# Linear Algebra standard matrix of transformation

I have done the work however I wanted to make sure my answers were correct.

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Problem (still getting the hang of mathjax I tried to make it as neat as possible):

Consider the linear transformation $$T:\mathbb R^2 \to \mathbb R^2$$ that first rotates a vector with $$\pi/4$$ radians clockwise and then projects onto the $$x_2$$ axis

(a) Find $$T\begin{pmatrix} 1 \\ 1 \\ \end{pmatrix}$$

(b) Find the standard matrix for $$T$$.

(c) Is $$T$$ onto (surjection)?

For (a) with some error and help I got to the answer of

$$\left[ \begin{array}{cc|c} 0\\ 0 \end{array} \right]$$

I am confused how I should be getting (b). Is my answer correct?

$$\left[ \begin{array}{cc|c} \cos(\pi/4)&-\sin(\pi/4)\\ \sin(\pi/4)&\cos(\pi/4) \end{array} \right]$$ times

$$\left[ \begin{array}{cc|c} 0&0\\ 0&1 \end{array} \right]$$

After I multiplied those two together I got

$$\left[ \begin{array}{cc|c} 0&\sqrt(2)/2\\ 0&\sqrt(2)/2 \end{array} \right]$$

Would this be the standard matrix for $$T$$?

Also can anyone lead me in the direction of how I can determine if this is a surjection or not?

Thank you!

$$\left[ \begin{array}{cc|c} \cos(\pi/4)&-\sin(\pi/4)\\ \sin(\pi/4)&\cos(\pi/4) \end{array} \right]$$
is the matrix for rotation by $$\frac\pi4$$ anti-clockwise, so it is not the right matrix.
Second, you then multiplied that matrix $$R$$ by the projection matrix $$P$$, but that means that the matrix you ended up with, $$R\cdot P$$, acts on the vector $$x$$ like so: $$(R\cdot P)\cdot x = R\cdot(P\cdot x)$$ which means you first project the vector, then you rotate it. This is incorrect.