The Dimension of The Null Space and Range

# The Dimension of The Null Space and Range

We have looked at the Null Space of a Linear Map and the Range of a Linear Map. Let $T \in \mathcal L (V, W)$. We have already proven that $\mathrm{null} (T)$ is a subspace of the domain $V$ and that $\mathrm{range} (T)$ is a subspace of the codomain $W$. We will now look at an extremely important theorem that relates the dimensions of the null space and range of the linear transformation $T$ to the dimension of the domain vector space $V$, provided that $V$ is a finite-dimensional vector space.

Theorem 1: If $T \in \mathcal L (V, W)$ and $V$ is a finite-dimensional vector space, then $\dim V = \dim ( \mathrm{null} (T)) + \dim (\mathrm{range} (T))$. |

**Proof:**Let $T \in \mathcal L (V, W)$ and let $V$ be a finite-dimensional vector space. Since $V$ is finite-dimensional and $\mathrm{null} (T)$ is a subspace of $V$, then we also have that $\mathrm{null} (T)$ is finite-dimensional. Let $\{ u_1, u_2, ..., u_m \}$ be a basis of $\mathrm{null} (T)$. Since this set of vectors is a linearly independent set, it can be extended to the set $\{ u_1, u_2, ..., u_m, v_1, v_2, ... v_n \}$ to form a basis of the finite-dimensional vector space $V$. Therefore $\dim V = m + n$ and $\dim ( \mathrm{null} (T)) = m$. We want to prove that $\dim ( \mathrm{range} (T)) = n$ to satisfy the equality hypothesized.

- Let $v \in V$. Since $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_n \}$ is a basis of $V$, then we can write $v = a_1u_1 + a_2u_2 + ... + a_mu_m + b_1v_1 + b_2v_2 + ... + b_nv_n$ where $a_1, a_2, ..., a_m, b_1, b_2, ..., b_n \in \mathbb{F}$. If we apply the linear transformation to both sides we obtain that:

\begin{align} T(v) = T(a_1u_1 + a_2u_2 + ... + a_mu_m + b_1v_1 + b_2v_2 + ... + b_nv_n) \\ \quad T(v) = a_1T(u_1) + a_2T(u_2) + ... + a_mT(u_m) + b_1T(v_1) + b_2T(v_2) + ... + b_nT(v_n) \end{align}

- Now notice that since $\{ u_1, u_2, ..., u_m \}$ is a basis of $\mathrm{null} (T)$ then this implies that $u_1, u_2, ..., u_m \in \mathrm{null} (T)$. Therefore $T(u_1) = T(u_2) = ... = T(u_m) = 0$, and so:

\begin{align} \quad T(v) = a_10 + a_20 + ... + a_m0 + b_1T(v_1) + b_2T(v_2) + ... + b_nT(v_n) \\ T(v) = b_1T(v_1) + b_2T(v_2) + ... + b_nT(v_n) \end{align}

- Therefore the set of vectors $\{ T(v_1), T(v_2), ..., T(v_n) \}$ spans $\mathrm{range} (T)$, that is $\mathrm{span} (T(v_1), T(v_2), ..., T(v_n)) = \mathrm{range} (T)$.

- We now need to show that the set $\{ T(v_1), T(v_2), ..., T(v_n) \}$ is linearly independent. Consider the following vector equation:

\begin{equation} c_1T(v_1) + c_2T(v_2) + ... + c_nT(v_n) = 0 \end{equation}

- Where $c_1, c_2, ..., c_n \in \mathbb{F}$. Then it follows that $T(c_1v_1 + c_2v_2 + ... + c_nv_n) = 0$, and so $(c_1v_1 + c_2v_2 + ... + c_nv_n) \in \mathrm{null} (T)$. Now since the set $\{ u_1, u_2, ..., u_m \}$ spans $\mathrm{null} (T)$, so for $d_1, d_2, ..., d_m \in \mathbb{F}$ we can write that:

\begin{align} c_1v_1 + c_2v_2 + ... + c_nv_n = d_1u_1 + d_2u_2 + ... + d_mu_m \\ \quad c_1v_1 + c_2v_2 + ... + c_nv_n - d_1u_1 - d_2u_2 - ... - d_mu_m = 0 \end{align}

- But the set $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_n \}$ is linearly independent (as it is a basis for $V$), which implies that $c_1 = c_2 = ... = c_n = d_1 = d_2 = ... = d_m = 0$. More importantly, $c_1 = c_2 = ... c_m = 0$. Therefore $\{ T(v_1), T(v_2), ..., T(v_n) \}$ is a linearly independent set of vectors.

- Since $\{ T(v_1), T(v_2), ..., T(v_n) \}$ spans $\mathrm{range} (T)$ and is linearly independent, then $\{ T(v_1), T(v_2), ..., T(v_n) \}$ is a basis for $\mathrm{range} (T)$, so $\dim ( \mathrm{range} (T)) = n$.

- Therefore $\dim V = \dim ( \mathrm{null} (T)) + \dim ( \mathrm{range} (T))$. $\blacksquare$

We will now look at a bunch of corollaries as a result of Theorem 1.

Corollary 1: If $T \in \mathcal L (V, W)$ and $V$ is a finite-dimensional vector space, then $\mathrm{range} (T)$ is finite-dimensional. |

**Proof:**If $V$ is finite-dimensional, then $\dim V = m$ for some $m \in \mathbb{N} \cup \{0 \}$. Since $\mathrm{null} (T)$ is a subspace of $V$ we have that $\mathrm{null} (T)$ is also finite-dimensional, so $\dim ( \mathrm{null} (T)) = n ≤ m$. By Theorem 1, it follows that $\dim ( \mathrm{range} (T)) = m - n$ and so $\mathrm{range} (T)$ is finite-dimensional. $\blacksquare$

Corollary 2: If $T \in \mathcal L (V, W)$ and $V$ and $W$ are finite-dimensional vector spaces where $\dim V > \dim W$, then $T$ is not injective. |

**Proof:**Suppose that $V$ and $W$ are finite-dimensional vector spaces such that $\dim V > \dim W$, and let $T \in \mathcal L (V, W)$. By Theorem 1 we have that $\dim V = \dim ( \mathrm{null} (T)) + \dim ( \mathrm{range} (T))$ and so rearranging these terms (while noting that $\dim V - \dim ( \mathrm{range} (T)) ≥ \dim V - \dim W$ since $\dim ( \mathrm{range} (T)) ≤ \dim W$) we get that:

\begin{align} \quad \dim ( \mathrm{null} (T)) = \dim V - \dim ( \mathrm{range} (T)) ≥ \dim V - \dim W > 0 \end{align}

- Therefore $\dim ( \mathrm{null} (T)) > 0$ and so $\mathrm{null} (T) \neq \{ 0 \}$ and $T$ is not injective. $\blacksquare$.

We will now look at a corollary that is analogous to that of corollary 2.

Corollary 3: If $T \in \mathcal L (V, W)$ and $V$ and $W$ are finite-dimensional vector spaces where $\dim V < \dim W$, then $T$ is not surjective. |

**Proof:**Suppose that $V$ and $W$ are finite-dimensional vector spaces such that $\dim V < \dim W$, and let $T \in \mathcal L (V, W)$. By Theorem 1 we have that $\dim V = \dim ( \mathrm{null} (T)) + \dim ( \mathrm{range} (T))$ and so rearranging these terms we get that:

\begin{align} \quad \dim (\mathrm{range} (T)) = \dim V - \dim (\mathrm{null} (T)) ≤ \dim V < \dim W \end{align}

- Therefore $\dim (\mathrm{range} (T)) < \dim W$ so $\mathrm{range} (T) \neq W$ so $T$ is not surjective. $\blacksquare$

Corollary 4: If $T \in \mathcal L (V, V)$ and $V$ is a finite-dimensional vector space then $T$ is injective if and only if $T$ is surjective. |

**Proof:**$\Rightarrow$ Suppose that $T$ is injective. Then $\mathrm{null} (T) = \{ 0 \}$ so $\dim (\mathrm{null} (T)) = 0$. By Theorem 1 we have that $\dim V = \dim (\mathrm{range} (T))$ which implies that $V = \mathrm{range} (T)$ (since $T : V \to V$), so $T$ is surjective.

- $\Leftarrow$ Suppose that $T$ is surjective. Then $\mathrm{range} (T) = V$, and so $\dim (\mathrm{range} (T)) = \dim V$. Therefore by Theorem 1, we have that $\dim (\mathrm{null} (T)) = 0$ which implies that $\mathrm{null} (T) = \{ 0 \}$, so $T$ is injective. $\blacksquare$