# Finding a basis given a transformation matrix

1. Oct 18, 2009

### enaktan90

1. The problem statement, all variables and given/known data
Let T : M2,2, --> M2,1 be the linear transformation given by

T ([a b; c d]) = [a-2b ; c-2d]
Fix bases B = { [1 0 ; 0 0], [ 0 1 ; 0 0], [0 0 ; 1 0], [0 0 ; 0 1]} and
C = { [1 ; 0], [0 ; 1]} for M2,2, and M2,1 respectively.

(a) Find the matrix [T]C,B of T with respecct to the bases B and C.

(b) Use the matrix from part (a) to find a basis for Ker(T)

(c) Find a basis D for M2,2 such that:

[T]C,D = [ 1 0 0 0 ; 0 1 0 0 ]

2. Relevant equations

above

3. The attempt at a solution

I HAVE MANAGED TO DO PART (A) AND (B). CLUELESS ABOUT PART (C)!

2. Oct 18, 2009

### HallsofIvy

Congratulations on doing (a) and (b)! Since you were able to do that, presumably, you know that you can take the linear transformation of each basis element in $M_{2,2}$ in turn, then write the result as a linear combination of the basis in $M_{2,1}$, the coefficients then being the columns.

Now you want to specify the bases that will give that particular matrix. That will be possible if and only if the given basis is "similar" to the matrix you got in (a).

Here's a pretty obvious way to do it: Since you want the last two columns to be zeroes, choose two independent matrices from the kernel of T (which would have to be 2 dimensional for this to be possible). Now, find two independent matrices that are not in the kernel so that T maps them into [1 0] and [0 1].

Since T([a, b; c, d])= [a- 2b, c- 2d] that means you are looking for 4 matrices such that
1) a- 2b= 1 and c- 2d= 0.
2) a- 2b= 0 and c- 2d= 1. And the matrices in (1) and (2) are independent.
3) a- 2b= 0 and c- 2d= 0.
4) a- 2b= 0 and c- 2d= 0. And the maricex in (3) and (4) are independent.