Linear Algebra: Idemponent matrix

[Amy-Lee]

In idemponent matrices:
Is there a value of r for which
(matrix) A= -1 -1
2 r

is idempotent? If so, what is the value of r and why.

I know that A^2=A, that's about all... How does one find r, if this is the only info given?

I have no idea how to approach this. Please help.

 

[CFDFEAGURU]

What is an indemponent matrix?

Thanks
Matt

 

[Amy-Lee]

If A is an n x n matrix, then A is called idempotent if A^2=A

 

[CFDFEAGURU]

Sorry, can't help you with that. But thanks for letting me no what idempotent means.

Thanks
Matt

 

[gabbagabbahey]

What is an indemponent matrix?

The simple definition is simply a matrix for which $A^2=A$.

In idemponent matrices:
Is there a value of r for which
(matrix) A= -1 -1
2 r

is idempotent? If so, what is the value of r and why.

I know that A^2=A, that's about all... How does one find r, if this is the only info given?

I have no idea how to approach this. Please help.

Well, if $$A=\begin{pmatrix}-1 & 1 \\ 2 &r \end{pmatrix}$$, what will $A^2$ be?...Setting $A^2=A$ should give you 4 equations (one for each component of the matrices), is there a value of $r$ that solves all 4 equations simultaneously?

 

[Amy-Lee]

A^2 = -1 -1
2 r

and -1 -1 = -1 -1
2 r 2 r ?

 

[Amy-Lee]

sorry for bad layout... still haven't figured out how to use advanced reply

 

[Preno]

Think about the determinant of an idempotent matrix.

 

[gabbagabbahey]

A^2 = -1 -1
2 r

and -1 -1 = -1 -1
2 r 2 r ?

No,

$$A=\begin{pmatrix}-1 & -1 \\ 2 &r \end{pmatrix}\implies A^2=\begin{pmatrix}-1 & -1 \\ 2 &r \end{pmatrix}\begin{pmatrix}-1 & -1 \\ 2 &r \end{pmatrix}$$

Carry out the matrix multiplication.

 

[Amy-Lee]

ok I get:
-1 1-r
-2+2r -2+r²

 

[gabbagabbahey]

Right, and for $A$ to be idempotent, that must also be equal to the matrix $A$...so you want to find an $r$ such that

$$\begin{pmatrix}-1 & 1-r \\ -2+2r &-2+r^2 \end{pmatrix}=\begin{pmatrix}-1 & -1 \\ 2 &r \end{pmatrix}$$

 

[Amy-Lee]

ok thank you... do I now have to reduce it to row-echelon form, by using the Gaussian Elimination?

 

[HallsofIvy]

You don't have to. If $A^2= A$ then you must have -1= -1, -1= 1-r, -2+2r= 2 and -2+r^2= r. Are there values of r that satisfy all of those and, if so, what are they?

 

[Amy-Lee]

r= 2; r=1/2; r=-1

 

[gabbagabbahey]

r= 2; r=1/2; r=-1

r=2 works, but the other two values do not satisfy all 4 equations. (For the two matrices to be equal, all 4 of there components must be equal)

 

[Amy-Lee]

r=2 works, but the other two values do not satisfy all 4 equations. (For the two matrices to be equal, all 4 of there components must be equal)

thank you so much for your help. One last question on the topic: is a square invertible idempotent matrix also the Identity matrix?

 

[gabbagabbahey]

thank you so much for your help. One last question on the topic: is a square invertible idempotent matrix also the Identity matrix?

Well, if $A^2=A$, what can you say about $\detA$?

 

[Amy-Lee]

Well, if $A^2=A$, what can you say about $\detA$?

I don't understand the last part of what you said

 

[gabbagabbahey]

Take the determinant of both sides od the equation $A^2=A$...what do you get?