Exercises — Dimension — basis cardinality

The standard basis is $\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\}$ — 4 vectors, independent and spanning. By the Dimension Theorem every basis has this same count, so $\dim {\mathbb{R}}^4=4$.

Basis $\{1,x,x^2,x^3\}$ has 4 elements. So $\dim P_3=4$. In general $\dim P_n=n+1$ (one slot for each power $x^0,\dots ,x^n$).

No. A basis must be independent, and by the exchange lemma any independent set in ${\mathbb{R}}^2$ has at most $\dim {\mathbb{R}}^2=2$ vectors. Three vectors cannot be independent, so this is not a basis. (Indeed $(2,3)=2(1,0)+3(0,1)$.)

No. Steinitz: independent ≤ spanning. Here spanning size is $5$, so any independent set has $\le 5$ vectors. Seven exceeds $5$ ⇒ they must be dependent.

Set $a(2,1)+b(1,3)=(0,0)$. Component-wise: $2a+b=0$ and $a+3b=0$. From the first, $b=-2a$; substitute: $a+3(-2a)=a-6a=-5a=0\Rightarrow a=0\Rightarrow b=0$. Only the trivial solution ⇒ independent. Two independent vectors in a 2-dimensional space also span it, so it is a basis — confirming $\dim {\mathbb{R}}^2=2$. See figure below.

Spanning: we recover each standard basis vector: $1=1$; $x=(1+x)-1$; $x^2=(1+x+x^2)-(1+x)$. Since $\{1,x,x^2\}$ spans $P_2$ and each is reachable, so does our set. Count: 3 elements $=\dim P_2$. A spanning set of size equal to the dimension is automatically a basis.

Solve $v_1=a{w}_1+b{w}_2$: $(2,3)=a(1,0)+b(1,1)=(a+b,b)$. So $b=3$, $a+b=2\Rightarrow a=-1$. Thus $v_1=-1\cdot w_1+3\cdot w_2.$ The coefficient on $w_1$ is $-1\ne 0$, so we may solve for $w_1$: $w_1=\frac{1}{-1}(v_1-3w_2)=-v_1+3w_2.$ Hence $w_1$ is expressible from $\{v_1,w_2\}$, so $\{v_1,w_2\}=\{(2,3),(1,1)\}$ still spans ${\mathbb{R}}^2$ (independence check: $\det \begin{array}{cc}2& 1\\ 3& 1\end{array}=2-3=-1\ne 0$). One $v$ swapped in, one $w$ kicked out, spanning size preserved. See figure.

Suppose every $c_j=0$. Then $v_{k+1}=b_1{v}_1+\cdots +b_k{v}_k$ — that is, $v_{k+1}$ is a combination of the earlier $v$'s. But $L=\{v_1,\dots ,v_m\}$ is independent, meaning no $v$ is a combination of the others. Contradiction. So some $c_j\ne 0$, guaranteeing a remaining $w$ is present to kick out. If it failed, we'd run out of $w$'s to swap and the size-$n$ spanning invariant would collapse — the whole bound $m\le n$ would be unproved. This is the single point where independence of $L$ enters (see Linear Independence).

• Direction 1: $B_1$ is independent, $B_2$ spans ⇒ lemma gives $p\le q$.
• Direction 2: $B_2$ is independent, $B_1$ spans ⇒ lemma gives $q\le p$.

Both hold because a basis is simultaneously independent and spanning, so each can play either role. From $p\le q$ and $q\le p$ we get $p=q$. That common value is $\dim V$.

Solve the constraint: pick $y,z$ free, then $x=-y-z$. So $(x,y,z)=(-y-z,y,z)=y(-1,1,0)+z(-1,0,1).$ The two vectors $(-1,1,0),(-1,0,1)$ span $W$ and are independent (neither is a scalar multiple of the other). So they form a basis and $\dim W=2$. Intuition: one linear equation removes one degree of freedom from a 3-dimensional space, leaving $3-1=2$. This is the Rank-Nullity Theorem in miniature: the map $T(x,y,z)=x+y+z$ has $1$-dimensional image and $2$-dimensional kernel $=W$.

We need total size $=\dim {\mathbb{R}}^3=3$, so exactly one more vector — provided it's not a combination of the two. Try $(0,0,1)$: is $\{(1,1,0),(1,0,1),(0,0,1)\}$ independent? Compute the determinant $\det \left(\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 0& 1\end{array}\right).$ Expand along the bottom row: $=1\cdot \det \left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)=1\cdot (1\cdot 0-1\cdot 1)=-1\ne 0.$ Independent ⇒ basis. Why exactly one? Any basis has cardinality $3$; we already have $2$ independent vectors, and by the exchange lemma we can always exchange to reach the full dimension without overshooting.

Rank-Nullity Theorem: $\dim (\ker T)+\dim (\text{im }T)=\dim (\text{domain})=5$. With $\dim \ker =2$, $\dim (\text{im }T)=5-2=3$. Since the codomain ${\mathbb{R}}^3$ has dimension $3$ and the image is a $3$-dimensional subspace of it, the image equals all of ${\mathbb{R}}^3$ (a subspace whose dimension equals the whole space is the whole space — that's basis-cardinality again). So yes, $T$ is surjective.

Its basis is the empty set $\varnothing$. The empty set is vacuously independent (there are no coefficients to constrain, so the only combination — the empty one — gives $\vec{0}$) and it spans $\{\vec{0}\}$ (the empty combination equals $\vec{0}$, the only vector present). Its cardinality is $0$, so $\dim \{\vec{0}\}=0$. Note $\{\vec{0}\}$ itself is not a basis: any set containing $\vec{0}$ is dependent since $1\cdot \vec{0}=\vec{0}$ is a nontrivial relation.

Take $V={\mathbb{R}}^n$ spanned by the standard basis (size $n$). The same standard basis is an independent set of size $n$. So an independent set can reach the spanning-set size exactly; the bound $m\le n$ is achieved with equality. Equality happens precisely when the independent set is also spanning — i.e. it's a basis.

Let $P$ be the space of all polynomials (any degree). The set $\{1,x,x^2,x^3,\dots \}$ is independent and spanning but infinite — no finite set spans $P$ (a finite set has a maximal degree $d$, so cannot reach $x^{d+1}$). Here $\dim P=\infty$. The exchange lemma still applies relatively: for any finite spanning set of any subspace $W\subseteq P$, independent sets in $W$ are bounded by that spanning size. The lemma's engine never breaks; only the assumption "some finite spanning set exists" does.

The change-of-basis matrix (see Coordinates and Change of Basis) has as many columns as the old basis size and as many rows as the new basis coordinate length — both equal the number of vectors in a basis. Because the Dimension Theorem forces both bases to have the same count $\dim V=3$, the matrix is $3\times 3$, hence square, hence potentially invertible (and it is invertible, since it maps one basis to another). Non-square would be impossible — that would mean two bases of different sizes, contradicting basis cardinality.

Recognition: dimension = size of a basis. Application: independent ≤ spanning. Analysis: one exchange preserves spanning because a nonzero coefficient lets us solve. Synthesis: combine with rank-nullity and extension. Mastery: empty-set basis of $\{\vec{0}\}$, tightness, and the infinite case.