# Find an Orthonormal Basis of the Given Two Dimensional Vector Space ## Problem 602

Let $W$ be a subspace of $\R^4$ with a basis
$\left\{\, \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \end{bmatrix} \,\right\}.$

Find an orthonormal basis of $W$.

(The Ohio State University, Linear Algebra Midterm) Add to solve later

## Solution.

Let
$\mathbf{v}_1= \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \end{bmatrix} .$ Note that they are not orthogonal as the dot product is
$\mathbf{v}_1\cdot \mathbf{v}_2= \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix}\cdot \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \end{bmatrix}=1\cdot0+ 0\cdot 1+ 1\cdot 1+1\cdot 1=2\neq 0.$

Let us first find an orthogonal basis for $W$ by the Gram-Schmidt orthogonalization process.

Let $\mathbf{w}_1:=\mathbf{v}_1$.
Next, let $\mathbf{w}_2:=\mathbf{v}_2+a\mathbf{v}_1$, where $a$ is a scalar to be determined so that $\mathbf{w}_1\cdot \mathbf{w}_2=0$.
(You may also use the formula of the Gram-Schmidt orthogonalization.)

As $\mathbf{w}_1$ and $\mathbf{w}_2$ is orthogonal, we have
\begin{align*}
0&=\mathbf{w}_1\cdot \mathbf{w}_2=\mathbf{v}_1\cdot(\mathbf{v}_2+a\mathbf{v}_1)\\
&=\mathbf{v}_1\cdot\mathbf{v}_2+a\mathbf{v}_1\cdot\mathbf{v}_1\\
&=2+3a.
\end{align*}
It follows that $a=-2/3$ and
$\mathbf{w}_2=\mathbf{v}_2-\frac{2}{3}\mathbf{v}_1=\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \end{bmatrix}-\frac{2}{3}\begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix}.$

Now, to avoid fractions in our computation, let us consider $3\mathbf{w}_2$, instead of $\mathbf{w}_2$. Note that the scaling does not change the orthogonality.
We have
$3\mathbf{w}_2=3\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \end{bmatrix}-2\begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix}=\begin{bmatrix} -2\\ 3\\ 1 \\1 \end{bmatrix}.$

Thus the set $\{\mathbf{w}_1, 3\mathbf{w}_2\}$ is an orthogonal basis for $W$.
However, the length of these vectors are not $1$ as we see
\begin{align*}
\|\mathbf{w}_1\|&=\sqrt{1^2+0^2+1^2+1^2}=\sqrt{3}\\
\|3\mathbf{w}_2\|&=\sqrt{(-2)^2+3^2+1^2+1^2}=\sqrt{15}.
\end{align*}

Now it suffices to normalize the vectors $\mathbf{w}_1, 3\mathbf{w}_2$ to obtain an orthonormal basis.
Therefore, the set
$\left\{\, \frac{1}{\sqrt{3}} \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix}, \, \frac{1}{\sqrt{15}}\begin{bmatrix} -2\\ 3\\ 1 \\1 \end{bmatrix}\,\right\}.$ is an orthonormal basis for $W$.

## Comment.

This is one of the midterm 2 exam problems for Linear Algebra (Math 2568) in Autumn 2017.

One common mistake is just to normalize the vectors by dividing them by their length $\sqrt{3}$.
The resulting vectors have length $1$, but they are not orthogonal.

Another mistake is that you just changed the numbers in the vectors so that they are orthogonal.
The issue here is that if you change the numbers randomly, then the new vectors might no longer belong to the subspace $W$.

The point of the Gram-Schmidt orthogonalization is that the process converts any basis for $W$ to an orthogonal basis for $W$.
The above solution didn’t use the full formula of the Gram-Schmidt orthogonalization. Of course, you may use the formula in the exam but you must remember it correctly.

## List of Midterm 2 Problems for Linear Algebra (Math 2568) in Autumn 2017 Add to solve later

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