Solution. The quickest test to apply first is whether the set contains \(0\); a subspace must, since it is closed under multiplication by the scalar \(0\).

(a)

Yes. The zero vector satisfies \(0+0+0=0\). If \(u\) and \(v\) both satisfy the condition then so does \(u+v\), because \((u_1+v_1)+(u_2+v_2)+(u_3+v_3)=0+0=0\), and so does \(\lambda u\), because \(\lambda u_1+\lambda u_2+\lambda u_3=\lambda \cdot 0=0\).

(b)

No. The zero vector gives \(0+0+0=0\neq 1\), so \(0\) is not in the set. (Equivalently, the sum of two such vectors has \(\alpha _1+\alpha _2+\alpha _3=2\).) The pattern is general: a condition “\(\text {linear}=0\)” gives a subspace, “\(\text {linear}=c\)” with \(c\neq 0\) does not.

(c)

No. Again \(0\) fails, since \(|0|>0\) is false. The set is also not closed under scalar multiplication: \(0\cdot u\) leaves it for any \(u\).

(d)

Yes. The conditions say \(\alpha _1=\alpha _2=\alpha _3\), which is a pair of homogeneous linear equations, so the same verification as in (a) applies.

Solution. (a), (c) and (d) are subspaces; (b) and (e) are not.

For (a), (c) and (d) the defining condition is of the form \(L(f)=0\) where \(L\) is a linear operation on functions — evaluation at a point in (a), the difference of two evaluations in (c), and integration in (d). Linearity gives everything at once: \[L(f+g)=L(f)+L(g)=0+0=0,\qquad L(\lambda f)=\lambda L(f)=\lambda \cdot 0=0,\] and \(L(0)=0\), so the zero function belongs. For (d) explicitly, \(\int _0^1(f+g)=\int _0^1 f+\int _0^1 g=0\).

For (b) and (e) the condition is \(L(f)=1\), and the zero function fails it. Two such functions add to give \(L=2\), so closure fails as well.

Solution.

(a)

Yes. \(a=b\) is a homogeneous linear condition on the entries: the zero matrix satisfies it, and if \(a=b\) and \(a'=b'\) then \(a+a'=b+b'\) and \(\lambda a=\lambda b\).

(b)

No. The zero matrix has \(a+b=0\neq 1\).

(c)

No, and this is the interesting one, because the zero matrix does satisfy \(\det A=0\) and the set is closed under scalar multiplication (\(\det (\lambda A)=\lambda ^2\det A=0\)). What fails is closure under addition: \[A=\begin {pmatrix}1&0\\0&0\end {pmatrix},\qquad B=\begin {pmatrix}0&0\\0&1\end {pmatrix}\] both have determinant \(0\), but \(A+B=I\) has determinant \(1\). The moral is that passing the zero-vector test is necessary, not sufficient — the determinant is not a linear function of the matrix.

Solution.

(a)

Independent. If \((8,-2)=k(4,1)\) then \(k=2\) from the first coordinate and \(k=-2\) from the second — impossible. (Equivalently \(\det \begin {pmatrix}4&8\\1&-2\end {pmatrix}=-16\neq 0\).)

(b)

Dependent, with the relation \[2(1,3,0)-(2,6,0)=(0,0,0),\] since the second vector is exactly twice the first.

(c)

Independent. Form the matrix with these as rows and take its determinant: \[\begin {vmatrix}3&-1&2\\7&1&-1\\-1&4&0\end {vmatrix}=69\neq 0,\] so the three vectors have rank \(3\).

(d)

Dependent, necessarily so: four vectors in \(\mathbb {R}^3\) can never be independent, since by the exchange lemma an independent set cannot exceed the size of the spanning set \(\{e_1,e_2,e_3\}\). Solving for an explicit relation gives \[3(1,1,0)+(2,0,1)-2(3,2,1)+(1,1,1)=(0,0,0),\] which is checked coordinatewise: \(3+2-6+1=0\), \(3+0-4+1=0\), \(0+1-2+1=0\).

Solution. They are not. \(\dim P_2(\mathbb {R})=3\), with basis \(\{1,x,x^2\}\), so four vectors cannot be independent — again the exchange lemma settles it before any computation.

The relation is easy to spot once looked for: \[f_1+f_2=(1-2x)+(2x-x^2)=1-x^2=f_3,\] that is, \[f_1+f_2-f_3+0\cdot f_4=0.\] Note that \(f_4\) does not appear: a dependent set need not have every vector involved in the relation. In fact \(\{f_1,f_2,f_4\}\) is independent, so the four vectors span all of \(P_2(\mathbb {R})\) and have rank \(3\).

Solution. A vector \(v\) lies in \(U\) exactly when \(v=\alpha u_1+\beta u_2\) has a solution.

\((1,0,0)\)

The equations are \[\alpha +3\beta =1,\qquad 2\alpha -5\beta =0,\qquad 3\alpha +\beta =0.\] From the third, \(\beta =-3\alpha \); substituting into the second gives \(2\alpha +15\alpha =17\alpha =0\), so \(\alpha =0\) and then \(\beta =0\). But that contradicts the first equation, which demands \(\alpha +3\beta =1\). There is no solution, so \((1,0,0)\notin U\).

\((5,-23,-9)\)

Now the equations are \[\alpha +3\beta =5,\qquad 2\alpha -5\beta =-23,\qquad 3\alpha +\beta =-9.\] From the first, \(\alpha =5-3\beta \); substituting into the second, \[2(5-3\beta )-5\beta =-23\implies 10-11\beta =-23\implies \beta =3,\] so \(\alpha =5-9=-4\). The third equation must be checked, since it was not used: \(3(-4)+3=-9\). It holds, so \[(5,-23,-9)=-4(1,2,3)+3(3,-5,1),\] and indeed \((-4+9,\,-8-15,\,-12+3)=(5,-23,-9)\).

\(U\) is a plane through the origin in \(\mathbb {R}^3\); the first vector lies off that plane and the second on it.

Solution. Two subspaces are equal when each contains the other, so it is enough to express each pair of generators in terms of the other pair.

\(T\subseteq S\)

\[(0,1,2,3)=\tfrac {1}{2}\big [(1,2,5,6)-(1,0,1,0)\big ],\] since \((1,2,5,6)-(1,0,1,0)=(0,2,4,6)\). For the other generator, solve \(\alpha (1,0,1,0)+\beta (1,2,5,6)=(2,-1,0,-3)\): the second coordinate gives \(2\beta =-1\), so \(\beta =-\frac 12\), and the first gives \(\alpha =2-\beta =\frac {5}{2}\). The remaining two coordinates then check, \(\frac 52-\frac 52=0\) and \(0-3=-3\), so \[(2,-1,0,-3)=\tfrac {5}{2}(1,0,1,0)-\tfrac {1}{2}(1,2,5,6).\]

\(S\subseteq T\)

Similarly \((1,0,1,0)=\frac {1}{2}(0,1,2,3)+\frac {1}{2}(2,-1,0,-3)\) and \((1,2,5,6)=\frac {5}{2}(0,1,2,3)+\frac {1}{2}(2,-1,0,-3)\), both verified coordinatewise.

Since each spanning set lies in the other’s span, \(S=T\). Both have dimension \(2\): stacking all four vectors into a matrix gives rank \(2\), which confirms that nothing was gained by taking the two pairs together.