Inverse matrix proof

Question

Let $A$ be a $n \times n$ matrix. Show that if $A^2=O$ then $A$ is singular, but $I−A$ is nonsingular and $(I−A)^{-1}=I+A$.

What I have tiedy: $(I-A)*(I+A)=I-A+A-A^2$

$=I-A^2$

$=I-0$ since A^2=0

$=I$

Therefore $I-A$ is nonsingular.

I am not sure if is complete and correct.

Your proof of the second part is the fine. For the first part, make use of the multiplicative property of the determinant. — lokodiz, Sep 21 '14 at 20:43
You don't need determinants here. If $Ax=0$ for all $x$, then clearly $A$ is singular. If not, then choose $x$ so that $Ax \neq 0$. Then $A (Ax) = 0$, hence $A$ is singular. — copper.hat, Sep 21 '14 at 20:55
If $A$ were invertible, multiplying both sides of $A^2 = 0$ by $A^{-1}$ would yield $A = 0$. — littleO, Sep 21 '14 at 21:34
See also: If $A$ is an $n \times n$ matrix such that $A^2=0$, is $A+I_{n}$ invertible? — Martin Sleziak, Jan 14 '20 at 14:55

score 0 · Accepted Answer · answered Sep 22 '14 at 00:15

0

$A^2 = O \implies \det(A)\det(A) = \det(A^2) = 0$ So $\det(A) = 0$ and $A$ is singular.

$(I-A)(I + A) = I + A - A - A^2 = I$ the same is true for $(I+A)(I-A)$ so $I - A$ is invertible with inverse $I + A$.

answered Sep 22 '14 at 00:15

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