ring $R$ such as : $$\forall a, b \in R, \ ab = 0 \implies a = 0 \lor b = 0$$ In such a ring, if $c \in R\setminus\{0\}$, we have : $$\forall a, b \in R, \ ac = bc \implies (a-b)c = 0 \implies a-b = 0 \lor c = 0 \implies a = b$$ ...which is the property you've used.

However, $\mathcal{M}_n(\mathbb{K})$ isn't an integral domain, for exemple : $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

Ahmed S. Attaalla's answer is a correct proof that $B$ is an invertible matrix.

Wirius
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  • Thanks ! He could have also done this step if $A^2$ was invertible, however, not only didn't he prooved that but it also was false : $(\det(A))^3 = \det(A^3) = 0$, so $\det(A) = 0$ and thus $\det(A^2) = \det(A)^2 = 0$. – Wirius Aug 22 '17 at 10:35
  • Basically, this answer is one third of the whole truth. the other two answers provide the rest :) – 5xum Aug 22 '17 at 10:36
  • Yes indeed ! :) – Wirius Aug 22 '17 at 10:38
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You can go further with your approach to make it work. One route through is to try to express the left-hand side as multiples of $B$ plus the original $I$ (you can do that with $A^2=A^2B$ so need to attack the $A$) $$A^2-A+I=B$$Multiply by $A^2$ as you did $$A^2=A^2B$$ Now also multiply by $A$ $$-A^2+A=AB$$Substitute the result for $A^2$ in this latest result $$-A^2B+A=AB; A=A^2B+AB$$ Now sub back into the original equation $$A^2B-A^2B-AB+I=B$$ or $$I=B+AB=IB+AB=(I+A)B$$

Mark Bennet
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