Skip to main content

Recitation 4A

Problem 1: Let b=[100]\mathbf{b}=\begin{bmatrix}-1 \\ 0 \\ 0\end{bmatrix} and let AA be the matrix [135501352444]\begin{bmatrix}1 & -3 & 5 & -5\\0 & 1 & -3 & 5\\2 & -4 & 4 & -4\end{bmatrix}. Is b\mathbf{b} in the range of the linear transformation xAx\mathbf{x}\mapsto A\mathbf{x}? Why or why not?

Solution: Because `b\mathbf{b} is in the range of the linear transformation xAx\mathbf{x}\mapsto A\mathbf{x}‘ is equivalent to say `b\mathbf{b} is a linear combination of the columns of AA‘. Therefore it is enough to check if the augmented matrix [135510135024440]\begin{bmatrix}1 & -3 & 5 & -5 & -1\\0 & 1 & -3 & 5 & 0\\2 & -4 & 4 & -4 & 0\end{bmatrix} is consistent. By row reductions, its echelon form is

[135510135000042]\begin{bmatrix}1 & -3 & 5 & -5 & -1\\0 & 1 & -3 & 5 & 0\\0 & 0 & 0 & -4 & 2\end{bmatrix} which is consistent. Therefore b\mathbf{b} is in the range of the linear transformation.

Problem 2: Let T:R2R2T: \mathbb{R}^2\to\mathbb{R}^2 be the transformation that first performs a horizontal shear that maps e2e_2 into e2e1/2e_2-e_1/2 and then reflects the result through the x2x_2-axis. Assuming TT is linear, find its standard matrix.

Solution: It is enough to find Te1Te_1 and Te2Te_2. Noticing that TT transforms e1e_1 to e1e_1 via the horizontal shear first and then transforms e1e_1 to e1-e_1 via the reflection, Te1=e1=[10]Te_1 = -e_1=\begin{bmatrix}-1\\ 0\end{bmatrix}. On the other hand, TT transforms e2e_2 to e2e1/2e_2 - e_1 / 2 via the horizontal shear first and then transforms e2e1/2e_2-e_1/2 to e2+e1/2e_2+e_1/2 via the reflection. Therefore Te2=e2+e1/2=[1/21]Te_2=e_2+e_1/2=\begin{bmatrix}1/2\\1\end{bmatrix}. Hence the standard matrix of TT is [11/201].\begin{bmatrix}-1 & 1/2\\0 & 1\end{bmatrix}.