Null space and range

Now that you know what linear transformations are, you're ready to delve deeper into their properties. Now, we're going to introduce two new sets intimately related to them: the null space and the image. Both arise naturally in the context of systems of equations.

Thanks to them, you will be able to reinterpret these systems, discover several relevant facts about them, and even analyze them geometrically. Here, you'll work with a linear transformation $T$ between two vector spaces $V$ and $W$.

Null space

Let's look at an $m\times n$ homogeneous system of linear equations. You already know that it can be represented as a matrix equation:

$$Ax=0$$But if you remember the linear transformation $L_A$ associated with $A$, which is given by $L_A(x)=A x$, then you can write the equation:

$$L_A(x)=0$$So, the solutions to $Ax=0$ are the vectors that $L_A$ maps to $0$. Well, it turns out that for any linear transformation $T$, this kind of vectors make up an important set, which is called the null space of $T$ (or kernel of $T$):

$$\ker(T) = \{v \in V: T(v) =0\}$$

The null space is full of surprises that you will discover now and in future topics. The first one is that it's not a simple set, but a subspace. In fact, it's quite straightforward to check this out.

Going back to the systems of linear equations, you've just shown that the solutions of the homogeneous system $Ax = 0$ are the null space of $L_A$! Now, let's say that:

$$A = \begin{pmatrix} 6&2&8 \\ 15&5&20 \end{pmatrix}$$

You can find the solutions to the system $Ax = 0$ by computing a row-echelon form of $A$. Then you'll find out that:$$\ker(L_A)=\{\lambda (-1,3,0) + \mu (-4, 0,3) : \, \lambda, \mu \in \mathbb{R}\}$$This means that $(-1,3,0)^T$ and $(-4, 0,3)^T$ are a basis, thus the null space of $L_A$ is a plane.

Null spaces properties

Transformations that send different vectors to different vectors are called injective. This means that the unique vector that you can map to $0$ is indeed $0$. In consequence $\ker(T) = \{ 0 \}$. So, injective transformations have the smallest possible null space.

It seems intuitive that the more vectors $T$ maps to $0$, the less injective it is. And this suggests that when the null space is as small as possible, $T$ is, in fact, injective. Well, the intuition is correct, and this is the second surprise of the null space:

This result means that, in order to prove that $T$ is injective, it's enough to show that $T(v)=0$ implies that $v=0$.

Do you remember that linear transformations preserve the structure of vector spaces? Well, the third surprise is that transformations whose null space is as small as possible preserve linear independence. More precisely, if $v_1, \dots, v_k$ are linearly independent and $T$ is injective, then $T(v_1), \dots, T(v_k)$ are linearly independent..

Range

Let's go back to systems of equations. When they are not homogeneous they look like this:

$$Ax=b$$But that's exactly $L_A(x) =b$. This means that it must be a vector $x$ whose image under $T$ is $b$. In general, if you put together the image under $T$ of all the vectors of $V$, then the result is a set called the range of $T$ (or image of $T$):

$$\mathrm{Im}(T) = \{ T(v) : \ v \in V \}$$The range of $T$ has a close relationship to its kernel. Both possess similar properties and actually complement each other.

To begin with, while the null space contains vectors of $V$, the range is a subset of $W$. And like the null space, it is a subspace.

If, for example, $T$ is a transformation from $\mathbb{R}^4$ into $\mathbb{R}^3$, then this means that it collapses all $\mathbb{R}^4$ into a point, a line, a plane or even all $\mathbb{R}^3$.

Now, the system $Ax=b$ has a solution if $L_A(x) =b$ which now means that $b \in \mathrm{Im}(L_A)$. Therefore, the fact that the system of equations $Ax=b$ has a solution is the same that $b$ is in $\mathrm{Im}(L_A)$! Let's continue with the matrix from the first example.

$$\begin{align*} L_A(x) &= xL_A(e_1) + y L_A(e_2 ) + zL_A(e_3) = xAe_1 +yAe_2 + zAe_3 \\ &=x \begin{pmatrix} 6 \\15 \end{pmatrix} +y \begin{pmatrix} 2 \\ 5 \end{pmatrix} +z \begin{pmatrix} 8 \\ 20 \end{pmatrix}\\ &=3x \begin{pmatrix} 2 \\5 \end{pmatrix} +y \begin{pmatrix} 2 \\ 5 \end{pmatrix} +4z \begin{pmatrix} 2 \\ 5 \end{pmatrix}\\ &= (3x+y+4z)\begin{pmatrix} 2 \\ 5 \end{pmatrix} \end{align*}$$

Then $(2, 5)^T$ generates the range of $L_A$, and so it is a line:

Range properties

Transformations that occupy the entire W space are called surjective. This means that $\mathrm{Im}(T) = W$. In other words, surjective transformations have the biggest possible range.

Linear transformations "preserve" spanning sets:

You can use this result to calculate the rank of a matrix. Since $L_A(e_i) = A(e_i)$ is the ith column of $A$, it follows immediately that the range of $L_A$ is the space spanned by the columns of $A$. The dimension of $\mathrm{Im}(L_A)$ is known as the rank of $A$.

In summary, injective transformations have the smallest null space and preserve linear independence. On the other hand, surjective transformations have the largest range and preserve the spanning sets.

Putting it all together, bijective transformations (injective and surjective simultaneously) preserve linearly independent spanning sets, in other words, they convert bases of V to bases of W! They are so important that they have their own name, isomorphisms.

Equilibrium

Up to this point, it seems pretty clear that the kernel and image tell us several things about the behavior of T and have similar properties.

However, their connection is so intimate that they balance each other: the smaller one is, the bigger the other. More precisely, as $\ker(T)$ is a subspace of $V$, it's clear that $\dim \ker(T) \leq \dim V$. Well, it turns out that $\dim \mathrm{Im}(T)$ is the number that complements $\dim \ker(T)$ in the sense that it is the exact amount that is missing to reach $\dim V$, in the sense that $\dim \, \ker(T) + \dim \, \mathrm{Im}(T) = \dim \, V$.

This is precisely the meaning of the last theorem, which is both a theoretical and a practical tool that you will use frequently in the future. It receives a name that is not at all modest: the dimension theorem (also known as the fundamental theorem of linear transformations).

As a first sample of the power of this theorem, you can show really easily that when $\dim(V)=\dim(W)$ it is equivalent that $T$ is injective, surjective and bijective!

Here are a couple of applications to systems of linear equations. It is usually quite easy to know the null space of $L_A$, so from the Dimension Theorem, you can quickly calculate the rank of $A$ and then determine if it is a line, a plane, or something else.

On the other hand, when the matrix is squared of $n\times n$, you can interpret its rank as a measure of how likely it is to find unique solutions. The larger the rank, the closer $\mathrm{Im}(L_A)$ is to $\mathbb{R}^n$, and when $rank=n$ we get $\mathrm{Im}(L_A) =\mathbb{R}^n$. From the last theorem, you know that this means that $L_A$ is bijective, so $L_A(x)=b$ is satisfied by a unique $x$. This simplifies to the fact that the system $Ax=b$ has a solution and that it is also unique.

Conclusion

• The null space of $T$ is $\ker(T) = \{v \in V: T(v) =0\}$.
• The solutions of a homogeneous system of linear equations are the null space of the matrix of the system.
• $T$ is injective if and only if $\ker(T)=\{0\}$.
• The range of $T$ is $\mathrm{Im}(T) = \{ T(v) : \ v \in V \}$.
• An inhomogeneous system of linear equations $Ax=b$ has a solution if and only if $b \in \mathrm{Im}(L_A)$.

• $T$ is surjective if and only if $\mathrm{Im}(T) = W$ .

• The dimension theorem states that $\dim \, V = \dim \, \ker(T) + \dim \, \mathrm{Im}(T)$