# Find an Orthonormal Basis of the Range of a Linear Transformation

## Problem 478

Let $T:\R^2 \to \R^3$ be a linear transformation given by

\[T\left(\, \begin{bmatrix}

x_1 \\

x_2

\end{bmatrix} \,\right)

=

\begin{bmatrix}

x_1-x_2 \\

x_2 \\

x_1+ x_2

\end{bmatrix}.\]
Find an orthonormal basis of the range of $T$.

(*The Ohio State University, Linear Algebra Final Exam Problem*)

Contents

## Solution.

Let $A$ be the matrix representation of the linear transformation of $T$.

That is,

\[A=\begin{bmatrix}

T(\mathbf{e}_1) & T(\mathbf{e}_2)

\end{bmatrix},\]
where

\[\mathbf{e}_1=\begin{bmatrix}

1 \\

0

\end{bmatrix}, \mathbf{e}_2=\begin{bmatrix}

0 \\

1

\end{bmatrix}\]
form the standard basis of the vector space $R^2$.

By the formula, we see that

\[T\left(\, \begin{bmatrix}

1 \\

0

\end{bmatrix} \,\right)

=\begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix}, T\left(\, \begin{bmatrix}

0 \\

1

\end{bmatrix}\,\right)=\begin{bmatrix}

-1 \\

1 \\

1

\end{bmatrix},\]
and thus the matrix $A$ for $T$ is

\[A=\begin{bmatrix}

1 & -1 \\

0 & 1 \\

1 &1

\end{bmatrix}.\]

Note that the range of $T$ is the same as the range of $A$.

We reduce the matrix $A$ by the elementary row operations as follows:

\begin{align*}

A=\begin{bmatrix}

1 & -1 \\

0 & 1 \\

1 &1

\end{bmatrix}

\xrightarrow{R_3-R_1}

\begin{bmatrix}

1 & -1 \\

0 & 1 \\

0 & 0

\end{bmatrix}

\xrightarrow{R_1+R_2}

\begin{bmatrix}

1 & 0 \\

0 & 1 \\

0 & 0

\end{bmatrix}.

\end{align*}

Since the both columns contain the leading $1$’s, we conclude that

\[\left\{\, \begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix}, \begin{bmatrix}

-1 \\

1 \\

1

\end{bmatrix} \,\right\}\]
is a basis of the range of $A$ by the leading $1$ method.

Note that the dot product of these basis vectors is $0$, hence they are already orthogonal.

Hence to obtain an orthonormal basis, we just need to normalize the length of these vectors to $1$.

In summary, an orthonormal basis of the range of $T$ is

\[\left\{\, \frac{1}{\sqrt{2}}\begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix}, \frac{1}{\sqrt{3}}\begin{bmatrix}

-1 \\

1 \\

1

\end{bmatrix} \,\right\}.\]

## Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).

The other problems can be found from the links below.

- Find All the Eigenvalues of 4 by 4 Matrix
- Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue
- Diagonalize a 2 by 2 Matrix if Diagonalizable
- Find an Orthonormal Basis of the Range of a Linear Transformation (This page)
- The Product of Two Nonsingular Matrices is Nonsingular
- Determine Wether Given Subsets in ℝ4 R 4 are Subspaces or Not
- Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials
- Find Values of $a , b , c$ such that the Given Matrix is Diagonalizable
- Idempotent Matrix and its Eigenvalues
- Diagonalize the 3 by 3 Matrix Whose Entries are All One
- Given the Characteristic Polynomial, Find the Rank of the Matrix
- Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$
- Determine Whether There Exists a Nonsingular Matrix Satisfying $A^4=ABA^2+2A^3$

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