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

1 A line through the origin ★★★

The zero vector satisfies \(0=3\cdot 0\), so \(\mathbf{0}\in W\). If \((x_1,3x_1)\) and \((x_2,3x_2)\) are in \(W\), their sum is \((x_1+x_2,\,3(x_1+x_2))\), which is in \(W\). For a scalar \(c\), \(c(x,3x)=(cx,3cx)\in W\). All three conditions hold, so \(W\) is a subspace. In fact \(W=\operatorname{span}\{(1,3)\}\), a line of dimension 1.

2 A half-plane ★★★

The vector \((1,0)\) belongs to \(H\), but \((-1)\cdot(1,0)=(-1,0)\) has a negative first coordinate, so it is not in \(H\). Closure under scalar multiplication fails, hence \(H\) is not a subspace (even though it contains \(\mathbf{0}\) and is closed under addition).

3 Two pairs of vectors ★★★

(a) Since \((2,4)=2(1,2)\), the pair is dependent.

(b) Neither is a multiple of the other. Equivalently, \(\det\begin{pmatrix}1&3\\2&1\end{pmatrix}=1-6=-5\neq0\). The pair is independent, hence a basis of \(\mathbb{R}^2\).

4 Counting dimensions ★★★

(a) 4. (b) 3, with basis \(1,x,x^2\). (c) \(2\cdot3=6\), one basis matrix per entry. (d) 3, with the three matrices that have a single 1 on the diagonal.

5 Is it linear? ★★★

(a) \(T(x,y)=\begin{pmatrix}2&1\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\) is a matrix product, so \(T\) is linear. \(T(1,2)=(2+2,\,1-2)=(4,-1)\).

(b) \(S(0,0)=(1,0)\neq(0,0)\), but every linear map sends \(\mathbf{0}\) to \(\mathbf{0}\). So \(S\) is not linear.

6 Rank-nullity quick check ★★★

The matrix has \(n=6\) columns, so nullity \(=6-3=3\). The columns have 4 entries, so the column space is a 3-dimensional subspace of \(\mathbb{R}^4\).

7 Coordinates in a basis ★★★

Solve \(a(1,0)+b(1,1)=(4,3)\): \(a+b=4\) and \(b=3\), so \(a=1\). Thus \([v]_B=(1,3)\). Check: \((1,0)+3(1,1)=(4,3)\).

8 Basis of a solution set ★★★

From \(x-z=0\) we get \(z=x\); then \(y=-x-z=-2x\). So \((x,y,z)=x(1,-2,1)\). The vector \((1,-2,1)\) is nonzero and spans \(W\), so it is a basis and \(\dim W=1\). Check: \(1-2+1=0\) and \(1-1=0\).

9 An affine plane ★★★

First way: \((0,0,0)\) gives \(0\neq 1\), so the zero vector is missing. Second way: \((1,0,0)\) and \((0,1,0)\) are in the set, but their sum \((1,1,0)\) has coordinate sum 2, so closure under addition fails.

10 Polynomial basis ★★★

Suppose \(a p_1+b p_2+c p_3=0\). Comparing coefficients of \(1\), \(x\), \(x^2\) gives \(a+c=0\), \(a+b=0\), \(b+c=0\). Adding the three equations gives \(a+b+c=0\), so \(b=0\), \(c=0\), \(a=0\). The three polynomials are independent. Since \(\dim P_2=3\), three independent vectors form a basis.

11 In the span or not? ★★★

We have \(au+cw=(a,\,a+c,\,a+2c)\).

(a) \(a=3\), \(a+c=5\) gives \(c=2\), and then \(a+2c=7\) matches. So \((3,5,7)=3u+2w\): yes.

(b) The same first two equations force \(a=3\), \(c=2\), but then \(a+2c=7\neq8\). The system is inconsistent: no.

12 Null and column space ★★★

Subtracting rows 1 and 2 from row 3 gives a zero row, so the echelon form is \(\begin{pmatrix}1&0&2\\0&1&-1\\0&0&0\end{pmatrix}\). Pivots are in columns 1 and 2, so the rank is 2 and \(\mathrm{Col}(B)\) has basis \((1,0,1)\), \((0,1,1)\). The free variable is \(x_3=t\); then \(x_1=-2t\) and \(x_2=t\), so \(\mathrm{Nul}(B)\) has basis \((-2,1,1)\). Check: \(2+1=3\) columns.

13 A rotation ★★★

\(T(e_1)=(0,1)\) and \(T(e_2)=(-1,0)\), so the matrix is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\). Then \(T(3,5)=(-5,3)\) and \(T(-5,3)=(-3,-5)\). Two quarter turns make a half turn, and indeed \((-3,-5)=-(3,5)\).

14 Kernel and range ★★★

\(T(x,y,z)=\mathbf{0}\) means \(x=y\) and \(y=z\), so \(\ker T=\{t(1,1,1)\}\) with basis \((1,1,1)\) and nullity 1. By rank-nullity the rank is \(3-1=2=\dim\mathbb{R}^2\), so \(T\) is onto. It is not one-to-one because the kernel is not \(\{\mathbf{0}\}\).

15 Trace-zero matrices ★★★

The zero matrix has trace 0. The trace is additive and \(\operatorname{tr}(cM)=c\operatorname{tr}(M)\), so \(W\) is closed under both operations: \(W\) is a subspace. A matrix \(\begin{pmatrix}a&b\\c&-a\end{pmatrix}\) in \(W\) equals \(a\begin{pmatrix}1&0\\0&-1\end{pmatrix}+b\begin{pmatrix}0&1\\0&0\end{pmatrix}+c\begin{pmatrix}0&0\\1&0\end{pmatrix}\). These three matrices span \(W\) and are independent (compare entries), so \(\dim W=3\), one less than \(\dim M_{2\times2}=4\).

16 The derivative as a linear map ★★★

Linearity: \((p+q)'=p'+q'\) and \((cp)'=cp'\). The kernel consists of polynomials with zero derivative, the constants: \(\ker D=\operatorname{span}\{1\}\), of dimension 1. Since \(\dim P_3=4\), the rank is \(4-1=3=\dim P_2\), so the range is all of \(P_2\) and \(D\) is onto. Directly: \(ax^2+bx+c=D\!\left(\tfrac{a}{3}x^3+\tfrac{b}{2}x^2+cx\right)\).

17 Coordinates in a new basis ★★★

\(P=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\), \(\det P=-2\), so \(P^{-1}=\dfrac{1}{-2}\begin{pmatrix}-1&-1\\-1&1\end{pmatrix}=\begin{pmatrix}\tfrac12&\tfrac12\\\tfrac12&-\tfrac12\end{pmatrix}\). Then \([v]_B=P^{-1}(7,3)=(5,2)\). Check: \(5(1,1)+2(1,-1)=(7,3)\).

18 Matrix in a basis ★★★

\(T(1,1)=(3,5)=4(1,1)-1(1,-1)\), so the first column of \([T]_B\) is \((4,-1)\). \(T(1,-1)=(-1,1)=0(1,1)-1(1,-1)\), so the second column is \((0,-1)\). Therefore \([T]_B=\begin{pmatrix}4&0\\-1&-1\end{pmatrix}\). Trace: \(4-1=3=1+2\). Determinant: \(-4-0=-4=1\cdot2-2\cdot3\). Both match.

19 Preserving independence ★★★

Suppose \(a(v_1+v_2)+b(v_2+v_3)+c(v_1+v_3)=\mathbf{0}\). Regrouping gives \((a+c)v_1+(a+b)v_2+(b+c)v_3=\mathbf{0}\). By independence, \(a+c=0\), \(a+b=0\), \(b+c=0\). Adding gives \(2(a+b+c)=0\), so \(a+b+c=0\), and subtracting each equation yields \(b=0\), \(c=0\), \(a=0\). The new family is independent.

20 A parameter and dependence ★★★

The vectors are dependent exactly when the determinant is zero: \(\begin{vmatrix}1&1&0\\1&2&1\\2&a&1\end{vmatrix}=1(2-a)-1(1-2)+0=3-a\). So \(a=3\). Then \((2,3,1)=(1,1,0)+(1,2,1)\), the relation \(v_1+v_2-v_3=\mathbf{0}\).

21 Recovering a map from a basis ★★★

Since \(e_1=\tfrac12[(1,1)+(1,-1)]\), \(T(e_1)=\tfrac12[(3,1)+(1,5)]=(2,3)\). Since \(e_2=\tfrac12[(1,1)-(1,-1)]\), \(T(e_2)=\tfrac12[(3,1)-(1,5)]=(1,-2)\). So \(A=\begin{pmatrix}2&1\\3&-2\end{pmatrix}\). Then \(T(4,-2)=(8-2,\,12+4)=(6,16)\). Check: \(A(1,1)=(3,1)\) and \(A(1,-1)=(1,5)\).