Practice: Basis and Dimension

(a) A basis for $\mathrm{span}(S)$. Put the vectors as columns and reduce:

Pivots in columns 1 and 3, so the basis is the original vectors there:

Dependences, read from the non-pivot columns: column 2 is $(2,0,0{)}^{\mathsf{T}}$, so $(2,2,0)=2(1,1,0)$ ✓. Column 4 is $(1,1,0{)}^{\mathsf{T}}$, so $(1,2,1)=(1,1,0)+(0,1,1)$ ✓. (b) A basis for $U$. Solve $2x-y+z=0$ for $y$: $y=2x+z$, with $x$ and $z$ free. The general element is

so $\{(1,2,0),(0,1,1)\}$ spans $U$; neither is a multiple of the other, so it is independent, and $\dim U=2$. A different basis: solve for $z$ instead, $z=y-2x$, with $x$ and $y$ free, giving $(x,y,y-2x)=x(1,0,-2)+y(0,1,1)$ and the basis $\{(1,0,-2),(0,1,1)\}$. Both vectors satisfy the equation, $2(1)-0+(-2)=0$ ✓ and $2(0)-1+1=0$ ✓, and they are independent. What the two have in common is their size: two elements each. They must agree, because any two bases of the same space have the same number of elements. Their spans are equal (both are $U$) even though no vector of one need appear in the other. (c) Is $\{1+x,x+x^2,1+x^2\}$ a basis of $P_2$? $\dim P_2=3$ and the set has 3 elements, so the counting shortcut applies: it suffices to check independence, and spanning then follows. Suppose $a(1+x)+b(x+x^2)+c(1+x^2)=0$. Collecting coefficients:

From the first $c=-a$; from the second $b=-a$; substituting into the third gives $-a-a=-2a=0$, so $a=0$, hence $b=c=0$. The set is independent, and being 3 independent vectors in a 3-dimensional space, it is a basis. Why spanning follows: if it did not span, some $p\in P_2$ would lie outside its span, and adjoining $p$ would give 4 independent vectors in a space of dimension 3, impossible by the exchange lemma. (d) Coordinates of $3+2x+x^2$ in $\{1,1+x,1+x+x^2\}$. Solve

Match from the highest power down. $x^2$: $c_3=1$. $x$: $c_2+c_3=2$, so $c_2=1$. Constant: $c_1+c_2+c_3=3$, so $c_1=1$. Coordinates $(1,1,1)$. Reconstruction: $1+(1+x)+(1+x+x^2)=3+2x+x^2$ ✓. (e) Why the two bases in (b) must have the same size. The exchange lemma says that in a space spanned by $n$ vectors, no independent set exceeds $n$ elements. Given two bases $B_1$ and $B_2$: $B_1$ spans and $B_2$ is independent, so $|B_2|\le |B_1|$; $B_2$ spans and $B_1$ is independent, so $|B_1|\le |B_2|$. Hence they are equal. Without that guarantee, "the dimension of $U$" would not name anything. The number would depend on which basis was chosen, so two people describing the same plane could report different dimensions and neither would be wrong. Dimension would become a property of a description rather than of the space, and every statement resting on it, that $n+1$ vectors in an $n$-dimensional space are dependent, that rank plus nullity is the dimension of the domain, would lose its meaning.