Linear Transformations (Simplified with Matrix Representation)

What if we don’t want to map from $R^{n}$ to $R^{m}$ but instead, we want to find a representation for any linear transformation between two finite-dimensional vector spaces. To do this, we must combine powers: eigenvectors and linear transformations.

Jenn (B.S., M.Ed.) of Calcworkshop® teaching linear transformations

Jenn, Founder Calcworkshop^®, 15+ Years Experience (Licensed & Certified Teacher)

If you may recall from our previous studies of transformations, if

$\begin{aligned} T : R^{n} & \to R^{m} \end{aligned}$

is a linear transformation, then there is an $m \times n$ matrix $A$ such that

$\begin{aligned} T (x) & = A x \end{aligned}$

for all $x$ in $R^{n}$ , where

$\begin{aligned} A & = [\begin{array}{llll} T (\vec{e_{1}}) & T (\vec{e_{2}}) & \dots & T (\vec{e_{n}}) \end{array}] \end{aligned}$

From Standard Basis to Coordinate Vectors

As we have seen, using the standard basis has been most convenient when performing a matrix transformation. But it comes up lacking when we want to map between vector spaces.

So, rather than using the standard basis, such that

$\begin{aligned} \vec{x} & \mapsto A \vec{x} \end{aligned}$

we will use the mapping

$\begin{aligned} \vec{u} & \mapsto D \vec{u} \end{aligned}$

which is essentially the same mapping just viewed from a different perspective.

How?

By representing vectors by their coordinate vectors for a basis!

Okay, let

$\begin{aligned} V & be an n -dimensional vector space, \\ W & be an m-dimensional vector space and \\ T & be the linear transformation from V to W \end{aligned}$

$\begin{aligned} T : V & \to W \end{aligned}$

Now, if

$\begin{aligned} B & = {{\vec{b}}_{1}, \dots, \vec{b_{n}}} \end{aligned}$

be an ordered basis for $V$ and

$\begin{aligned} C & = {\vec{c_{1}}, \dots, \vec{c_{n}}} \end{aligned}$

be the ordered basis for W, then we can express each vector $\vec{x}$ in $V$ as a linear combination

$\begin{aligned} \vec{x} & = r_{1} \vec{b_{1}} + \dots + r_{n} \vec{b_{n}} \end{aligned}$

for unique scalar weights $r$ .

Thus, we can represent the vector $\vec{x}$ in $V$ by its coordinate vector in $R^{n}$ as

$\begin{aligned} _{B} & = [\begin{array}{c} r_{1} \\ ⋮ \\ r_{n} \end{array}] \end{aligned}$

Likewise, we can represent each vector $\vec{y}$ in $W$ by its coordinate vector

$\begin{aligned} _{C} & in R^{m} \end{aligned}$

Matrix Representation of a Linear Transformation

Therefore, if we want to find a $m \times n$ matrix $M$ such that the linear transformation $T$ is equivalent to the coordinate vectors of $M$ then

$\begin{aligned} _{C} & = M [\vec{x}]_{B} \end{aligned}$

where

$\begin{aligned} M & = [{[T ({\vec{b}}_{1})]}_{C} {[T ({\vec{b}}_{2})]}_{C} \dots {[T (\vec{b_{n}})]}_{C}] \end{aligned}$

The matrix $M$ is a matrix representation of $T$ and is called the matrix for $T$ relative to the basis $B$ and $C$ .

Alright, this is great and all, but how does this really work? I’m confused.

Here’s how this work.

The Power of Diagonalization

In a problem involving $R^{n}$ , a linear transformation $T$ typically appears first as a matrix transformation. But if $A$ is diagonalizable, then there is a basis $B$ for $R^{n}$ consisting of eigenvectors of $A$ . In this case, the B-matrix for $T$ is diagonal. Therefore, by diagonalizing $A$ we are in essence finding a diagonal matrix representation of $\vec{x} \mapsto \vec{A x}$

So, if $A = P D P^{- 1}$ , as we learned in our Diagonalization video, where D is a diagonal $n \times n$ matrix and if $B$ is the basis for $R^{n}$ formed from the columns of $P$ , which represent the eigenvectors of $A$ , then $D$ is the B-matrix for the transformation.

Example: Finding the B-Matrix with Eigenvectors

For example, suppose $T : R^{2} \to R^{2}$ by

$\begin{aligned} T (\vec{x}) & = A \vec{x} \end{aligned}$

where

$\begin{aligned} A & = [\begin{array}{cc} 2 & 7 \\ 1 & - 4 \end{array}] \end{aligned}$

Let’s find a basis B for $R^{2}$ with the property that the $B$ -matrix for $T$ is a diagonal matrix.

First, we know that

$\begin{aligned} A & = P D P^{- 1} \end{aligned}$

so we will need to find our D matrix.

$\begin{aligned} \det [\begin{array}{cc} 2 - λ & 7 \\ 1 & - 4 - λ \end{array}] & = 0 \\ (2 - λ) (- 4 - λ) - 7 & = 0 \\ λ^{2} + 2 λ - 15 & = 0 \\ (λ - 3) (λ + 5) & = 0 \\ λ & = 3, - 5 \\ D & = [\begin{array}{cc} 3 & 0 \\ 0 & - 5 \end{array}] \end{aligned}$

Now we will find our $P$ matrices consisting of the eigenvectors of $A$ .

$\begin{aligned} λ & = 3 \\ [\begin{array}{ccc} 2 - 3 & 7 & 0 \\ 1 & - 4 - 3 & 0 \end{array}] & = [\begin{array}{ccc} - 1 & 7 & 0 \\ 1 & - 7 & 0 \end{array}] \\ \sim [\begin{array}{ccc} 1 & - 7 & 0 \\ 0 & 0 & 0 \end{array}] & \Rightarrow \vec{x} = x_{2} [\begin{array}{c} 7 \\ 1 \end{array}] \\ λ & = - 5 \\ [\begin{array}{ccc} 2 - (- 5) & 7 & 0 \\ 1 & - 4 - (- 5) & 0 \end{array}] & = [\begin{array}{lll} 7 & 7 & 0 \\ 1 & 1 & 0 \end{array}] \\ \sim [\begin{array}{lll} 1 & 1 & 0 \\ 0 & 0 & 0 \end{array}] & \Rightarrow \vec{x} = x_{2} [\begin{array}{c} - 1 \\ 1 \end{array}] \\ P & = [\begin{array}{cc} 7 & - 1 \\ 1 & 1 \end{array}] \end{aligned}$

Now, all that’s left is to recognize that the D matrix is the B-matrix in disguise!!!

It’s true! The D matrix is the B-matrix for $T$ when P represents the eigenvectors of $A$ .

Awesome, right?

Example: Finding the B-Matrix with Given Matrices

Additionally, we can use A and P matrices to help us find our B-matrix.

Assume

$\begin{aligned} A & = [\begin{array}{cc} 4 & - 9 \\ 4 & 8 \end{array}] \end{aligned}$

and

$\begin{aligned} \vec{b_{1}} & = [\begin{matrix} 3 \\ 2 \end{matrix}] \end{aligned}$

and

$\begin{aligned} \vec{b_{2}} & = [\begin{matrix} 2 \\ 1 \end{matrix}] \end{aligned}$

Let’s find our B-matrix.

Well, we’re given A and $P$ , so what we need to find is $D$ , which is really our B-matrix for the linear transformation.

$\begin{aligned} A & = [\begin{array}{cc} 4 & - 9 \\ 4 & 8 \end{array}] \end{aligned}$

and

$\begin{aligned} P & = [\begin{array}{ll} {\vec{b}}_{1} & \vec{b_{2}} \end{array}] = [\begin{array}{ll} 3 & 2 \\ 2 & 1 \end{array}] \end{aligned}$

And if we resolve

$\begin{aligned} A & = P D P^{- 1} \end{aligned}$

for $D$ , which is our B-matrix, we get

$\begin{aligned} D & = P^{- 1} A P \end{aligned}$

Therefore

$\begin{aligned} B - matrix & = P^{- 1} A P \\ = \underset{P^{- 1}}{\underset{⏟}{\frac{1}{- 1} [\begin{array}{cc} 1 & - 2 \\ - 2 & 3 \end{array}]}} \underset{A}{\underset{⏟}{[\begin{array}{cc} 4 & - 9 \\ 4 & 8 \end{array}]}} \underset{P}{\underset{⏟}{[\begin{array}{ll} 3 & 2 \\ 2 & 1 \end{array}]}} \\ = [\begin{array}{cc} - 2 & 1 \\ 0 & - 2 \end{array}] \end{aligned}$

See, not so bad!

Next Steps

In this lesson, you will:

Understand how diagonalization applies to linear transformation using coordinate vectors
Learn how the matrix of a linear transformation associates with ordered bases $B$ and $C$
Discover how the matrix for $T$ relative to $B$ (B-Matrix) is used for transforming polynomials
Make connections with the diagonal matrix representation and similar matrices, along with eigenvalues (D-Matrix or B-Matrix) and eigenvectors (P matrix or basis)

Dive in and start exploring!

Video Tutorial w/ Full Lesson & Detailed Examples

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