30 - Properties of Rank and Kernel

Properties of Rank and Kernel / Null Space

Let $V,W$ be vector spaces of $V$ over $\mathbf{F}$
Let $T:V\to W$ be linear

1. $\ker T$ is a subspace of $V$ and $\mathrm{Im}T$ is a subspace of $W$

2. $T$ is injective if and only if $\ker T=\{0_V\}$

3. If $A$ is a spanning set of $V$, then $T(A)$ is a spanning set of $\mathrm{Im}T$

4. If $V$ is finite-dimensional, then $\ker T$ and $\mathrm{Im}T$ are finite-dimensional

Proof (Image)

![[property-of-rank-and-kernel.png]]

Equivalent Statements for Rank and Nullity corollary

Let $V$ be a finite-dimensional vector space
Let $T:V\to V$ be linear

1. $T$ is invertible

2. $\mathrm{rank}(T)=\dim V$

3. $\mathrm{nullity}(T)=0$

Proof

$(1)⟹(3)$
Assume that $T$ is invertible.
Then $T$ is injective and hence surjective, so $\mathrm{Im}T=V$ hence $\mathrm{rank}(T)=\dim V$

$(2)⟹(3)$
Assume that $\mathrm{rank}(T)=\dim V$ so by Rank-Nullity Theorem, then $\mathrm{nullity}(T)=0$

$(3)⟹(1)$
Assume that $\mathrm{nullity}(T)=0$ so that $\ker T=\{0_V\}$ then $T$ is injective
Also, by Rank-Nullity, $\mathrm{rank}(T)=\dim V$ and $\mathrm{Im}T\le V$ so $\mathrm{Im}T=V$, so $T$ is surjective
So $T$ is bijective, so $T$ is invertible

Dimension Inequality for Linear Maps lemma

Let $V,W$ be vector spaces, with $V$ finite-dimensional
Let $T:V\to W$ be linear and $U\le V$ then

$$\dim U-\mathrm{nullity}(T)\le \dim T(U)\le \dim U$$

In particular, if $T$ is injective then $\dim T(U)=\dim U$

Proof

Let $S:U\to W$ be the restriction of $T$ to $U$ (that is, $S(u)=T(u)$ for all $u\in U$)
Then $S$ is linear, and $\ker S\le \ker T$ so $\mathrm{nullity}(S)\le \mathrm{nullity}(T)$
And $\mathrm{Im}S=T(U)$
By Rank-Nullity Theorem then

$$\begin{array}{rl}\dim T(U)& =\dim \mathrm{Im}S=\dim U-\mathrm{nullity}(S)\le \dim U\\ & \\ \dim T(U)& =\dim U-\mathrm{nullity}(S)\ge \dim U-\mathrm{nullity}(T)\end{array}$$

Note that if $T$ is injective than $\mathrm{nullity}T=0$, so $\dim T(U)=\dim U$