Lecture 17 - Continuing Matrices

Recall:

Where $V, W$ are finite dimensional, and $v_{1}, . . ., v_{n}$ is a basis for $V$ and likewise $w_{1}, . . ., w_{m}$ is a basis for $W$ . We constructed our transformation matrix via:

\begin{aligned} T (v_{1}) & = A_{11} w_{1} + A_{21} w_{2} + \dots + A_{m 1} w_{m} \\ T (v_{2}) & = A_{12} w_{1} + A_{22} w_{2} + \dots + A_{m 2} w_{m} \\ ⋮ \\ T (v_{n}) & = A_{1 n} w_{1} + A_{2 n} w_{2} + \dots + A_{m n} w_{m} \end{aligned}

creating our matrix representation of a map:

M (T, {v_{1}, . . ., v_{n}}, {w_{1}, . . ., w_{m}}) = [\begin{matrix} A_{11} & A_{12} & \dots & A_{1 n} \\ A_{21} & A_{22} & \dots & A_{2 n} \\ ⋮ & ⋱ \\ A_{m 1} & A_{m 2} & \dots & A_{m n} \end{matrix}]

namely the first equation $T (v_{1}) = . . .$ becomes the first column of the $M$ matrix we get. Note that we get $A_{i j}$ where:

$i$ is the row
$j$ is the column
$i$ represents the $w_{i}$ we are scaling and adding with the other vectors
We get a $m \times n$ matrix, where $m = dim (W)$ and $n = dim (V)$ . Note that this has to be this way, since the input of the matrix needs to be the number of columns of our matrix, which is $n$ , which is the dimensional size of our input. Hence, we have to do the flipping over the diagonal like we did before.

Let's look at an example of a different linear transformation.

Let $T : R^{3} \to R^{3}$ , be defined by reflection across the plane $2 x - y + z = 0$ .

We won't do the proof itself of additivity and homogeneity, but consider the following:

But let's find *a* matrix representation of $T$. Notice that matrix depends on the constructions from the bases given, ie $\mathcal{B}$ and $\mathcal{C}$ so we can choose any bases.

Thus, we could choose $e_{1}, e_{2}, e_{3}$ as our bases for $B, C$ . But we'd have to find the reflection of each over the plane, which is a lot of work. Instead, we could just choose the normal vector $\vec{n}$ and two vectors in the plane $v_{1}, v_{2}$ .

So let's not use the standard basis. Can we find two, non-parallel, non-zero vectors on the plane? Let's just choose one vector $v_{1} = ⟨ 1, 2, 0 ⟩$ . Let's also just choose another vector $v_{2} = ⟨ 2, 5, 1 ⟩$ . These vectors are on the plane, and clearly non-parallel since they're LI.

Thus, choose $\vec{n} = ⟨ 2, - 1, 1 ⟩$ as our last basis vector, so choose $B = {\vec{n}, v_{1}, v_{2}}$ . Notice that our domain and co-domain are the same, so we can use $B$ for both. Thus, look at the transformations for all our vectors:

\begin{aligned} T (\vec{n}) & =__\vec{n} +__v_{1} +__v_{2} \\ T (\vec{v_{1}}) & =__\vec{n} +__v_{1} +__v_{2} \\ T (\vec{v_{2}}) & =__\vec{n} +__v_{1} +__v_{2} \end{aligned}

We know the general geometry of this, so we can fill our blanks intuitively:

\begin{aligned} T (\vec{n}) & = - 1 \vec{n} + 0 v_{1} + 0 v_{2} \\ T (\vec{v_{1}}) & = 0 \vec{n} + 1 v_{1} + 0 v_{2} \\ T (\vec{v_{2}}) & = 0 \vec{n} + 0 v_{1} + 1 v_{2} \end{aligned}

Thus extract the matrix:

M (T, B, B) = [\begin{matrix} - 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]

Notice that the order here matters! The column number indicates which basis vector number in the range we map to. $B$ is an ordered list, not a set, so the order here is important.

One last thing of note is that, had we choose a different basis, like the standard basis ${e_{1}, e_{2}, e_{3}}$ , we could still find the matrix representation of our map:

M (T, {e_{1}, e_{2}, e_{3}}, {e_{1}, e_{2}, e_{3}}) \neq previous matrix

However, these matrices do have similar properties. In this way, these matrices are similar matrices.

Something Else

Consider:

We won't prove this, but $S \circ T$ is a linear map here. So if we have some $V = v_{1}, . . ., v_{n}$ as a basis for $V$ , and $W = w_{1}, . . ., w_{p}$ is a basis for $W$ , and $U = u_{1}, . . ., u_{m}$ is a basis for $U$ . Let's say we have $M (T, V, W)$ . We also have $M (S, W, U)$ . Let's denote the matrices $A, B$ correspondingly.

Suppose we want to compute $M (S \circ T, V, U)$ . How would we do this? Let's calculate $S \circ T (v_{j})$ for some $v_{j} \in V$ :

\begin{aligned} (S \circ T) (v_{j}) & = S (T (v_{j})) \\ = S (\sum_{k = 1}^{p} B_{k j} {\vec{w}}_{k}) \\ = \sum_{k = 1}^{p} B_{k j} S ({\vec{w}}_{k}) \\ = \sum_{k = 1}^{p} B_{k j} \sum_{i = 1}^{m} A_{i k} \vec{u_{i}} \\ = \sum_{i = 1}^{m} (\sum_{k = 1}^{p} A_{i k} B_{k j}) {\vec{u}}_{i} \end{aligned}

So we can just define the matrix $C$ such that the $i j$ -th entry is $\sum_{k = 1}^{p} A_{i k} B_{k j}$ . This correspondingly is an $m \times n$ matrix, and this definition shouldn't suprise you since this is the matrix multiplication definition!!!!