Vector Spaces and Linear Transformations: math test solutions, College – download the PDF

1 A subspace of R^3 / 4 pts

1. \(\mathbf{0}\in W\) (0.5 pt). If \(u,v\in W\), adding the equations shows \(u+v\in W\) (1 pt); multiplying by \(c\) shows \(cu\in W\) (0.5 pt).
2. \(y=2x+z\), so \((x,y,z)=x(1,2,0)+z(0,1,1)\) (1 pt). The two vectors are independent and span \(W\); the dimension is 2 (1 pt).

2 Dependence / 3 pts

Form the matrix with these vectors as rows or columns (1 pt). Its determinant is \(1(3+1)-0+2(-2-0)=0\) (1 pt). They are dependent, and \((0,-1,3)=2(1,0,2)-(2,1,1)\) (1 pt).

3 Null space and column space / 4 pts

Reduced form \(\begin{pmatrix}1&-1&0&-2\\0&0&1&1\\0&0&0&0\end{pmatrix}\) (1 pt). With \(x_2=s\), \(x_4=t\): \(x_1=s+2t\), \(x_3=-t\), so \(\mathrm{Nul}(M)\) has basis \((1,1,0,0)\), \((2,0,-1,1)\) (1 pt). Pivots in columns 1 and 3, so \(\mathrm{Col}(M)\) has basis \((1,2,1)\), \((2,5,3)\) (1 pt). Rank 2 + nullity 2 = 4 columns (1 pt).

4 A linear map and its kernel / 3 pts

\(A=\begin{pmatrix}1&1&-1\\0&2&1\end{pmatrix}\) (1 pt). \(2y+z=0\) gives \(z=-2y\), and \(x=z-y=-3y\); a basis of \(\ker T\) is \((-3,1,-2)\) (1 pt). Rank \(=3-1=2\), so \(T\) is onto (1 pt).

5 Change of basis / 4 pts

1. \(P=\begin{pmatrix}3&1\\1&1\end{pmatrix}\), \(\det P=2\), \(P^{-1}=\tfrac12\begin{pmatrix}1&-1\\-1&3\end{pmatrix}\) (1 pt).
2. \([v]_B=\tfrac12(7-5,\,-7+15)=(1,4)\) (1 pt).
3. \(A=\begin{pmatrix}2&0\\0&1\end{pmatrix}\), \(AP=\begin{pmatrix}6&2\\1&1\end{pmatrix}\) (1 pt), and \([T]_B=P^{-1}AP=\begin{pmatrix}\tfrac52&\tfrac12\\-\tfrac32&\tfrac12\end{pmatrix}\) (1 pt).

6 True or false? / 2 pts

(a) False (1 pt): the \(x\)-axis and \(y\)-axis are subspaces, but \((1,0)+(0,1)=(1,1)\) lies in neither. (b) True (1 pt): \(\dim\mathbb{R}^3=3\), and a family of more than 3 vectors is dependent.