Exercises — Rank-nullity theorem — proof

WHAT we do: plug into $\mathrm{rank}+\mathrm{nullity}=\dim V$. WHY: the theorem is an equation with three slots; given two, the third is forced. The domain is ${\mathbb{R}}^7$ so $\dim V=7$. Then $\mathrm{rank}(T)=\dim V-\mathrm{nullity}(T)=7-3=\boxed{4}.$ Notice the codomain ${\mathbb{R}}^4$ never entered — we balanced against the domain ${\mathbb{R}}^7$.

WHAT we do: bound the rank from both sides. The image sits inside the codomain, so $\mathrm{rank}\le \dim W=9$. But the image is also fed only by $5$ input directions, so $\mathrm{rank}=\dim V-\mathrm{nullity}\le \dim V=5$. Since $7>5$, it is impossible. Answer: False. The tightest cap is $\mathrm{rank}\le \min (\dim V,\dim W)=\min (5,9)=5$.

WHAT we do: row-reduce to count pivots (that count is the rank — see Rank of a Matrix (row vs column space)). WHY: each pivot marks one independent surviving direction of the column space = image. Subtract row 1 from row 3: $R_3\to R_3-R_1=(0,1,1,1)$. That equals row 2, so $R_3\to R_3-R_2=(0,0,0,0)$. $\left(\begin{array}{cccc}1& 2& 0& 1\\ 0& 1& 1& 1\\ 0& 0& 0& 0\end{array}\right).$ Two nonzero pivot rows $\Rightarrow \mathrm{rank}=2$. By the theorem, with $\dim V=4$: $\mathrm{nullity}=4-2=\boxed{2},\mathrm{rank}=\boxed{2}.$

Kernel: $p^{\prime }=0$ forces $p$ constant, so $\ker T=\{\text{constants}\}$, a $1$-dimensional space $\Rightarrow \mathrm{nullity}=1$. Image: differentiating the basis $\{1,x,x^2,x^3,x^4\}$ gives $\{0,1,2x,3x^2,4x^3\}$, which spans $P_3$ (degree $\le 3$). So the image is $P_3$, $\dim =4\Rightarrow \mathrm{rank}=4$. Check: $4+1=5=\dim P_4$. ✓ The theorem holds on abstract spaces, not just ${\mathbb{R}}^n$.

WHAT we do: turn nullity $0$ into injectivity, then use rank–nullity for surjectivity. Nullity $0\Rightarrow$ injective: $\ker T=\{0\}$ (dimension $0$ means only the zero vector). If $T(a)=T(b)$ then $T(a-b)=0$ so $a-b\in \ker T=\{0\}$, giving $a=b$. Injective. Injective $\Rightarrow$ surjective here: by the theorem $\mathrm{rank}=\dim V-0=n$. The image is an $n$-dimensional subspace of the $n$-dimensional codomain ${\mathbb{R}}^n$, so the image is all of ${\mathbb{R}}^n$. Surjective. Injective + surjective $=$ bijective, hence invertible. See Invertible Linear Maps and Isomorphisms. $■$ Key idea: for a square situation ($\dim V=\dim W$), injective $\iff$ surjective $\iff$ invertible — rank–nullity glues these together.

WHAT we do: identify the solution space of $Ax=0$ as $\ker T$ where $T:{\mathbb{R}}^6\to {\mathbb{R}}^3$. The number of unknowns is $6=\dim V$. Pivots $=\mathrm{rank}=2$, so $\text{free variables}=\dim V-\mathrm{rank}=6-2=\boxed{4}.$ The solution space is exactly the kernel, so its dimension is the nullity $=4$ (see Solving Linear Systems — free variables count). Free-variable count and nullity are the same number.

WHAT we do: $T\circ S$ first lands in $\mathrm{im}S$ (a $3$-dim space inside ${\mathbb{R}}^4$), then $T$ acts on it. Upper bound: the composite's image is $T$ applied to $\mathrm{im}S$, which cannot be bigger than either $\mathrm{im}S$ or $\mathrm{im}T$. So $\mathrm{rank}(T\circ S)\le \min (\mathrm{rank}S,\mathrm{rank}T)=\min (3,2)=2.$ Lower bound: restricting $T$ to the $3$-dimensional $\mathrm{im}S$, rank–nullity on this restriction gives $\mathrm{rank}(T\circ S)\ge \mathrm{rank}(S)-\mathrm{nullity}(T)=3-(4-2)=3-2=1.$ So $\boxed{1\le \mathrm{rank}(T\circ S)\le 2}$, and both values are achievable — the data alone cannot pin it. (The claim in the prompt is a deliberate misdirection; the honest answer is the range.)

WHAT we do: $T^2=0$ means every output of $T$ is killed by the next $T$, i.e. $\mathrm{im}T\subseteq \ker T$. WHY: for any $v$, $T(T(v))=0$, so $T(v)\in \ker T$; that's exactly $\mathrm{im}T\subseteq \ker T$. Taking dimensions of nested subspaces: $\mathrm{rank}(T)=\dim (\mathrm{im}T)\le \dim (\ker T)=\mathrm{nullity}(T).$ Now apply rank–nullity, $\mathrm{rank}+\mathrm{nullity}=4$. Since $\mathrm{rank}\le \mathrm{nullity}$, $2\mathrm{rank}\le \mathrm{rank}+\mathrm{nullity}=4\Rightarrow \mathrm{rank}\le 2.■$ Because $T\ne 0$ also $\mathrm{rank}\ge 1$, so $\mathrm{rank}\in \{1,2\}$.

WHAT we do: build a clever linear map whose kernel and image are the pieces we want. Define $T:U\times W\to V,T(u,w)=u-w.$ This is linear, and its domain $U\times W$ has dimension $\dim U+\dim W$ (concatenate a basis of $U$ with a basis of $W$; see Dimension of a Vector Space). Image of $T$: all differences $u-w$ with $u\in U,w\in W$. Every element of $U+W$ can be written $u+w=u-(-w)$ and $-w\in W$, so $\mathrm{im}T=U+W$. Thus $\mathrm{rank}T=\dim (U+W)$. Kernel of $T$: $u-w=0$ means $u=w$, so the common vector lies in $U\cap W$. The kernel is $\{(x,x):x\in U\cap W\}$, which is isomorphic to $U\cap W$, giving $\mathrm{nullity}T=\dim (U\cap W)$. Apply rank–nullity on $T$: $\underset{\text{rank}}{\underbrace{\dim (U+W)}}+\underset{\text{nullity}}{\underbrace{\dim (U\cap W)}}=\underset{\dim (U\times W)}{\underbrace{\dim U+\dim W}}.$ Rearrange to get the formula. $■$ The lesson: to prove a dimension identity, engineer a map so the identity becomes rank + nullity = domain.

From $\dim (U+W)=\dim U+\dim W-\dim (U\cap W)=9-\dim (U\cap W)$. The sum $U+W$ lives in ${\mathbb{R}}^6$, so $\dim (U+W)\le 6$. Therefore $9-\dim (U\cap W)\le 6\Rightarrow \dim (U\cap W)\ge 3.$ The smallest intersection dimension is $\boxed{3}$, attained when $U+W={\mathbb{R}}^6$. Cross-check via $T$: domain dimension $=4+5=9$, image $\le \dim {\mathbb{R}}^6=6$, so $\mathrm{nullity}=9-\mathrm{rank}\ge 9-6=3$, matching $\dim (U\cap W)\ge 3$.