# Math 31AH: Lecture 11

We met linear transformations for the first time back in Lecture 2, and have encountered them a handful of times in subsequent lectures. In this lecture, we begin a systematic study of linear transformation that will take up the rest of the course. As with the study of any new species, there will be a lot of classifying: we will divide up linear transformations into various classes defined by certain common features which to a large extent determine their behavior. Before classifying linear transformations into various special species and families, let us consider the ecosystem of all linear transformations as a collective whole.

Let $\mathbf{V}$ and $\mathbf{W}$ be vector spaces, and let $\mathrm{Hom}(\mathbf{V},\to\mathbf{W})$ be the set of all linear transformations $T \colon \mathbf{V} \to \mathbf{W}.$ This seemingly strange notation stems from the fact that is a fancy alternative name for linear transformations is homomorphisms, which roughly means
“same shape” in Greek.

Theorem 1: $\mathrm{Hom}(\mathbf{V},\mathbf{W})$ is a vector space.

Proof: In order to promote $\mathrm{Hom}(\mathbf{V},\mathbf{W})$ from simply a set to a vector space, we need to give a rule for adding and scaling linear transformations. These rules are simply and natural. If $T_1,T_2 \in \mathrm{Hom}(\mathbf{V},\mathbf{W})$ linear transformations and $a_1,a_2 \in \mathbb{R}$ are scalars we define a new linear transformation $a_1T_1 + a_2T_2$ by

$[a_1T_1+a_2T_2]\mathbf{v} := a_1T_1\mathbf{v} + a_2T_2\mathbf{v} \quad \forall \mathbf{v} \in \mathbf{V}.$

One must check first that the function from $\mathbf{V}$ to $\mathbf{W}$ so defined satisfies the linear transformation axioms, and then that the set $\mathrm{Hom}(\mathbf{V},\mathbf{W})$ equipped with these notions of addition and scalar multiplication satisfies the vector spaces axioms. This is left as an exercise to the reader (it is a long but easy exercise which you should do at least once).

— Q.E.D.

Now let us start asking questions about the vectors in $\mathrm{Hom}(\mathbf{V},\mathbf{W}),$ i.e. considering what properties a linear transformation $T$ from $\mathbf{V}$ to $\mathbf{W}$ may or may not have. Given $T \in \mathrm{Hom}(\mathbf{V},\mathbf{W}),$ one of the first things we would like to know is whether it is injective and/or surjective. These questions reduce to questions about certain subspaces of $\mathbf{V}$ and $\mathbf{W}$ associated to $T.$

Definition 1: The kernel of $T$ is the subset of $\mathbf{V}$ defined by

$\mathrm{Ker}T = \{\mathbf{v} \in \mathbf{V} \colon T(\mathbf{v}) = \mathbf{0}_\mathbf{W}\},$

and the image of $T$ is the subset of $\mathbf{W}$ defined by

$\mathrm{Im}T = \{ \mathbf{w} \in \mathbf{W} \colon \exists\ \mathbf{v} \in \mathbf{V} \text{ such that } T\mathbf{v} = \mathbf{w}\}.$

You can think of the kernel and image of $T$ as the solution sets to two different equations involving $T.$ The kernel in terms of certain vector equations associated to $T.$ First, the kernel of $T$ is the set of solutions to the equation

$T\mathbf{v}=\mathbf{0}_\mathbf{W},$

where $\mathbf{v}$ is an unknown vector in $\mathbf{V}.$ Second, the image of $T$ is the set of all $\mathbf{w} \in \mathbf{W}$ such that the equation

$T\mathbf{v}=\mathbf{w}$

has a solution.

Proposition 1: The kernel of $T$ is a subspace of $\mathbf{V}$, and the image of $T$ is a subspace of $\mathbf{W}.$

Proof: Since $T$ is a linear transformation, it is by definition the case that $T\mathbf{0}_\mathbf{V} = \mathbf{0}_\mathbf{W},$ so that $\mathbf{0}_\mathbf{V} \in \mathrm{Ker}T.$ Now let $\mathbf{v}_1,\mathbf{v}_2 \in \mathrm{Ker}T$ be any two vectors in the kernel of $T,$ and let $a_1,a_2 \in \mathbb{R}$ be any two scalars. We then have

$T(a_1\mathbf{v}_1 + a_2\mathbf{v}_2) = a_1T\mathbf{v}_1+a_2T\mathbf{v}_2=a_1\mathbf{0}_\mathbf{W} + a_2\mathbf{0}_\mathbf{W} = \mathbf{0}_\mathbf{W},$

which shows that $\mathrm{Ker}T$ is closed under taking linear combinations. Hence, $\mathrm{Ker}T$ is a subspace of $\mathbf{V}.$

Now consider the image of $T.$ Since $T$ is a linear transformation, it is by definition the case that $T\mathbf{0}_\mathbf{V} = \mathbf{0}_\mathbf{W},$ so that $\mathbf{0}_\mathbf{W} \in \mathrm{Im}T.$ Now let $\mathbf{w}_1,\mathbf{w}_2 \in \mathrm{Im}T$ be any two vectors in the image of $T.$ Then, by definition of $\mathrm{Im}T,$ there exist vectors $\mathbf{v}_1,\mathbf{v}_2 \in \mathbf{V}$ such that $T\mathbf{v}_1=\mathbf{w}_1,\ T\mathbf{v}_2=\mathbf{w}_2.$ So, for any scalars $a_1,a_2 \in \mathbb{R},$ we have that

$a_1\mathbf{w}_1+a_2\mathbf{w}_2 = a_1T\mathbf{v}_1+a_2T\mathbf{v}_2 = T(a_1\mathbf{v}_1+a_2\mathbf{v}_2),$

which shows that $\mathrm{Im}T$ is closed under taking linear combinations.

— Q.E.D.

Proposition 2: The linear transformation $T$ is injective if and only if $\mathrm{Ker}T = \{\mathbf{0}_\mathbf{V}\},$ and surjective if and only if $\mathrm{Im}T = \mathbf{W}.$

Proof: First we consider the relationship between the injectivity of $T$ and its kernel. Suppose that $\mathrm{Ker}T = \{\mathbf{0}_\mathbf{V}\}.$ Let $\mathbf{v}_1,\mathbf{v}_2 \in \mathbf{V}$ be vectors such that $T\mathbf{v}_1=T\mathbf{v}_2.$ This is equivalent to

$T\mathbf{v}_1-T\mathbf{v}_2=T(\mathbf{v}_1-\mathbf{v}_2)=\mathbf{0}_\mathbf{W},$

which says that $\mathbf{v}_1-\mathbf{v}_2 \in \mathrm{Ker}T.$ Since the only vector in $\mathrm{Ker}T$ is $\mathbf{0}_\mathbf{V},$ this forces $\mathbf{v}_1-\mathbf{v}_2=\mathbf{0}_\mathbf{V},$ which means that $\mathbf{v}_1=\mathbf{v}_2,$ and we conclude that $T$ is injective. Conversely, suppose we know that $T$ is injective. Let $\mathbf{v}_1,\mathbf{v}_2 \in \mathrm{Ker}T$ be any two vectors in the kernel of $T.$ Then $T\mathbf{v}_1=T\mathbf{v}_2=\mathbf{0}_\mathbf{W}.$ Since $T$ is injective, this forces $\mathbf{v}_1=\mathbf{v}_2,$ which says that any two vectors in $\mathrm{Ker}T$ are equal to one another. But this means that any vector in $\mathrm{Ker}T$ is equal to $\mathbf{0}_\mathbf{V},$ and we conclude that $\mathrm{Ker}T=\{\mathbf{0}_\mathbf{V}\}.$

Now let us consider the relationship between the surjectivity of $T$ and its image. Suppose that $\mathrm{Im}T=\mathbf{W}.$ This is exactly what it means for the function $T$ to be surjective: every element in the codomain of $T$ is in fact in the range of $T.$ Conversely, suppose we know that $T$ is surjective. By definition of surjectivity, this means that $\mathrm{Im}T=\mathbf{W}.$

— Q.E.D.

We can use the above propositions to gain some insight into linear equations, meaning equations of the form

$T \mathbf{v}=\mathbf{w},$

where $T$ is a given linear transformation from $\mathbf{V}$ to $\mathbf{W},$ $\mathbf{w}$ is a given vector in $\mathbf{W},$ and $\mathbf{v}$ is an unknown vector in $\mathbf{V}.$

Theorem 1: The solution set of the above linear equation is $\mathbf{v}_0+\mathrm{Ker}T,$ i.e. the set of of all vectors in $\mathbf{V}$ of the form $\mathbf{v}_0 + \mathbf{k},$ where $\mathbf{v}_0$ is any particular vector such that $T\mathbf{v}_0=\mathbf{w},$ and $\mathbf{k} \in \mathrm{Ker}T.$

Proof: We begin by first consider the homogenous equation associated to the linear equation we want to solve, which is

$T\mathbf{v}=\mathbf{0}_\mathbf{W}.$

By definition, the solution set of this homogeneous equation is $\mathrm{Ker}T.$ Now, if $\mathbf{v}_0 \in \mathbf{V}$ is any vector such that $T\mathbf{v}_0 = \mathbf{w},$ then we also have

$T(\mathbf{v}_0+\mathbf{k})= T\mathbf{v}_0+T\mathbf{k}=\mathbf{w}+\mathbf{0}_\mathbf{W} = \mathbf{w},$

which shows that all vectors in $\mathbf{v}_0+\mathrm{Ker}T$ are solutions of the equation $T\mathbf{v}=\mathbf{w}.$

Conversely, suppose that $\mathbf{v}_1$ is a vector in $\mathbf{V}$ such that $T\mathbf{v}_1=\mathbf{w}.$ Then, we have

$T(\mathbf{v}_1-\mathbf{v}_0) = T\mathbf{v}_1-T\mathbf{v}_0=\mathbf{w}-\mathbf{w}=\mathbf{0}_\mathbf{W},$

which shows that $\mathbf{v}_1-\mathbf{v}_0$ is a vector in $\mathrm{Ker}T.$ That is, $\mathbf{v}_1-\mathbf{v}_0=\mathbf{k}$ for some $\mathbf{k} \in \mathrm{Ker}T,$ or equivalently $\mathbf{v}_1=\mathbf{v}_0+\mathbf{k}.$ This shows that all solutions of $T\mathbf{v}=\mathbf{W}$ belong to the set $\mathbf{v}_0 + \mathrm{Ker}T.$

— Q.E.D.

We can now make the following statements about solutions of the linear equation $T\mathbf{v}=\mathbf{w}$:

• The equation has a solution if and only if $\mathbf{w} \in \mathrm{Im}T$.
• If the equation has a solution, this solution is unique if and only if $\mathrm{Ker}T = \{\mathbf{0}_\mathbf{V}\}.$
• If the equation has a solution but the solution is not unique, then the equation has infinitely many solutions.

Exercise 1: Arrange the above information into a tree diagram which shows the relationship between the various cases.

In Lecture 12, we will think in more detail about the structure of the vector space $\mathrm{Hom}(\mathbf{V},\mathbf{W})$ in the case that $\mathbf{V}$ and $\mathbf{W}$ are finite-dimensional. In the finite-dimensional case, we shall see that everything we might want to know can be expressed in the language of matrices.