The Dimension of The Null Space and Range Examples 2

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 $V$ be a finite-dimensional vector space and define a linear map $T \in (V, V)$ where $\mathrm{dim} \mathrm{null} (T) = 1$ and $\mathrm{dim} \mathrm{range} (T) = 4$.

By the dimension formula above, we must have that:

(2)

\begin{align} \quad \mathrm{dim} (V) = 1 + 4 = 5 \end{align}

Let $\{ v_1, v_2, v_3, v_4, v_5 \}$ be a basis of $V$. Then for every vector $v \in V$ we have that $v = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 + a_5v_5$. Define the linear map $T$ by:

(3)

\begin{align} \quad T(v_1) = v_1 \\ \quad T(v_2) = v_2 \\ \quad T(v_3) = v_3 \\ \quad T(v_4) = v_4 \\ \quad T(v_5) = 0 \end{align}

Such a linear map exists as seen on the Linear Maps Defined by Bases page. We have that:

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\begin{align} \quad T(v) = T(a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 + a_5v_5) \\ \quad T(v) = a_1T(v_1) + a_2T(v_2) + a_3T(v_3) + a_4T(v_4) + a_5T(v_5) \\ \quad T(v) = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 + a_5(0) \\ \quad T(v) = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 \end{align}

Notice that $T(v) = 0$ if and only if $a_1 = a_2 = a_3 = a_4 = 0$, and so:

(5)

\begin{align} \quad \mathrm{null} (T) = \{ v = a_5v_5 \in V : a_5 \in \mathbb{F} \} \end{align}

Note that $\mathrm{null} (T) = \mathrm{span} (v_5)$ and $\mathrm{dim} (\mathrm{span} (v_5)) = 1$, so $\mathrm{null} (T) = 1$.

Now we will show that $\mathrm{dim} ( \mathrm{range} (T)) = 4$. Let $w \in \mathrm{range} (T)$. Then there exists a vector $v = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 + a_5v_5 \in V$ such that:

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\begin{align} \quad T(v) = T(a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 + a_5v_5) = a_1T(v_1) + a_2T(v_2) + a_3T(v_3) + a_4 T(v_4) + a_5T(v_5) \\ \quad = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 + a_5(0) = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 = w \end{align}

Thus we have that:

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\begin{align} \quad \mathrm{range} (T) = \{ v = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 \in V : a_1, a_2, a_3, a_4 \in \mathbb{F} \} \end{align}

Notice that $\mathrm{range} (T) = \mathrm{span} (v_1, v_2, v_3, v_4)$ and that $\mathrm{dim} (\mathrm{span} (v_1, v_2, v_3, v_4)) = 4$ so $\mathrm{dim} (\mathrm{range} (T)) = 4$

Example 2

Let $n \in \mathbb{N}$. Find a linear map $T : \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ where $\mathrm{dim} (\mathrm{null} (T)) = \mathrm{dim} (\mathrm{range} (T))$.

Note that $\mathrm{dim} (\mathbb{R}^{2n}) = 2n$. Let $\{ v_1, v_2, ..., v_n, v_{n+1}, v_{n+2}..., v_{2n} \}$ be a basis of $V$ and define $T$ by:

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\begin{align} \quad T(v_1) = v_1 \\ \quad T(v_2) = v_2 \\ \quad \quad \vdots \quad \quad \\ \quad T(v_n) = v_n \\ \quad T(v_{n+1}) = 0 \\ \quad T(v_{n+2}) = 0 \\ \quad \quad \vdots \quad \quad \\ \quad T(v_{2n}) = 0 \end{align}

Now since $\{ v_1, v_2, ..., v_n, v_{n+1}, v_{n+2} ..., v_{2n} \}$ is a basis of $V$, then for every vector $v \in V$ there exists scalars in $\mathbb{F}$ such that:

(9)

\begin{align} \quad v = a_1v_1 + a_2v_2 + ... + a_nv_n + a_{n+1}v_{n+1} + a_{n+2}v_{n+2} + ... + a_{2n}v_{2n} \end{align}

Apply the linear map $T$ to both sides of the equation above to get that:

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\begin{align} \quad T(v) = a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n) + a_{n+1}T(v_{n+1}) + a_{n+2} T(v_{n+2}) + ... + a_{2n}T(v_{2n}) \\ \quad T(v) = a_1v_1 + a_2v_2 + ... + a_nv_n \end{align}

We then have that:

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\begin{align} \quad \mathrm{null} (T) = \{ a_{n+1}v_{n+1} + a_{n+2}v_{n+2} + ... + a_{2n}v_{2n} \in V : a_{n+1}, a_{n+2}, ..., a_{2n} \in \mathbb{F} \} \end{align}

As we can see, $\mathrm{null} (T) = \mathrm{span} (v_{n+1}, v_{n+2}, ..., v_{2n})$ and $\mathrm{dim} (\mathrm{span} (v_{n+1}, v_{n+2}, ..., v_{2n})) = n$.

Furthermore we have that:

(12)

\begin{align} \quad \mathrm{range} (T) = \{ a_1v_1 + a_2v_2 + ... + a_nv_n \in V : a_1, a_2, ..., a_n \in \mathbb{F} \} \end{align}

As we can see, $\mathrm{range} (T) = \mathrm{span} (v_1, v_2, ..., v_n)$ and $\mathrm{dim} (\mathrm{span} (v_1, v_2, ..., v_n)) = n$.

Therefore $T$ is a linear map such that $\mathrm{dim} (\mathrm{null} (T)) = n = \mathrm{dim} (\mathrm{range} (T))$.

Example 3

Let $n \in \mathbb{N}$. Show that there exists no linear map $T \in \mathcal ( \mathbb{R}^{2n-1}, \mathbb{R}^{2n-1})$ such that $\mathrm{dim} (\mathrm{null} (T)) = \mathrm{dim} (\mathrm{range} (T))$.

We see that $\mathrm{dim} (\mathbb{R}^{2n-1} ) = 2n - 1$ which is an odd number. Now suppose that $\mathrm{dim} (\mathrm{null} (T)) = \mathrm{dim} (\mathrm{range} (T))$. Then by the dimension formula from the top of this page, we have that:

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\begin{align} \quad \mathrm{dim} (\mathbb{R}^{2n-1}) = \mathrm{dim} (\mathrm{null}(T)) + \mathrm{dim} (\mathrm{range} (T)) \\ \quad 2n - 1 = 2 \mathrm{dim}( \mathrm{null} (T)) \end{align}

The equation above implies that $\mathrm{dim} (\mathrm{null} (T)) = \frac{2n - 1}{2}$, however, $\frac{2n - 1}{2}$ is not a whole number, and so there exists no linear map $T \in \mathcal ( \mathbb{R}^{2n-1}, \mathbb{R}^{2n-1})$ such that $\mathrm{dim} (\mathrm{null} (T)) = \mathrm{dim} (\mathrm{range} (T))$.