The Dimension of The Null Space and Range Examples 1

# The Dimension of The Null Space and Range Examples 1

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 examples applying this formula.

## Example 1

Let $U$, $V$, and $W$ be finite-dimensional vector spaces, and suppose that $S \in \mathcal L (V, W)$ and $T \in \mathcal L (U, V)$. Let $ST \in \mathcal L (U, W)$. Show that $\mathrm{dim} (\mathrm{null} (ST)) ≤ \mathrm{dim} ( \mathrm{null} (S)) + \mathrm{dim} ( \mathrm{null} (T))$.

Using the dimension formula above we have that:

(2)
\begin{align} \quad \mathrm{dim} (\mathrm{null} (ST)) = \mathrm{dim} (U) - \mathrm{dim} (\mathrm{range} (ST)) \end{align}

Since $\mathrm{dim} (U) = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{range} (T))$ we can substitute this into the equation above to get that:

(3)
\begin{align} \quad \mathrm{dim} (\mathrm{null} (ST)) = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{range} (T)) - \mathrm{dim} (\mathrm{range} (ST)) \end{align}

Now notice that $\mathrm{dim} (\mathrm{range}(T)) ≤ \mathrm{dim} (V)$ and so:

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

Since $\mathrm{dim} (V) = \mathrm{dim} (\mathrm{null} (S)) + \mathrm{dim} (\mathrm{range} (S))$, we can substitute this into the equation above to get:

(5)
\begin{align} \quad \mathrm{dim} (\mathrm{null} (ST)) ≤ \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{null} (S)) + \mathrm{dim} (\mathrm{range} (S)) - \mathrm{dim} (\mathrm{range} (ST)) \end{align}

Note that $\mathrm{dim} (\mathrm{range} (S)) - \mathrm{dim} (\mathrm{range} (ST)) ≤ 0$ though since $S$ maps elements $v \in V$ to elements $S(v) \in W$, and $ST$ maps elements $T(u) \in V$ to elements $S(T(u)) \in W$ and so $\mathrm{range} S \subseteq \mathrm{range} (ST)$ which implies that $\mathrm{dim} ( \mathrm{range} (S)) ≤ \mathrm{dim} ( \mathrm{range} (ST))$, so:

(6)
\begin{align} \quad \mathrm{dim} (\mathrm{null} (ST)) ≤ \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{null} (S)) \end{align}

## Example 2

Let $V$ be a finite-dimensional vector space. Suppose that $V = \mathrm{null}(T) + \mathrm{range}(T)$. Prove that $\mathrm{null} (T) \cap \mathrm{range} (T) = \{ 0 \}$.

If we apply the dimension formula given above, we have that:

(7)
\begin{align} \quad \mathrm{dim} (V) = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{range} (T)) \end{align}

Furthermore, from The Dimension of a Sum of Subspaces we have that:

(8)
\begin{align} \quad \mathrm{dim} (\mathrm{null} (T) + \mathrm{range}(T)) = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{range} (T)) - \mathrm{dim} (\mathrm{null} (T) \cap \mathrm{range} (T)) \end{align}

Since $V = \mathrm{null} (T) + \mathrm{range} (T)$, this implies that the two equations above are equal, and so:

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

The equation above implies that $\mathrm{dim} (\mathrm{null} (T) \cap \mathrm{range} (T)) = 0$, so $\mathrm{null} (T) \cap \mathrm{range} (T) = \{ 0 \}$.