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Show that if a matrix $\mathbf{A}$ satisfy $\mathbf{A}^2 - \mathbf{A} + \mathbf{E} = \mathbf{0}$ then matrix $\mathbf{A}$ is regular.

(Note that $\mathbf{E}$ denotes identity matrix.)

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up vote 5 down vote accepted


1) For any $n \times n$ matrix $A$ if there exists an $n\times n$ matrix $B$ such that $AB=I$, the identity matrix, then $A$ is invertible.

2) Can you use the identity in the question that $A$ satisfies to find an appropriate $B$?

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Following your hints - I'll get $\mathbf{A}(\mathbf{E} - \mathbf{A}) = \mathbf{E}$ which means that $\mathbf{A}^{-1} = \mathbf{E} - \mathbf{A}$. I still cannot see why it needs to be regular. – OukiDouki Feb 1 '13 at 7:15
since you just found that A has an inverse matrix. Which is the same as saying that it is regular. – Ittay Weiss Feb 1 '13 at 7:49
True, the fact that the equation can be expressed as $\mathbf{A}(\mathbf{E}-\mathbf{A}) = \mathbf{0}$ implies the existence of inverse matrix. What can I say ... :-) Thanks – OukiDouki Feb 1 '13 at 8:04
you are welcome – Ittay Weiss Feb 1 '13 at 8:39
... in my last comment should be $\mathbf{A}(\mathbf{E} - \mathbf{A}) = \mathbf{E}$ ... – OukiDouki Feb 1 '13 at 9:00

If $A$ is not regular, there is a non zero vector $v$ such that $Av=0$.

Then $0=0v=(A^2-A+E)v=A^2v-Av+v=v$, which is absurd.

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Why $ \mathbf{A}^2 \mathbf{v} - \mathbf{A} \mathbf{v} = \mathbf{0}$ ? – OukiDouki Feb 1 '13 at 7:21
I'm maybe asking stupid questions, but I'm missing some important point. – OukiDouki Feb 1 '13 at 7:25
@OukiDouki because $Av=$ by assumption and hence also $A^2v=A\cdot Av=A0=0$. More generally, if $p$ is a polynom such that $p(A)=0$ and $A$ is not regular, then we must have $p(0)=0$ (the proof of which is just as for this special case). – Hagen von Eitzen Feb 1 '13 at 7:49

The essential point is that $\mathbf A$ satifies a polynomial equation in which $\mathbf E$ occurs (with nonzero coefficient). Just isolate this $\mathbf E$ as one side of the equation, factor an $\mathbf A$ form the other side, and you have an explicit description of the inverse of $\mathbf A$; this always works. Here the equation becomes $$ \mathbf E=\mathbf A-\mathbf A^2 = \mathbf A(\mathbf E-\mathbf A) $$ so $\mathbf A^{-1}=\mathbf E-\mathbf A$.

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