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If $\bf I$ is an $n \times n$ identity matrix and $\bf A$ is any $n \times n$ matrix such that $\bf{A}^{2} = \bf{A}$, then which one of the following is true?

a) $(\bf{I}+\bf{A})^{2} = \bf{I} + \bf{A}$

b) $(\bf{I}-\bf{A})^{2} = \bf{I} + \bf{A}$

c) $(\bf{I}+\bf{A})^{2} = \bf{I} - \bf{A}$

d) $(\bf{I}-\bf{A})^{2} = \bf{I} - \bf{A}$

If $\bf I$ is an $n \times n$ identity matrix and $\bf A$ is any $n \times n$ matrix such that $\bf{A}^{3}$ is the zero matrix, then which one of the following is true?

a) $(\bf{I}-\bf{A})^{-1} = \bf{I} + \bf{A} + \bf{A}^{2}$

b) $(\bf{I}-\bf{A})^{-1} = \bf{I} + \bf{A}$

c) $(\bf{I}-\bf{A})^{-1} = \bf{I} - \bf{A}$

d) $(\bf{I}-\bf{A})^{-1} = 1 - \bf{A} - \bf{A}^{2}$

Would I be correct in choosing d) for the first question? I am not sure how to use the definition of an inverse for the second question.

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Hi, I edited the question and it now reads correctly, my apologies for the mistake –  user47779 Nov 1 '12 at 11:20
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What do you get when you multiply matrices $(I-A)(I-A)$? What do you get if you multiply $(I-A)(I+A+A^2)$? –  Martin Sleziak Nov 1 '12 at 11:22
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You are given a number of alternatives, each of the form $X^{-1}=Y$. But $X^{-1}=Y$ means $XY=1$, which should tell you a way to test them. –  Gerry Myerson Nov 1 '12 at 12:04
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@Martin Sleziak, I get (I - A) if I multiply matrices (I - A)(I - A) and the identity matrix if I multiply (I - A)(I + A + A^2) but I'm not sure how to evaluate (I - A)^-1. –  user47779 Nov 1 '12 at 22:56

1 Answer 1

You solved the first part correctly. You have $(I-A)^2=(I-A)(I-A)=I-2A+A^2$. If you know that $A^2=A$, you get $(I-A)^2=(I-A)(I-A)=I-2A+A^2=I-A$.


If $X$ is an $n\times n$-matrix, then the inverse of $X$ is a $n\times n$-matrix $Y$ such that $$XY=YX=I$$ and it is denoted by $X^{-1}$.

Not every matrix must have an inverse - such matrix is called invertible.

In fact, if we have two square matrices $n\times n$, it suffices to verify one of the equalities $XY=I$ and $YX=I$. If one of these equalities holds, then the other one must hold too and $Y$ is the inverse of $X$. See If $AB = I$ then $BA = I$.

In the second part you are asked to find $(I-A)^{-1}$, i.e., the inverse of $I-A$ and you are given some choices. So all you have to do is to try whether some of these choices gives $I$ after multiplication by $I-A$.

Indeed $$(I-A)(I+A+A^2)=I+A+A^2-A-A^2-A^3=I-A^3.$$ If $A^3=0$, then we get $(I-A)(I+A+A^2)=I$ and $(I-A)^{-1}=(I+A+A^2)$.

(As mentioned before, you should either compute $(I+A+A^2)(I-A)=I$ too or you should refer to the result that $XY=I$ implies $YX=I$ for square matrices.)


Perhaps it is useful to remind you that multiplication of matrices is not commutative in general. Therefore if you want, for example, $(A+B)^2$, you get $(A+B)^2=(A+B)(A+B)=A^2+AB+BA+B^2$, which is different from $A^2+2AB+B^2$ if $AB\ne BA$. (In you example we worked with matrices $I$ and $A$. These matrices do commute: $IA=AI=A$. So in this case the computations are simpler.)


Another interesting fact: The second problem is in fact a special case of Neumann series: $$(I-A)^{-1}=\sum_{k=0}^\infty A^k = I+A+A^2+\dots,$$ if the series on the RHS converges.

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