Subspaces, Span and Linear Independence: Practice with Solutions

The successful sums only check selected cases. Closure must hold for every pair of members and every real scalar.

The zero vector satisfies both equations. Now take arbitrary $a,b\in W$ and $\alpha,\beta\in\mathbb R$, and put $c=\alpha a+\beta b$. Since $a,b\in W$,

$$\begin{aligned}c_2&=\alpha(a_1+a_3)\\&\quad+\beta(b_1+b_3)\\&=c_1+c_3,\end{aligned}$$

$$\begin{aligned}c_4&=\alpha(2a_1-a_3)\\&\quad+\beta(2b_1-b_3)\\&=2c_1-c_3.\end{aligned}$$

Both required equations hold, so $c\in W$. Together with the zero check, this proves that $W$ is a subspace.

Common error: assuming that $c\in W$ before checking its equations. Membership is the conclusion you must establish.

If you got stuck: revisit the worked start. Replace its single parameter condition with the two conditions defining $W$, then explain where each was used.

The zero vector is in $W$ and gives $0^2+0^2=0\leq1$, so it is in $D$.

For $e=(1,1,0,2)$, the equations of $W$ give $1=1+0$ and $2=2(1)-0$. Also $1^2+0^2=1$, so $e\in D$.

However, $2e=(2,2,0,4)$ gives $2^2+0^2=4>1$. Thus $2e\notin D$, even though $2e\in W$. Closure under scalar multiplication fails. Since $e+e=2e$, addition also fails: both summands belong to $D$, but their sum does not. Therefore $D$ is not a subspace.

Common error: stopping after the zero check, or choosing a starting vector that does not belong to $D$. The extra restriction must be checked as well as the equations of $W$.

If you got stuck: return to the counterexample checklist and write three lines: starting membership, operation, failed membership. Only one failed closure condition was needed; this example supplies two.

Every linear combination has the form

$$\begin{gathered}su+rv\\=(s,s+r,r,2s-r).\end{gathered}$$

Its second coordinate is the first plus the third, and its fourth is twice the first minus the third. Thus every combination is in $W$: $\operatorname{span}\{u,v\}\subseteq W$.

Conversely, if $x\in W$, its defining equations give

$$x=\begin{pmatrix}x_1\\x_1+x_3\\x_3\\2x_1-x_3\end{pmatrix}.$$

$$x=x_1u+x_3v.$$

Every member of $W$ is therefore in the span. The two inclusions establish equality.

For $p$, choose coefficients 2 and $-1$:

$$\begin{aligned}2u-v&=(2,1,-1,5)\\&=p.\end{aligned}$$

The vector $q$ satisfies its second-coordinate condition, but its fourth coordinate would have to be $2(2)-(-1)=5$, not 4. Hence $q\notin W$ and cannot be a combination of $u,v$.

Common error: showing that two generators belong to $W$ and stopping. Because $W$ is a subspace, that proves their span is contained in $W$; it does not prove that the span reaches every vector in $W$.

If you got stuck: review linear combinations and span. For a geometric three-coordinate example, use when three vectors span only a plane.

The relation

$$u-2v-w=\mathbf0$$

has coefficients $(1,-2,-1)$, which are not all zero. The triple is dependent.

The pairs are nevertheless independent:

• For $au+bv=\mathbf0$, coordinates 1 and 3 give $a=0$ and $b=0$.
• For $au+bw=\mathbf0$, those coordinates give $a+b=0$ and $-2b=0$, hence $a=b=0$.
• For $av+bw=\mathbf0$, they give $b=0$ and $a-2b=0$, again forcing both coefficients to zero.

Pairwise independence rules out a dependence involving only one pair. It does not rule out a relation involving all three vectors. The pair $(u,v)$ is a basis of $W$, by its independence here and its spanning property in question 3. Adding $w$ leaves the span unchanged and makes the list dependent.

Common error: replacing “the only zero combination is the trivial one” with “the vectors look different.” Also, a nontrivial relation means that at least one coefficient is nonzero; it does not require every coefficient to be nonzero.

If you got stuck: revisit the definition of independence, then the two-vector check in the fundamentals lesson. Write the zero-combination equation before judging the list.

Expanding gives

$$\begin{aligned}ca_t+db_t&=(c+td)u\\&\quad+(tc+d)v.\end{aligned}$$

Case $t\notin\{-1,1\}$. A zero combination requires $c+td=0$ and $tc+d=0$, since $u,v$ are independent. Subtraction gives $(1-t^2)d=0$. The factor is nonzero, so $d=0$, then $c=0$. Thus $(a_t,b_t)$ is independent.

For an arbitrary target $Au+Bv\in W$, solve the same equations with right sides $A,B$:

$$\begin{aligned}c&=\frac{A-tB}{1-t^2},\\d&=\frac{B-tA}{1-t^2}.\end{aligned}$$

These coefficients exist for every real $A,B$. Direct substitution gives $c+td=A$ and $tc+d=B$, so the pair spans $W$ and is a basis of $W$.

Case $t=1$. Both vectors equal $u+v$. The relation $a_1-b_1=\mathbf0$ proves dependence, and the span is the line $\operatorname{span}\{u+v\}$.

Case $t=-1$. We have $a_{-1}=u-v$ and $b_{-1}=-(u-v)$. The relation $a_{-1}+b_{-1}=\mathbf0$ proves dependence, and the span is $\operatorname{span}\{u-v\}$.

Neither exceptional line is all of $W$. For example, if $u=s(u+v)$ or $u=s(u-v)$, matching the $v$-coefficient forces $s=0$, while matching the $u$-coefficient forces $s=1$, a contradiction. Thus the pair is independent and spans $W$ exactly when $t\ne\pm1$.

Common error: dividing by $1-t^2$ and forgetting the values that make it zero. Another is concluding “spans $\mathbb R^4$” when the calculation only reaches $W$. Every combination still satisfies the two constraints of $W$.

If you got stuck: review solving and checking a system. If the equations were easy but their meaning was unclear, return to span in question 3 and independence in question 4.

Independence gives uniqueness. Suppose

$$\begin{aligned}b&=c_1v_1+\cdots+c_kv_k,\\b&=d_1v_1+\cdots+d_kv_k.\end{aligned}$$

Subtracting gives

$$\begin{aligned}\mathbf0&=(c_1-d_1)v_1+\cdots\\&\quad+(c_k-d_k)v_k.\end{aligned}$$

Independence forces $c_i-d_i=0$ for every $i$, so the coefficient lists agree. Existence was supplied by $b$ being in the span.

A unique representation gives independence. If the list were dependent, there would be coefficients $h_1,\ldots,h_k$, not all zero, with $h_1v_1+\cdots+h_kv_k=\mathbf0$. Adding that relation to the representation with coefficients $c_i$ gives a second representation with coefficients $c_i+h_i$. At least one coefficient changes, contradicting uniqueness. Therefore the list is independent.

In fact, varying $s$ over the real numbers and adding $s(h_1,\ldots,h_k)$ to one coefficient list produces infinitely many distinct representations when a nontrivial relation exists.

For the concrete list, let the coefficient of $w$ be $r$. Since

$$\begin{aligned}&cu+dv+rw\\&=(c+r)u\\&\quad+(d-2r)v,\end{aligned}$$

independence of $u,v$ shows that this equals $p=2u-v$ exactly when $c+r=2$ and $d-2r=-1$. Thus all representations are

$$\begin{aligned}p&=(2-r)u\\&\quad+(-1+2r)v+rw,\\r&\in\mathbb R.\end{aligned}$$

For instance, $r=0$ gives $2u-v$, while $r=1$ gives $u+v+w$. Both evaluate to $(2,1,-1,5)$. The derivation also proves that no other coefficient lists were missed.

A target outside the span has no representation. This is why the statement began with $b$ in the span. Uniqueness refers to coefficients relative to the specified ordered list; changing the list can change the coordinates of the same vector.

Common error: proving only one direction of “if and only if,” or confusing nonunique coefficients with different target vectors.

If you got stuck: return to the worked example separating existence from uniqueness. First subtract its two concrete representations, then repeat the argument with symbols.