If Matrix $A^3 = 0$ then what will be $I+A+A^2$
Andrew Mclaughlin 
Matrix $A$ is a square matrix such that $$A^3=0$$ then what would be value of $$I+A+A^2=?$$ such that $I$ is unit matrix of same order as $A$
I firstly supposed sum to be $X$ that is $X=I+A+A^2$ then $$AX=A+A^2+A^3$$ $$AX=A+A^2$$ and multiplying both sides with $A^{-1}$ we get $$K=I+A$$ but my answer was incorrect , what is correct solution to this problem?
$\endgroup$ 34 Answers
$\begingroup$$$(I+A+A^2)(I-A)=I-A^3=I$$
Hence $I+A+A^2=(I-A)^{-1}$.
$\endgroup$ 0 $\begingroup$If $A^3=0$, then $A^4=0$, $A^5=0$, and so on, so $I+A+A^2=I+A+A^2+A^3+A^4+\cdots$. The infinite sum $I+A+A^2+A^3+A^4+\cdots$ looks a lot like an infinite geometric series. There’s a formula for the sum of an infinite geometric series (of real numbers, if the ratio has absolute value less than $1$), $\frac{a}{1-r}=a+ar+ar^2+\cdots$ so maybe it would be interesting to play with that formula in this case:
$$I+A+A^2+A^3+A^4+\cdots \overset{?}{=}\frac{I}{I-A}.$$
Given that $I+A+A^2+A^3+A^4+\cdots = I+A+A^2$, this becomes
$$I+A+A^2 \overset{?}{=}\frac{I}{I-A}.$$
Even if you don’t believe this makes any sense, it might lead you to look at whether
$$\left(I+A+A^2\right)\left({I-A}\right) =I.$$
It does, so you’ve discovered that when $A^3=0$, $I+A+A^2=\left(I-A\right)^{-1}$.
$\endgroup$ $\begingroup$Try to use the remarkable identity $A^3 - I^3$ then $A^3 - I^3 = (A-I)(A^2+A+I)$ and because $A^3-I^3 = -I$ we get by multiplying both sides (from the right) by $(A-I)^{-1}$ $A^2+A+I = -(A-I)^{-1}$
$\endgroup$ $\begingroup$Since $A^3 = 0$, the eigenvalues of the matrix $A$ are zero. Thus $I-A$ is invertible and we have $(I-A)(I+A+A^2) = I$. So $(I-A)^{-1}=(I+A+A^2)$.
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