I have a question in my exam which is:
Let :v = [3,1] and T : R2 -> R2 denote the transformation such that T(x) is the vector in span{v} nearest to x
Let :0<a<pi/2 denote the angle between the positive x-axis and the vector v. Let R: R2->R2 denote the transformation such that R(x) is the vector x rotated by the angle a clockwise
Find the Linear transformation R and T (Dot product and Differentiation is not allowed)
I know that R is just a clockwise matrix, but for T, the answer says it is found by producing an orthogonal vector w=[-1,3] to form a new basis and then produce a change-of-basis matrix P = $$\begin{bmatrix} 3 & -1 \\ 1 & 3 \\ \end{bmatrix} $$ Then since T(x) is in span(v), so the transformation performed on the new basis = B = $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix} $$
Then by T=PB(P^-1) we can find out the answer A = $$\begin{bmatrix} 0.9 & 0.3 \\ 0.3 & 0.1 \\ \end{bmatrix} $$
But I have 2 questions neither the professor nor TA answer me
- What if I choose any other arbitrary vector to form the new basis instead of using the orthogonal one? says w=[0,3]
I did some try before asking, but the answer is wrong. I guess that happened on the Matrix B transformation performed on a new basis, probably because the basis is no longer orthogonal. But isn't the T simply project x on v when the basis is changed? which means no height and any other vectores, so the matrix shouldn't be same????
- TA says there is actually another easier method, he just hint us T(x) = [ ? ? ]
I am extremely confused with this so-called easier form, I referred to David Lay, linear algebra Ch5.4, I guess he wanted to say is the transformation form x -> Ax, which can be calculated by putting a basis in the transformation and attaining a new basis vector, then form a new transformation matrix. Is that the so-called easier form? How does it work??
Really sorry for my ignorance and bothering your time