Vector Spaces and Linear Operators: Difficult Problems with Solutions

Solution:
The vectors [tex]f_1=2e_1+e_2[/tex] and [tex]f_2=e_1-e_2[/tex] form a basis of [tex]V[/tex], because their coordinate determinant is nonzero:
[tex]\begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix}=-3\ne 0[/tex]
hence the operator is uniquely determined by [tex]p(f_1)[/tex] and [tex]p(f_2)[/tex]. Write [tex]a=\alpha f_1+\beta f_2[/tex] and compare the coefficients of [tex]e_1[/tex] and [tex]e_2[/tex]:
[tex]\begin{cases} 2\alpha+\beta=2 \\ \alpha-\beta=-1 \end{cases}\Rightarrow \alpha=\frac{1}{3},\quad \beta=\frac{4}{3}[/tex]
By linearity
[tex]p(a)=\frac{1}{3}p(f_1)+\frac{4}{3}p(f_2)=\frac{1}{3}(e_1+3e_2)+\frac{4}{3}(2e_1+3e_2)=3e_1+5e_2[/tex]
Check with the matrix of [tex]p[/tex]: from [tex]e_1=\frac{f_1+f_2}{3}[/tex] and [tex]e_2=\frac{f_1-2f_2}{3}[/tex] we get [tex]p(e_1)=e_1+2e_2[/tex], [tex]p(e_2)=-e_1-e_2[/tex], and
[tex]\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \end{bmatrix}=\begin{bmatrix} 3 \\ 5 \end{bmatrix}[/tex]

Solution:
The vectors [tex]g_1, g_2[/tex] form a basis, since [tex]\begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix}=1\ne 0[/tex]. The columns of the matrix are the coordinates of [tex]f(g_1)[/tex] and [tex]f(g_2)[/tex] with respect to [tex]g_1, g_2[/tex]. We have
[tex]f(g_1)=f(2, -1)=(6-5, \ 8+7)=(1, 15),\qquad f(g_2)=f(-1, 1)=(-3+5, \ -4-7)=(2, -11)[/tex]
From [tex]\alpha g_1+\beta g_2=(1, 15)[/tex]:
[tex]\begin{cases} 2\alpha-\beta=1 \\ -\alpha+\beta=15 \end{cases}\Rightarrow \alpha=16,\quad \beta=31[/tex]
From [tex]\alpha g_1+\beta g_2=(2, -11)[/tex]:
[tex]\begin{cases} 2\alpha-\beta=2 \\ -\alpha+\beta=-11 \end{cases}\Rightarrow \alpha=-9,\quad \beta=-20[/tex]
Hence the matrix is
[tex]\begin{bmatrix} 16 & -9 \\ 31 & -20 \end{bmatrix}[/tex]
Equivalently, it equals [tex]G^{-1}AG[/tex], where [tex]A=\begin{bmatrix} 3 & 5 \\ 4 & -7 \end{bmatrix}[/tex] is the matrix of [tex]f[/tex] in the standard basis and [tex]G=\begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}[/tex] is the transition matrix.

Solution:
The vectors [tex]a_1,a_2,a_3[/tex] are linearly dependent; checking coordinate by coordinate:
[tex]2a_1-a_2+3a_3=(6-12+6, 2+7-9, 10-4-6, 8-11+3)=(0, 0, 0, 0)[/tex]
Hence, if [tex]v=\alpha a_1+\beta a_2+\gamma a_3[/tex], then adding [tex]t(2a_1-a_2+3a_3)=\vec{0}[/tex] gives another representation for every [tex]t\ne 0[/tex]. So [tex]v[/tex] has either no representation or infinitely many, and it suffices that [tex]v[/tex] be a linear combination at all. Since [tex]a_2=2a_1+3a_3[/tex], this means [tex]v=\alpha a_1+\gamma a_3[/tex] for some numbers [tex]\alpha,\gamma[/tex]. The first two coordinates give
[tex]\begin{cases}3\alpha+2\gamma=8\\ \alpha-3\gamma=-1\end{cases}\Rightarrow \alpha=2,\ \gamma=1[/tex]
The third and fourth coordinates must then match as well:
[tex]5\times 2-2\times 1=8=x+6,\quad 4\times 2+1\times 1=9=x+7\Rightarrow x=2[/tex]
For [tex]x\ne 2[/tex] the vector [tex]v[/tex] is not a linear combination of [tex]a_1,a_2,a_3[/tex] at all, while for [tex]x=2[/tex] it has infinitely many representations, for example
[tex]v=2a_1+0\times a_2+a_3=0\times a_1+a_2-2a_3[/tex]

Solution:
Let [tex]E[/tex] be the identity matrix of order [tex]3[/tex] and
[tex]P=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix},\qquad Q=\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}[/tex]
Every matrix of [tex]W[/tex] equals [tex]a_1E+a_2P+a_3Q[/tex], so [tex]W[/tex] is the linear span of [tex]E,P,Q[/tex], a subspace of [tex]M_3(\mathbb{R})[/tex]. The matrices [tex]E,P,Q[/tex] have their ones in different positions, so they are linearly independent. A symmetric matrix is determined by its three diagonal entries and its three entries above the diagonal. Hence
[tex]\dim W=3,\qquad \dim U=6[/tex]
Intersection: denote the entries of a matrix by [tex]m_{ij}[/tex]. A matrix of [tex]W[/tex] is symmetric exactly when
[tex]m_{12}=m_{21},\ m_{13}=m_{31},\ m_{23}=m_{32}\iff a_2=a_3,\ a_3=a_2,\ a_2=a_3\iff a_2=a_3[/tex]
Hence [tex]U\cap W[/tex] consists of the matrices [tex]a_1E+a_2(P+Q)[/tex]; since [tex]E[/tex] and [tex]P+Q[/tex] are linearly independent, [tex]\dim(U\cap W)=2[/tex]. By the Grassmann formula
[tex]\dim(U+W)=\dim U+\dim W-\dim(U\cap W)=6+3-2=7[/tex]

Solution:
The formula indeed defines an inner product. It is linear in each argument, and after expanding one sees that it is symmetric in [tex]P[/tex] and [tex]Q[/tex]:
[tex](P,Q)=4a_0b_0+a_1b_1+a_2b_1+a_1b_2+2a_2b_2[/tex]
Moreover, for [tex]P\ne 0[/tex]
[tex](P,P)=4a_0^2+a_1^2+2a_1a_2+2a_2^2=4a_0^2+(a_1+a_2)^2+a_2^2\gt 0[/tex]
For [tex]P_1[/tex] we have [tex]a_2=1,\ a_1=1,\ a_0=0[/tex], and for [tex]Q_1[/tex] we have [tex]b_2=-1,\ b_1=1,\ b_0=-1[/tex]. Hence
[tex](P_1,Q_1)=4\times 0\times (-1)+(1+1)\times 1+(1+2)\times (-1)=-1[/tex]
[tex](P_1,P_1)=4\times 0^2+(1+1)^2+1^2=5,\qquad (Q_1,Q_1)=4\times (-1)^2+(1-1)^2+(-1)^2=5[/tex]
Therefore the cosine of the angle [tex]\varphi[/tex] between them is
[tex]\cos\varphi=\frac{(P_1,Q_1)}{\sqrt{(P_1,P_1)}\times \sqrt{(Q_1,Q_1)}}=\frac{-1}{\sqrt{5}\times \sqrt{5}}=-\frac{1}{5}[/tex]
(With the usual dot product of the coefficient vectors one would wrongly get [tex]0[/tex].)

Solution:
The map [tex]f[/tex] sends [tex]\mathbb{R}_3[x][/tex] into itself (the degree does not increase) and is linear:
[tex](\alpha p+\beta q)(x+1)-(\alpha p+\beta q)(x)=\alpha\left(p(x+1)-p(x)\right)+\beta\left(q(x+1)-q(x)\right)[/tex]
Images of the basis [tex]1,x,x^2,x^3[/tex]:
[tex]f(1)=0,\quad f(x)=1,\quad f(x^2)=2x+1,\quad f(x^3)=3x^2+3x+1[/tex]
The image is spanned by these images, i.e. by the polynomials [tex]1[/tex], [tex]2x+1[/tex], [tex]3x^2+3x+1[/tex]. They have different degrees [tex]0[/tex], [tex]1[/tex] and [tex]2[/tex], so they are linearly independent and form a basis of [tex]\operatorname{Im} f[/tex]:
[tex]\operatorname{rank} f=\dim\operatorname{Im} f=3\qquad\left(\operatorname{Im} f=\mathbb{R}_2[x]\right)[/tex]
Check with the rank-nullity theorem: if [tex]p\in\ker f[/tex], then [tex]p(x+1)=p(x)[/tex], so [tex]p(k)=p(0)[/tex] for all integers [tex]k\ge 0[/tex]. Then the polynomial [tex]p(x)-p(0)[/tex] has infinitely many roots, so it is the zero polynomial and [tex]p[/tex] is constant. Thus
[tex]\dim\ker f=1,\qquad \dim\operatorname{Im} f=\dim\mathbb{R}_3[x]-\dim\ker f=4-1=3[/tex]