Linear Transformations in Bases

Given a linear transformation $T: V \to W$, we can completely capture all of its information by just knowing what the image is of a basis set for $V$.

Lemma. Suppose $S,T : V \to W$ are two linear transformation between $k$-vector space and that $v_i$ for $i \in B$ is a basis for the vector space $V$. If $S(v_i) = T(v_i)$ for all $i \in B$, then $S(v) = T(v)$ for any $v \in V$.

Proof. (Expand to view)

For any $v \in V$, we can write it as a sum $v = \sum_{i \in B} c_i v_i$ with only finitely many of the $c_i \neq 0$. Since $S$ is a linear transformation, we have $S(v) = \sum_{i \in B} c_i S(v_i)$ From our assumption, $\sum_{i \in B} c_i S(v_i) = \sum_{i \in B} c_i T(v_i)$ Using the fact tht $T$ is a linear transformation, we get $\sum_{i \in B} c_i T(v_i) = T(v)$ So we can conclude that $S(v) = T(v)$. ■

The previous lemma tells us that two linear transformations which do the same thing on a basis are equal as functions. The set of values $T(v_i)$ for $i \in B$ completely captures all information about $T$.

Given a linear transformation $T$, we can ask: how special are the values $T(v_i) \in W$? Do they have to some special property to arise as the image a linear transformation applied to a basis? Or can they be any possible set of vectors in $W$ indexed by $B$?

It turns out there is no condition. Given such a collection of vectors in $W$, we can build a unique linear transformation from it.

Lemma. Let $V$ and $W$ be $k$-vector spaces and assume we have a basis $v_i$ for $i \in B$ of $V$. Given any choice of vectors $w_i \in W$ for $i \in B$, there is a unique linear transformation $T: V \to W$ with $T(v_i) = w_i ~\text{for all } i \in B.$

Proof. (Expand to view)

Remember that with a basis, we can write any $v \in V$ uniquely as a linear combination of the $v_i$: $v = \sum_{i \in B} c_i v_i$ for a unique choice of scalars $c_i \in k$ and only finitely many are nonzero. We define $\begin{aligned} T : V & \to W \\ v & \mapsto \sum_{i \in B} c_i w_i \end{aligned}$ Since the $c_i$ are uniquely determined by $v$ and the $v_i$, this is a well-defined function.

We need to check it is a linear transformation. First notice that $cv = c\left(\sum_{i \in B} c_i v_i \right) = \sum_{i \in B} (cc_i) v_i$ Since this is one linear combination expressing $cv$ in terms of $v_i$ and there is a unique one, this must be it. By definition, $T(cv) = \sum_{i \in B} (cc_i) w_i$ Pulling out the $c$ gives $\sum_{i \in B} (cc_i) w_i = c\left(\sum_{i \in B} c_i w_i \right) = cT(v)$ So $T$ commutes with scalar multiplication.

Now, we turn to addition. We write $v = \sum_{i \in B} c_i v_i \\ v^\prime = \sum_{i \in B} c_i^\prime v_i$ Then $v + v^\prime = \sum_{i \in B} (c_i+c_i^\prime) v_i$ By definition of $T$, we have $T(v + v^\prime) = \sum_{i \in B} (c_i + c_i^\prime) w_i$ Expanding this out gives $\sum_{i \in B} (c_i + c_i^\prime) w_i = \sum_{i \in B} c_i w_i + \sum_{i \in B} c_i^\prime w_i = T(v) + T(v^\prime)$ So $T$ also commutes with addition. Thus, $T$ is a linear transformation.

We have made one linear transformation with $T(v_i) = w_i$ for all $i \in B$. The previous lemma tells us it is uniquely determined by this property. ■

So a linear transformation is completely determined by the images of a basis and for any choice of images of the basis we can extend the choice to a linear transformation with that action on the basis.

Note that linear transformations make no reference to any basis in their definition but introducing a basis in their presence allows us great insight. If we take a basis for $W$, we arrive back at a familiar friend.

Definition. Let $T: V \to W$ be a linear transformation. Given bases $v_i, i \in B$ and $w_j, j \in C$, then we can uniquely write $T(v_i) = \sum_{j \in C} A_{ji} w_j$ for some scalars $A_{ij} \in k$. This is called the matrix representation of $T$ in the bases $B$ and $C$.

If $\dim V = n$ and $\dim W = m$, then we get a familiar $m \times n$ matrix $A$.

Lemma. Let $V$ and $W$ be $k$-vector spaces with $\dim V = n$ and $\dim W = m$. A linear transformation is completely determined by its matrix representation. Moreover, for each $m \times n$ matrix $A$, there is unique linear transformation $T: V \to W$ whose matrix representation $A$.

Proof. (Expand to view)

Fix bases $v_1,\ldots,v_n$ for $V$ and $w_1,\ldots,w_m$ for $W$. If $S,T : V \to W$ have the same matrix representations, we have $S(v_i) = T(v_i)$ and thus $S = T$ by a previous lemma.

For a matrix $A$, we know there is a unique linear transformation $T: V \to W$ with $T(v_i) = \sum_{j \in C} A_{ji} w_j$ from a previous lemma. ■

So the matrix representation completely determines a linear transformation. But, take care because the matrix representation of $T$ depends very much on which bases we choose for $V$ and $W$.

Example. Consider the matrix $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$ As we have seen, this determines a linear transformation $\begin{aligned} T: \mathbb{R}^3 & \to \mathbb{R}^2 \\ \mathbf{v} & \mapsto A \mathbf{v} \end{aligned}$ If we take the standard bases for both $\mathbb{R}^3$ and $\mathbb{R}^2$, then matrix representation of $T$ is exactly $A$.

But what happens if we take different bases? Let’s take the basis $v_1 := \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, v_2 := \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, v_3 := \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ for $\mathbb{R}^3$ and $w_1 := \begin{pmatrix} 1 \\ -1 \end{pmatrix}, w_2 := \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ for $\mathbb{R}^2$.

To determine the matrix representation in our a new bases, we first compute $Av_1 = \begin{pmatrix} 1 \\ 4 \end{pmatrix}, \ Av_2 = \begin{pmatrix} 3 \\ 7 \end{pmatrix}, \ Av_3 = \begin{pmatrix} 6 \\ 15 \end{pmatrix}$

Now, we need to express each $Av_i$ as linear combination of the $v_i$. Doing this is equivalent to solving the linear systems $\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \mathbf{c}_i = Av_i$ Solving this gives $\mathbf{c}_1 = \frac{1}{2} \begin{pmatrix} -3 \\ 5 \end{pmatrix}, \ \mathbf{c}_2 = \frac{1}{2} \begin{pmatrix} -6 \\ 12 \end{pmatrix}, \ \mathbf{c}_3 = \frac{1}{2} \begin{pmatrix} -9 \\ 21 \end{pmatrix}$ So our matrix representation is $\frac{1}{2} \begin{pmatrix} -3 & -6 & -9 \\ 5 & 12 & 21 \end{pmatrix}$

This is the straightforward way to proceed for computing the matrix representation of a linear transformation $\begin{aligned} T: \mathbb{R}^n & \to \mathbb{R}^m \\ \mathbf{v} & \mapsto A\mathbf{v} \end{aligned}$ but there is a more efficient way. We had to solve three linear systems with the same coefficient matrix but different inhomogeneous terms. It is better to compute the inverse of $D = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ Then $\mathbf{c}_i = D^{-1} A v_i$

We can interpret this in terms change of basis matrices.

The matrix $D$ represents writing the basis $\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ in terms of the standard basis $\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \ \begin{pmatrix} 0 \\ 1 \end{pmatrix}$

Similarly, we have the matrix $C$ which represents writing the basis $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ in terms of the standard basis $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ Note that $Av_i = (AC)_i$ Thus, the matrix representation of $A$ in the new bases is given by the product $D^{-1} A C$

More generally, we have the following way to related two matrix representations of $T$.

Lemma. Let $T: V \to W$ be a linear transformation between finite dimensional vector spaces. Assume we have bases $v_1,\ldots, v_n$ and $v_1^\prime, \ldots, v_n^\prime$ of $V$ and bases $w_1,\ldots,w_m$ and $w_1^\prime,\ldots,w_m^\prime$. Assume $A$ is the matrix representation of $T$ in the bases $v_1,\ldots, v_n$ and $w_1,\ldots,w_m$. Then the matrix representation of $T$ in the bases $v_1^\prime, \ldots, v_n^\prime$ for $V$ and $w_1^\prime,\ldots,w_m^\prime$ of $W$ is given by the product $D^{-1} A C$ where $C$ is the change of basis matrix from $v_1,\ldots, v_n$ to $v_1^\prime, \ldots, v_n^\prime$ and $D$ is the change of basis matrix from $w_1,\ldots,w_m$ to $w_1^\prime,\ldots,w_m^\prime$.

Proof. (Expand to view)

To find the matrix representation of $T$ for $v_1^\prime, \ldots, v_n^\prime$ and $w_1^\prime,\ldots,w_m^\prime$, we write $Tv_i^\prime$ as linear combination of the $w_1^\prime,\ldots,w_m^\prime$ for each $i$ and record the coefficients in a matrix.

We know that if we do this with $v_1, \ldots, v_n$ and $w_1,\ldots,w_m$, then we get $A$ as our matrix.

We have $v_j^\prime = \sum_{i=1}^n C_{ij} v_i$
and $w_j = \sum_{i=1}^m D_{ij}^{-1} w_i^\prime$ If we apply $T$ to $v_j^\prime$, we have $T(v_j^\prime) = T\left( \sum_{i=1}^n C_{ij} v_i \right) = \sum_{i=1}^n C_{ij} T(v_i)$ Then, using $A$ we have $\begin{aligned} T(v_j^\prime) & = \sum_{i=1}^n C_{ij} \left( \sum_{l=1}^m A_{li}w_l \right) \\ & = \sum_{i=1}^n C_{ij} \left( \sum_{l=1}^m A_{li} \left(\sum_{s=1}^m D_{sl}^{-1} w_s^\prime \right) \right) \end{aligned}$ So the coefficient in the expansion of $T(v_j^\prime)$ for $w_s^\prime$ is $\sum_{l=1}^m \sum_{i=1}^n D_{sl}^{-1}A_{li}C_{ij} = (D^{-1}AC)_{sj}$ which proves the claim. ■

Some things to keep in mind:

One linear transformation can have many matrix representations! It all depends on the bases.
One matrix can give many different linear transformations! It all depends on the bases.

Two matrices that differ by changing bases represent the same linear transformation and therefore share many features in common. In particular, if we have any idea or property we can derive from the definition of a linear transformation alone (no bases allowed), then these two matrices should share that idea or property.

We will look at what some of these properties can be soon. But to discuss these relationships properly we need a new notion relating vector spaces: isomorphisms