Proof of square and identity matricies

Question

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...

Answer 1

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?

Answer 2

For the first one:

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

Answer 3

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 $0.$