The matrix of a linear transformation

The matrix of a linear transformation is a matrix for which \(T(\vec{x}) = A\vec{x}\), for a vector \(\vec{x}\) in the domain of T. This means that applying the transformation T to a vector is the same as multiplying by this matrix. Such a matrix can be found for any linear transformation T from \(R^n\) to \(R^m\), for fixed value of n and m, and is unique to the transformation. In this lesson, we will focus on how exactly to find that matrix A, called the standard matrix for the transformation.

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How to find the matrix of a linear transformation

In order to find this matrix, we must first define a special set of vectors from the domain called the standard basis. The big concept of a basis will be discussed when we look at general vector spaces. For now, we just need to understand what vectors make up this set.

The standard basis for \(R^2\) is:

\(\vec{e_1} = \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}\) , \(\vec{e_2} = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}\)

The standard basis for \(R^3\) is:

\(\vec{e_1} = \begin{bmatrix} 1 \\ 0 \\ 0\\ \end{bmatrix}\) , \(\vec{e_2} = \begin{bmatrix} 0 \\ 1 \\ 0\\ \end{bmatrix}\) , \(\vec{e_3} = \begin{bmatrix} 0 \\ 0 \\ 1\\ \end{bmatrix}\)

See the pattern? We can define the standard basis like this for any \(R^n\).

The standard matrix of a transformation \(T:R^n \rightarrow R^m\) has columns \(T(\vec{e_1})\), \(T(\vec{e_2})\), … , \(T(\vec{e_n})\), where \(\vec{e_1}\),…,\(\vec{e_n}\) represents the standard basis. That is:

\(T(\vec{x}) = A \vec{x} \iff A = \left[T(\vec{e_1})\;\; T(\vec{e_2})\;\; \cdots \;\; T(\vec{e_n})\right]\)

Therefore, to find the standard matrix, we will find the image of each standard basis vector. This is shown in the following example.

Example

Find the standard matrix for the transformation T where:
\(T\left(\begin{bmatrix} x_1 \\ x_2\\ x_3\\ \end{bmatrix}\right) = \begin{bmatrix} x_1 – x_2 \\ 2x_3\\ \end{bmatrix}\)

Solution

T takes vectors with three entries to vectors with two entries. Therefore:

\(T:R^3 \rightarrow R^2\)

So, the domain of T is \(R^3\). To find the columns of the standard matrix for the transformation, we will need to find:

\(T(\vec{e_1})\), \(T(\vec{e_2})\), and \(T(\vec{e_3})\)

Using the given rule for T:

\(\begin{align}T(\vec{e_1}) &= T\left(\begin{bmatrix} 1 \\ 0\\ 0\\ \end{bmatrix}\right)\\ &= \begin{bmatrix} 1 – 0 \\ 2(0)\\ \end{bmatrix}\\ &= \begin{bmatrix} 1 \\ 0\\ \end{bmatrix}\end{align}\)

\(\begin{align}T(\vec{e_2}) &= T\left(\begin{bmatrix} 0 \\ 1\\ 0\\ \end{bmatrix}\right)\\ &= \begin{bmatrix} 0 – 1 \\ 2(0)\\ \end{bmatrix}\\ &= \begin{bmatrix} -1 \\ 0\\ \end{bmatrix}\end{align}\)

\(\begin{align}T(\vec{e_3}) &= T\left(\begin{bmatrix} 0 \\ 0\\ 1\\ \end{bmatrix}\right)\\ &= \begin{bmatrix} 0 – 0 \\ 2(1)\\ \end{bmatrix}\\ &= \begin{bmatrix} 0 \\ 2\\ \end{bmatrix}\end{align}\)

Using these as our columns, the standard matrix for T is:

\(A = \begin{bmatrix} 1 & -1 & 0\\ 0 & 0 & 2\\ \end{bmatrix}\)

Checking our answer

We can easily check that we have a matrix which implements the same mapping as T. If we are correct, then:

\(T(\vec{x}) = A\vec{x}\)

So let’s check! We just need to verify that when we plug in a generic vector \(\vec{x}\), that we get the same result as when we apply the rule for T.

\(\begin{align} A\vec{x} &= \begin{bmatrix} 1 & -1 & 0\\ 0 & 0 & 2\\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix}\\ &= x_1\begin{bmatrix}1\\0\\ \end{bmatrix} + x_2\begin{bmatrix}-1\\0\\ \end{bmatrix} + x_3\begin{bmatrix}0\\2\\ \end{bmatrix}\\ &= \begin{bmatrix}x_1 – x_2\\ 2x_3\\ \end{bmatrix}\end{align}\)

Compare this to the rule for T from the problem:

\(T\left(\begin{bmatrix} x_1 \\ x_2\\ x_3\\ \end{bmatrix}\right) = \begin{bmatrix} x_1 – x_2 \\ 2x_3\\ \end{bmatrix}\)

These are the same, so we have in fact found the matrix where \(T(\vec{x}) = A\vec{x}\).

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Summary

For any linear transformation T between \(R^n\) and \(R^m\), for some \(m\) and \(n\), you can find a matrix which implements the mapping. This means that multiplying a vector in the domain of T by A will give the same result as applying the rule for T directly to the entries of the vector. There is only one standard matrix for any given transformation, and it is found by applying the matrix to each vector in the standard basis of the domain.

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