The Dimension of The Null Space and Range Examples 3

# The Dimension of The Null Space and Range Examples 3

Recall from The Dimension of The Null Space and Range page that if $T$ is a linear map from $V \to W$ and $V$ is finite-dimensional then we have the following formula relating the dimension of $V$ to the dimension of the the null space of $T$ and the dimension of the range of $T$:

(1)
\begin{align} \quad \mathrm{dim} (V) = \mathrm{dim} (\mathrm{null}(T)) + \mathrm{dim} (\mathrm{range} (T)) \end{align}

We will now look at some more examples applying this formula.

## Example 1

Determine whether there exists a linear map $T \in \mathcal (\mathbb{F}^6, \mathbb{F}^2)$ such that $\mathrm{null}(T) = \{ (x_1, x_2, x_3, x_4, x_5, x_6) \in \mathbb{F}^6 : x_1 = x_2 = x_3, x_4, x_5 = 2x_6 \}$.

Suppose that such a linear map exists. We note that $\mathrm{dim} (\mathbb{F}^6) = 6$. Furthermore, note that:

(2)
\begin{align} \quad \mathrm{null} (T) = \{ (x_1, x_1, x_1, x_4, x_5, x_5) : x_1, x_4, x_5 \in \mathbb{F} \} = \mathrm{span} ((1, 1, 1, 0, 0, 0), (0, 0, 0, 1, 0, 0), (0, 0, 0, 0, 1, 1)) \end{align}

We note that no smaller set of vectors can span $\mathrm{null} (T)$ so $\mathrm{dim} (\mathrm{null} (T)) = \mathrm{dim} (\mathrm{span} ((1, 1, 1, 0, 0, 0), (0, 0, 0, 1, 0, 0), (0, 0, 0, 0, 1, 1)) )= 3$.

If we apply the dimension formula we see that:

(3)
\begin{align} \quad \mathrm{dim} (\mathbb{F}^3) = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{range} (T)) \\ \quad 6 = 3 + \mathrm{dim} (\mathrm{range}(T)) \\ \quad 3 = \mathrm{dim} (\mathrm{range}(T)) \end{align}

Therefore we must have that $\mathrm{dim} (\mathrm{range}(T)) = 3$. However, $\mathrm{range}(T)$ is a subspace of the codomain space $\mathbb{F}^2$ and $\mathrm{dim} (\mathbb{F}^2) = 2$, so any subspace $U$ of $\mathbb{F}^2$ must be such that $\mathrm{dim} (U) ≤ \mathrm{dim} (\mathbb{F}^2)$.

Therefore since $3 > 2$ we have a contradiction and so there exists no linear map $T \in \mathcal (\mathbb{F}^6, \mathbb{F}^2)$ such that $\mathrm{null}(T) = \{ (x_1, x_2, x_3, x_4, x_5, x_6) \in \mathbb{F}^6 : x_1 = x_2 = x_3, x_4, x_5 = 2x_6 \}$.

## Example 2

Let $V$ and $W$ be finite-dimensional vector spaces. Prove that there exists a linear map $T \in \mathcal L(V, W)$ that is injective if and only if $\mathrm{dim} (V) ≤ \mathrm{dim} (W)$.

$\Rightarrow$ Suppose that $T$ is injective. Then $\mathrm{dim} (\mathrm{null} (T)) = 0$ and so by the dimension formula above we have that:

(4)
\begin{align} \quad \mathrm{dim} (V) = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{range}(T)) = \mathrm{dim} (\mathrm{range} (T)) ≤ \mathrm{dim} (W) \end{align}

Therefore $\mathrm{dim} (V) ≤ \mathrm{dim} (W)$.

$\Leftarrow$ Now suppose that $\mathrm{dim} (V) = n ≤ m = \mathrm{dim} (W)$. Let $\{ v_1, v_2, ..., v_n \}$ be a basis of $V$ and let $\{ w_1, w_2, ..., w_m \}$ be a basis of $W$. We can define a linear map $T \in \mathcal L (V, W)$ by:

(5)

For every vector $u = a_1v_1 + a_2v_2 + ... + a_nv_n \in V$ we have that:

(6)
\begin{align} \quad T(v) = T(a_1v_1 + a_2v_2 + ... + a_nv_n) \\ \quad T(v) = a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n) \\ \quad T(v) = a_1w_1 + a_2w_2 + ... + a_nw_n \end{align}

Now suppose that $T(u) = T(v)$ where $v = b_1v_1 + b_2v_2 + ... + b_nv_n$. Then we have that:

(7)
\begin{align} \quad T(u) = T(v) \\ \quad a_1w_1 + a_2w_2 + ... + a_nw_n = b_1w_1 + b_2w_2 + ... + b_nw_n \\ \quad (a_1 - b_1)w_1 + (a_2 - b_2)w_2 + ... + (a_n - b_n)w_n = 0 \end{align}

Since $\{w_1, w_2, ..., w_n \}$ is linearly independent set of vectors (since this set of vectors is a subset of the basis $\{ w_1, w_2, ..., w_n, ..., w_m \}$ of $W$) we must have that:

(8)

Therefore $a_1 = b_1$, $a_2 = b_2$, …, $a_n = b_n$. Therefore $u = v$ so $T$ is injective.

## Example 3

Let $V$ and $W$ be finite-dimensional vector spaces. Prove that there exists a linear map $T \in \mathcal L(V, W)$ that is surjective if and only if $\mathrm{dim} (V) ≥ \mathrm{dim} (W)$.

$\Rightarrow$ Suppose that $T$ is surjective. Then $\mathrm{range} (T) = W$ and so $\mathrm{dim} (\mathrm{range} (T)) = \mathrm{dim} (W)$. Using the dimension formula and we have that:

(9)
\begin{align} \quad \mathrm{dim} (V) = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{range} (T)) \\ \quad \mathrm{dim} (V) = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (W) \end{align}

Since $\mathrm{dim} (\mathrm{null} (T)) ≥ 0$ we must have that $\mathrm{dim} (V) ≥ \mathrm{dim} (W)$ as desired.

$\Leftarrow$ Now suppose that $\mathrm{dim} (V) = n ≥ m = \mathrm{dim} (W)$. Let $\{ v_1, v_2, ..., v_n\}$ be a basis of $V$ and let $\{ w_1, w_2, ..., w_m \}$ be a basis of $W$. Extend the set of vectors $\{ w_1, w_2, ..., w_m \}$ to $\{ w_1, w_2, ..., w_m, w_{m+1}, ..., w_n \}$. This set of vectors still spans $W$ but is not longer linearly independent.

Now define a linear map $T \in \mathcal L (V, W)$ by:

(10)
Notice that for any vector $w \in W$, we have that $w = b_1w_1 + b_2w_2 + ... + b_mw_m$ since $\{ w_1, w_2, ..., w_m \}$ is a basis of $W$. So for any vector $w \in W$, there exists a vector $v = b_1v_1 + b_2v_2 + ... + b_mv_m \in V$ and:
Therefore $\mathrm{range} (T) = W$ and so $T$ is surjective.