**Problem 1:** Let T: \mathbb{R}^2\to\mathbb{R}^2 be the transformation that rotates each point in \mathbb{R}^2 about the origin through an angle \phi, with counterclockwise rotation for a positive angle. Assuming such a transformation is linear, find the standard matrix A of this transformation.

**Solution:** It is enough to find T(e_1), T(e_2) where e_1 and e_2 are the columns of the identity matrix. According to the description of T, we have T(e_1)=\begin{pmatrix}\cos\phi \\ \sin\phi\end{pmatrix}, T(e_2)=\begin{pmatrix}-\sin\phi \\ \cos\phi\end{pmatrix}. Therefore the standard matrix A=\begin{pmatrix}T(e_1) & T(e_2)\end{pmatrix}=\begin{pmatrix}\cos\phi & -\sin\phi \\ \sin\phi & \cos\phi\end{pmatrix}.

**Problem 2:** Let T be the linear transformation whose standard matrix is A=\begin{pmatrix}1 & -4 & 8 & 1\\ 0 & 2 & -1 & 3 \\ 0 & 0 & 0 & 5\end{pmatrix} Does T maps \mathbb{R}^4 onto \mathbb{R}^3? Is T a one-to-one mapping?

**Solution:** Because the standard matrix is already an echelon form with each row containing a pivot position, its columns span \mathbb{R}^3. In other words, T maps \mathbb{R}^4 onto \mathbb{R}^3. On the other hand, since not every columns is a pivot column, Tx=0 has a non-trivial solution. In other words, T is not one-to-one.

**Problem 3:** Let T(x_1, x_2)=(3x_1+x_2,5x_1+7x_2, x_1+3x_2). Show that T is a one-to-one linear transformation. Does T map \mathbb{R}^2 onto \mathbb{R}^3?

**Solution:** First we shall find the standard matrix of T. Let x_1=1 and x_2=0. We obtain T(e_1)=T \begin{pmatrix}1\\0\end{pmatrix}= \begin{pmatrix}3 \\ 5 \\ 1\end{pmatrix}. Let x_1=0 and x_2=1. We obtain T(e_2)=T \begin{pmatrix}0 \\ 1\end{pmatrix}= \begin{pmatrix}1 \\ 7 \\ 3\end{pmatrix}. Therefore the standard matrix of T is A=\begin{pmatrix}3 & 1 \\ 5 & 7 \\ 1 & 3\end{pmatrix} Since the echelon form of A contains two pivot columns, T is one-to-one. On the other hand, since A has three rows, not every row has a pivot position. Therefore T does not map \mathbb{R}^2 onto \mathbb{R}^3. In general, a linear transformation never maps \mathbb{R}^m onto \mathbb{R}^n if m < n.