*Question 1* (a) Note that $(1,0,0)^{T} \in \mathbb{R}^3$ but $(1,0,0)^{T}\not\in W$ since $1+2\times{0}-3\times {0}=1\neq 0$ hence $W\neq \mathbb{R}^{3}$. $W$ is a subspace of $\mathbb{R}^3$ so $\dim W \leq \dim \mathbb{R}^{3}=3$ with equality iff $W=\mathbb{R}^{3}$. Since $W\neq \mathbb{R}^{3}$, $\dim W < 3$. (b) For all $\underline{v}=(x,y,z)\in W$, $x=3z-3y$ so $\underline{v}=\begin{pmatrix} 3z -2y \\ y \\ z\end{pmatrix} = z \begin{pmatrix} 3 \\ 0 \\ 1\end{pmatrix} +y\begin{pmatrix} -2 \\ 1 \\ 0\end{pmatrix}$. Also $(3z-2y,y,z)^{T}=\underline{0}$ iff $y=z=0$ hence $(3,0,1)^{T}, (-2,1,0)^{T}$ are linearly independent and span $W$. *Question 2* (a) By linearity, $\begin{align} \varphi(2x^{3}-3x+2)&=2\varphi(x^{3})-3\varphi(x)+2\varphi(1)\\&= 2\begin{pmatrix}-1\\3\end{pmatrix}-3\begin{pmatrix}-1\\1\end{pmatrix}+2\begin{pmatrix}1\\0\end{pmatrix}\\&=\begin{pmatrix}7\\3\end{pmatrix}. \end{align}$ (b) If $f(x)=1$ then $f'(x)=0$ so $\psi(1)=(1,0)$. If $f(x)=x$ then $f'(x)=1$ so $\psi(x)=(-1,1)$. If $f(x)=x^{2}$ then $f'(x)=2x$ so $\psi(x^{2}) =(1,-2)$. If $f(x)=x^{3}$ then $f'(x)=3x^{2}$ so $\psi(x^{3})=(-1,3)$. Since $\psi$ and $\varphi$ are equal on the basis of $V$, $\psi=\varphi$ by Prop 5.17. (c) Clearly $\varphi$ is surjective (for example $(1,0), (-1,1)$ form a basis of $\mathbb{R}^{2}$) so $\text{Im } \varphi=\mathbb{R}^{2}$ (d) Using Rank-Nullity Formula, $\dim \ker \varphi=\dim V-\dim \text{Im }\varphi=4-2=2$. *Question 3* (a) Take $v\in V$ then $\exists!\lambda_{i}\in\mathbb{R}$ such that $v=\lambda_{1} +\lambda_{2}x+\lambda_{3}x^{2}$. Then by linearity, $\varphi(\lambda_{1}+\lambda_{2}(x-a)+\lambda_{3}(x-a)^{2})=\lambda_{1}+\lambda_{2}x+\lambda x^{2}=v$. Hence $\varphi$ is surjective. Take $w\in V$ then $\exists!\mu_{i}\in\mathbb{R}$ such that $w=\mu_{1}+\mu_{2}x+\mu_{3}x^{2}$. Suppose $\varphi(v)=\varphi(w)$, by linearity $\begin{align} 0&=\varphi(v-w)\\&=\varphi(\lambda_{1}-\mu_{1}+(\lambda_{2}-\mu_{2})x+(\lambda_{3}-\mu_{3})x^{2})\\&=\lambda_{1}-\mu_{1}+(\lambda_{2}+\mu_{2})x +(\lambda_{3}-\mu_{3})(x+a)^2 \end{align}$hence $\lambda_{i}=\mu_{i}$ for $i=1,2,3$ so $v=w$. So $\varphi$ is injective and therefore a linear isomorphism. (b) The matrix of $\pi$ with respect to the basis $1,x,x^{2}$ is $\begin{pmatrix} 1 & a & a^{2} \\ 0 & 1 & 2a \\ 0 & 0 & 1 \end{pmatrix}$ *Question 4* (a) Suppose $\lambda_{1}\cos (x) +\lambda_{2}\cos(2x)\equiv0$. Setting $x=\frac{\pi}{2}$ gives $\lambda_{2}=0$ and $x=\pi$ gives $\lambda_{1}=0$. So $\cos(x),\cos(2x)$ are linearly independent hence they form a basis for $W$ and $\dim W=2$. (b) Consider the linear map $\begin{align} \varphi:W&\to \mathbb{R} \\ f &\mapsto f(10) \end{align} $Note that $U=\ker \varphi$. Now $\varphi$ is clearly surjective (for example $\cos(10)\neq 0$) so $\dim \text{Im }\varphi =1$. Therefore $\dim U = \dim W - \dim \text{Im }\varphi = 2-1 = 1. $ (c) Take $f,g\in U_{2}$. Then $(f+g)(10)=1+1=2$. So $f+g\not \in U_{2}$ which shows that $U_{2}$ is not a subspace. *Question 5* (a) Suppose $V,W$ are vector spaces where $V$ is finite dimensional. If $\varphi:V\to W$ is a linear map then $\dim V =\dim \ker \varphi + \dim \text{Im } \varphi$ (b) If $a,b,c,d=0$ then $\dim \text{Colspan }A=1$. In this case, $\dim \text{Im }L_{A}=1$ and $\dim \ker L_{A}=3$. (c) If $a=1$ and $b,c,d=0$ then $\dim\text{Colspan }A=2$. In this case $\dim \text{Im }L_{A}=2$ and $\dim \ker L_{A}=2$. (d) If $a=d=1$ and $b=c=0$ then $\dim \text{Colspan } A=3$. In this case $\dim \text{Im }L_{A}=3$ and $\dim \ker L_{A}=1$.