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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 finitedimensional 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)
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)
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)
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 finitedimensional 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)
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)
Now suppose that $T(u) = T(v)$ where $v = b_1v_1 + b_2v_2 + … + b_nv_n$. Then we have that:
(7)
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 finitedimensional 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)
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:
(11)
Therefore $\mathrm{range} (T) = W$ and so $T$ is surjective.
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