Using function T_A(v) = Av to transform 2 vectors

[leej72]

Homework Statement

Let u1 = [1 1]^T and u2 = [0 -1]^T. Find a 2 x 2 real matrix A so that the function T_A is a map from ℝ^2 to ℝ^2, given by multiplication by A,

T_A := Av,

sends T_A(u1) = v1 and T_A(u2) = v2 where v1 = [cosθ sinθ]^T and v2 = [-sinθ cosθ]^T. Explain/justify your work.

Homework Equations

The Attempt at a Solution

in the first part of the question, we are asked to prove that v1 and v2 form an orthonormal basis for ℝ^2. At first I thought that we would use Gram-Schmidt process but the two vectors are already orthonormal. So basically I am clueless as to where to start/proceed.

 

[leej72]

I attached the pdf file of the question, it is question #1b.

 

[damabo]

we must have $A.u_1=v_1=(cos(theta),sin(theta))^T$, and $A.u_2=v_2=(-sin(theta),cos(theta))$
In other words, a_11 + a_12 = cos(theta) ; a_21 + a_22 = sin(theta) ; - a_12 = -sin(theta) ; - a_22 = cos(theta) .

 

[damabo]

well, first you prove that v_1 and v_2 are orthonormal:
normal: sqrt(cos²(theta) + sin²(theta))=1 in both cases.
ortho: if [.,.] is the inproduct in R, then [v_1,v_2]= (-sin(th)cos(th)+sin(th)cos(th))=0.
then you prove that they are a basis of R^2. which means 1. linearly independent 2. span{(v_1),(v_2)}=R^2.

 

[leej72]

we must have $A.u_1=v_1=(cos(theta),sin(theta))^T$, and $A.u_2=v_2=(-sin(theta),cos(theta))$
In other words, a_11 + a_12 = cos(theta) ; a_21 + a_22 = sin(theta) ; - a_12 = -sin(theta) ; - a_22 = cos(theta) .

Thanks a lot for helping me out, I guess I was overcomplicating the question because we are covering different topics, mainly Gram-Schmidt process so I thought we would have to incorporate that into the question.