Can someone check my workings, linear matrix question

[jasmeetcolumbia98]

Matrix D is defined to be idempotent if D^2 = D Idempotent means if you multiply a matrix by itself it will be equal to the original matrix A Matrix is non singular is its determinent does not equal 0 Could I show this by creating a 2x2 matrix D with elements (1,2,3,4) where D^2 = D*D then I could use ad-bc to show that its determinant isn't equal to 0 (1*4)-(2*3)=-2 det(D) =/= 0 (is not equal to 0) Is this correct? Sorry this isn't in latex form

 

[topsquark]

Matrix D is defined to be idempotent if D^2 = D Idempotent means if you multiply a matrix by itself it will be equal to the original matrix A Matrix is non singular is its determinent does not equal 0 Could I show this by creating a 2x2 matrix D with elements (1,2,3,4) where D^2 = D*D then I could use ad-bc to show that its determinant isn't equal to 0 (1*4)-(2*3)=-2 det(D) =/= 0 (is not equal to 0) Is this correct? Sorry this isn't in latex form

Well, there is no idempotent 2x2 matrix with elements 1, 2, 3, 4, but otherwise yes. Find an idempotent matrix and just calculate its determinant. However, consider this... applying an idempotent matrix twice changes nothing. What does that mean for the determinant? It's an easy proof but you should play with the numbers a bit to get a "feel" for it.

-Dan

 

[jasmeetcolumbia98]

Thanks so much for your reply
If the following was a 2x2 matrix
(4,-1,12,-3) is an idempotent matrix using ad-bc its determinant is 0
multiplying this matrix by itself would equal (4,-1,12,-3)
det(D)=(4*-3)-(-1*12)=0
I'm a little confused because det(D) = 0 which suggests it's a singular matrix contradicting the question

I thought about saying D^2 =D if D is idempotent which means
(D^2) -D=D(D-I)
If D(D-1) this would have an eigenvalue of 0 so D would be singular
if D has to be non singular if we state z=D(D-1)
z' = D-1 this means D=I which also means D is both non singular and idempotent
Is this right?
Thank you so much for your help, I really appreciate it!

 

[HallsofIvy]

Thanks so much for your reply
If the following was a 2x2 matrix
(4,-1,12,-3) is an idempotent matrix using ad-bc its determinant is 0
multiplying this matrix by itself would equal (4,-1,12,-3)
det(D)=(4*-3)-(-1*12)=0
I'm a little confused because det(D) = 0 which suggests it's a singular matrix contradicting the question

I thought about saying D^2 =D if D is idempotent which means
(D^2) -D=D(D-I)

This is very confused! D(D-I) is what? I presume you meant to say that D(D-I)= 0.

If D(D-1) this would have an eigenvalue of 0 so D would be singular

First, after "if" you need a statement and "D(D-1)" is not a statement because it has no verb! You mean "If D(D-1)= 0". Yes, D would be singular which it clearly is since its determinant is 0.

if D has to be non singular if we state z=D(D-1)

Two "if"s without a "then"! I don't know what you are trying to say. If you meant
"D has to be non singular if we state z= D(D- 1)", no it doesn't. It would if z is not the 0 matrix but where did this "z" come from?

z' = D-1 this means D=I which also means D is both non singular and idempotent
Is this right?
Thank you so much for your help, I really appreciate it!

No, saying "z'= D- 1" doesn't say anything about "D= I". If "0= D-I" then D is the identity but I have no Idea where z' came from!

 

[topsquark]

I'm a little confused because det(D) = 0 which suggests it's a singular matrix contradicting the question

Good! That means you are thinking. Let's take a look at some real idempotent matrices.

Let
[math]D = \left ( \begin{matrix} a & b \\ c & d \end{matrix} \right )[/math]
Then
[math]D^2 = \left ( \begin{matrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{matrix} \right ) = \left ( \begin{matrix} a & b \\ c & d \end{matrix} \right )[/math]
So we need to solve the system of equations
[math]\begin{cases} a = a^2 + bc \\ b = ab + bd \\ c = ac + cd \\ d = bc + d^2 \end{cases}[/math]
There are many ways to play with these, I'll let you go ahead and solve some if you like.

Let's take
[math]D = \left ( \begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix} \right )[/math]This is idempotent, as you can check. It also has a determinant of 0.

Another. [math]D = \left ( \begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix} \right )[/math]. Again, det(D) = 0.

But is this all that there is? Notice that [math]\left ( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right )[/math] is also idempotent and has determinant 1. Are there any other possibilities?

Here's a brief rundown of the result. The only way we can have an invertible idempotent matrix is for the main diagonal to be all 1s. If you work with it you will find that the only idempotent invertible matrices are the n x n identities. So there are idempotent matrices with a non-zero determinant but they are only the identities. (You can see this physically. Any idempotent matrix is a projection matrix. When we project a vector onto a surface we lose information (the "height" in the space perpendicular to the space you are projecting onto) so you can't "undo" it, ie. it has no inverse, ie. it has determinant 0.) There are a number of holes in my argument. Feel free to plug those holes... it will be a good exercise.

-Dan