Vector Spaces and Linear Operators: Problems with Solutions

Solution:
Since [tex]g[/tex] is linear, [tex]g(e_1+e_2)=g(e_1)+g(e_2)[/tex] and [tex]g(e_1+e_2+e_3)=g(e_1+e_2)+g(e_3)[/tex]. Hence
[tex]g(e_2)=-1-1=-2,\qquad g(e_3)=2-(-1)=3[/tex]
Using linearity once more,
[tex]g(e_1+2e_2+3e_3)=g(e_1)+2g(e_2)+3g(e_3)=1+2\times (-2)+3\times 3=6[/tex]

Solution:
The solution set of a homogeneous linear system in [tex]n[/tex] unknowns is a subspace of [tex]\mathbb{R}^n[/tex] of dimension [tex]n-r[/tex], where [tex]r[/tex] is the rank of the coefficient matrix. Here [tex]n=7[/tex]: the unknowns [tex]x_1, x_2, x_3, x_4, x_6[/tex] do not appear in the equations, so they are free. The two equations are proportional:
[tex]4x_5-10x_7=2(2x_5-5x_7),\qquad 6x_5-15x_7=3(2x_5-5x_7)[/tex]
so the system is equivalent to the single equation [tex]2x_5-5x_7=0[/tex], and [tex]r=1[/tex] (not [tex]2[/tex]). Hence
[tex]\dim U=7-1=6[/tex]
A basis of [tex]U[/tex]: [tex]e_1, e_2, e_3, e_4, e_6[/tex] and [tex](0, 0, 0, 0, 5, 0, 2)[/tex].

Solution:
Every subspace contains the zero vector [tex](0, 0)[/tex]. Substituting it:
[tex]U_1:\ 3\times 0+2\times 0=0\ne 1,\qquad U_4:\ 0\ne 3\times 0+2=2[/tex]
so [tex]U_1[/tex] and [tex]U_4[/tex] are not subspaces (they are lines that do not pass through the origin).
[tex]U_2[/tex] is the solution set of the homogeneous equation [tex]3x-2y=0[/tex]. If [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] belong to [tex]U_2[/tex], then for all [tex]\alpha,\beta\in\mathbb{R}[/tex]
[tex]3(\alpha x_1+\beta x_2)-2(\alpha y_1+\beta y_2)=\alpha(3x_1-2y_1)+\beta(3x_2-2y_2)=0[/tex]
so [tex]U_2[/tex] is closed under linear combinations and is a subspace. For [tex]U_3[/tex] simplify the equation:
[tex]4y+2=2x+2\iff x-2y=0[/tex]
This equation is again homogeneous, so by the same argument [tex]U_3[/tex] is a subspace (the line through the origin with direction vector [tex](2, 1)[/tex]). Therefore exactly [tex]U_2[/tex] and [tex]U_3[/tex] are subspaces.

Solution:
The Gram-Schmidt process gives
[tex]v_1=a_1,\quad v_2=a_2-\frac{\langle a_2,v_1\rangle}{\langle v_1,v_1\rangle}v_1,\quad v_3=a_3-\frac{\langle a_3,v_1\rangle}{\langle v_1,v_1\rangle}v_1-\frac{\langle a_3,v_2\rangle}{\langle v_2,v_2\rangle}v_2[/tex]
First [tex]v_1=(1, 0, 1)[/tex] and [tex]\langle v_1,v_1\rangle=2[/tex]. Since [tex]\langle a_2,v_1\rangle=-1+0-1=-2[/tex],
[tex]v_2=(-1, 1, -1)-\frac{-2}{2}(1, 0, 1)=(0, 1, 0),\qquad \langle v_2,v_2\rangle=1[/tex]
Next, [tex]\langle a_3,v_1\rangle=2+0+1=3[/tex] and [tex]\langle a_3,v_2\rangle=1[/tex], so (the projection onto [tex]v_2[/tex] must not be forgotten)
[tex]v_3=(2, 1, 1)-\frac{3}{2}(1, 0, 1)-(0, 1, 0)=\left(\frac{1}{2}, 0, -\frac{1}{2}\right)[/tex]
Check: [tex]\langle v_1,v_2\rangle=0[/tex], [tex]\langle v_1,v_3\rangle=\frac{1}{2}-\frac{1}{2}=0[/tex], [tex]\langle v_2,v_3\rangle=0[/tex].

Solution:
Step [tex]1[/tex]: [tex]0v=\vec{0}[/tex] for every [tex]v\in V[/tex]. Since [tex]0=0+0[/tex] in [tex]K[/tex], distributivity with respect to the addition of scalars gives
[tex]0v=(0+0)v=0v+0v[/tex]
Add the opposite vector [tex]-(0v)[/tex] to both sides and use associativity and the axioms [tex]w+(-w)=\vec{0}[/tex], [tex]w+\vec{0}=w[/tex]:
[tex]\vec{0}=0v+(-(0v))=(0v+0v)+(-(0v))=0v+\left(0v+(-(0v))\right)=0v+\vec{0}=0v[/tex]
Step [tex]2[/tex]: by the axiom [tex]1v=v[/tex], distributivity and Step [tex]1[/tex],
[tex]v+(-1)v=1v+(-1)v=\left(1+(-1)\right)v=0v=\vec{0}[/tex]
so [tex](-1)v[/tex] is an opposite vector of [tex]v[/tex].
Step [tex]3[/tex]: the opposite vector is unique. If [tex]v+w=\vec{0}[/tex] and [tex]v+w'=\vec{0}[/tex], then by the axioms (commutativity, associativity, zero vector)
[tex]w=w+\vec{0}=w+(v+w')=(w+v)+w'=\vec{0}+w'=w'+\vec{0}=w'[/tex]
Applying this to [tex]w=(-1)v[/tex] and [tex]w'=-v[/tex] gives [tex](-1)v=-v[/tex].