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So I am given this to prove:

If D is a square matrix and $D^2=D$, where I is the identity matrix, show that:

i) $(I-D)^2=I-D$

ii) $2D-I$ is invertible

My approach for i) was just to brute force is by plugging in a $2$ $X$ $2$ square matrix $\begin{vmatrix} a & b \\ c & d \end{vmatrix} $

and a $2$ $X$ $2$ identity matrix and just hopefully equating both sides after I multiplied the resultant but I think if I do that I have a loss of generality so I am not sure if that approach will work.

I'm really stuck on ii) though. A matrix is said to be invertible if one matrix along with another matrix are commutative and equal the identity matrix but again, if I just brute force this, I think i'll lose generality so I don't think that will work either.

I was hoping to get some guidance on this...

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    Don't brute force it – 2017-02-03
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    How would I go about it then? – 2017-02-03
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    Think about the two Matrices as Linear mappings and what $D^2=D$ means for the linear mapping. – 2017-02-03
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    @SubhashisChakraborty Even if you proved it for a 2x2 matrix, that doesn't prove it in the general case. – 2017-02-03
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    Yeah I thought as much haha... I will attempt this a little more. I thought because I used general variables it would be okay but I see my mistake. – 2017-02-03
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    $D\in M_n(k)$ is some square matrix. $n=2$ is not really a general variable, now, is it? – 2017-02-03
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    Yes you are right. – 2017-02-03

3 Answers 3

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For i), you need to show it in any dimension, so a $2\times 2$ case isn't sufficient. Just multiply it out! $$(I-D)^2 = I^2 -ID-DI+D^2$$ and simplify using the properties of $I$ and $D,$ that $ID=DI=D$ and $D^2=D.$

To show that $2D-I$ is invertable, you need to find an inverse for it, i.e. a matrix $M$ such that $(2D-I)M = I.$ Hint: try multiplying $(2D-I)(aD-bI)$ using the properties above and then figure out what $a$ and $b$ need to be for the product to equal $I.$ When you multiply out and simplify you'll get a term proportional to $I$ and a term proportional to $D$ so set $a$ and $b$ so that the coefficient of $I$ is $1$ and the coefficient of $D$ is $0.$

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    I got the first part now. For the second part, I get a=2 and b=1. I think this makes sense since I have just multiplied the original matrix by a matrix that was in proportion to itself. Since I found a second matrix which makes the original commutative, I can now conclude that that matrix is indeed invertible. IS That the correct thought? – 2017-02-03
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    It's just really funny since the inverse is itself in the end. – 2017-02-03
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    @SubhashisChakraborty I'm not sure what you mean by "makes the original commutative". You mean "multiplies the original to equal the identity"? Anyway, it appears you got $a=2$ and $b=1$ which does mean that the inverse is itself. This isn't a coincidence. You will learn that when $D=D^2,$ this means $D$ is a projection onto a subspace and $I-D$ projects onto its orthogonal complement. So $2D-I = D- (I-D)$ just mirror reverses the orthogonal complement while leaving the subspace $D$ projects onto alone. Doing this twice is two mirror reversals which is the same as doing nothing. – 2017-02-03
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    Yeah that' what I meant. I got it now. It's okay. Okay that's pretty neat. Linear algebra is just really abstract so it's hard for me to wrap my head around it sometimes. – 2017-02-03
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For the first one:

$$(I-D)^2 = I^2 -ID - DI +D^2 = I- 2D +D = I- D$$

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For (i), distribute the product $$ (I-D)^2 = (I-D)(I-D) = I(I-D) - D(I-D) = \\ I^2 - 2D + D^2 = I - D $$ For (ii), you may similarly distribute $$ (2D - I)^2 $$ what does this get you? What can you conclude?

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    for (i) why does the $-2D$ got to 0? I can't reason tat part out. – 2017-02-03
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    @SubhashisChakraborty I don't understand how you came to that conclusion. What I left out was $$ -2D + D^2 = -2D + D = -D $$ – 2017-02-03
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    Oops I forgot that D was a square matrix so I can just add on the terms. Oh my gosh... – 2017-02-03
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    Okay for the second one, if I square it, I just get back the identity matrix. Don't I have to pick a different matrix in order to prove this though? – 2017-02-03
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    Nope. You've just shown that $(2D - I)$ is its own inverse. – 2017-02-03
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    Is that possible? A matrix's inverse can be itself? The identity matrix comes to mind, but other than that. – 2017-02-03
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    Thanks by the way. I think I got it. – 2017-02-03
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    Yes, it's possible. An example: $$ \pmatrix{1&0\\0&-1} $$ – 2017-02-03
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    Very interesting. Thanks a lot. – 2017-02-03