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The matrix is:

cos(pi/3) 0

sin(pi/3) 0

I have no clue what to put. No solution?

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The preimage is a set, not necessarily just a single vector. – Jonathan Oct 3 '12 at 3:30
up vote 1 down vote accepted

Assuming the vector you're given is $(y_1,y_2)^\intercal$, try writing out the multiplication $$\left( \begin{array}{cc} \cos{\pi/3} & 0 \\ \sin{\pi/3} & 0 \end{array} \right)\left( \begin{array}{c} x_1 \\ x_2 \end{array} \right)=\left( \begin{array}{c} y_1 \\ y_2 \end{array} \right)$$ and describe the set of vectors $(x_1,x_2)^\intercal$ which satisfy that criteria.

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I found x1 and x2. Are you saying I now need to find them under T? – Grace C Oct 3 '12 at 4:45
Or would x1 and x2 alone be a preimage? – Grace C Oct 3 '12 at 4:47
When you multiply out the LHS you get two equations, $\cos{\pi/3}x_1=y_1$ and $\sin{\pi/3}x_1=y_2$. These two equations only depend on $x_1$, so assuming a solution exists, the preimage will be the set of matrices with first entry as the value of $x_1$ given by those equations and any second entry you want (because it doesn't matter what $x_2$ you pick, it just gets multiplied by 0 anyway). – Alexander Gruber Oct 3 '12 at 21:06

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