Finding a basis of the image of a linear transformation

[WednesdayBass]

Homework Statement

Let Ψ: Mat_2x2(R) -> Mat_2x2(R) be defined as:
[a,b;c,d] -> [a+b, a-c; a+c, b-c]
Find a basis for the image of Ψ.

Homework Equations

None, AFAIK.

The Attempt at a Solution

I started by using the standard basis, B, for Mat_2x2(R) to get ^B [with u in Mat_2x2(R)] as [a;b;c;b] and [Ψ]^B_B as:
[1,1,0,0;
1,0,-1,0;
1,0,1,0;
0,1,-1,0]

And tried to use Gaussian elimination on that but that just gives me an inconsistent result.

 

[FanofAFan]

won't your standard basis be v₁ = ([tex]\stackrel{1}{0}[/tex][tex]\stackrel{0}{0}[/tex]) and so on for v₂, v₃, v₄

 

[WednesdayBass]

won't your standard basis be v₁ = ([tex]\stackrel{1}{0}[/tex][tex]\stackrel{0}{0}[/tex]) and so on for v₂, v₃, v₄

Yeah, I've used that basis to get the coordinates of the matrix u ([a,b;c,d]) and find the matrix which you multiply that coordinate vector by to get the new matrix. I have no idea if that's heading in the right direction or not...

 

[FanofAFan]

Is this a problem in a book... also I'm not sure what you are doing wrong but isn't it (a+b)v₁ + (a-c)v₂ + (a+c)v₃ + (b-c)v₄

 

[WednesdayBass]

No, it's not from a book, it's from a worksheet I've been given.

(a+b)v₁ + ...
is the coordinates for Ψ(u) in the standard basis of Mat_2x2(R). So the coordinate vectors are ^B = [a;b;c;d] and [Ψ(u)]^B = [a+b;a-c;a+c;b-c] which is where I got the 4x4 matrix.