Using Gram-Schmidt Orthogonalization, Find an Orthogonal Basis for the Span

Problems and solutions in Linear Algebra

Problem 716

Using Gram-Schmidt orthogonalization, find an orthogonal basis for the span of the vectors $\mathbf{w}_{1},\mathbf{w}_{2}\in\R^{3}$ if
\[
\mathbf{w}_{1}
=
\begin{bmatrix}
1 \\ 0 \\ 3
\end{bmatrix}
,\quad
\mathbf{w}_{2}
=
\begin{bmatrix}
2 \\ -1 \\ 0
\end{bmatrix}
.
\]

 
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Solution.

We apply Gram-Schmidt orthogonalization as follows. The first step is to define $\mathbf{u}_{1}=\mathbf{w}_{1}$. Before defining $\mathbf{u}_{2}$, we must compute
\begin{align*}
\mathbf{u}_{1}^{T}\mathbf{w}_{2}
&=
\mathbf{w}_{1}^{T}\mathbf{w}_{2}
=
\begin{bmatrix}
1 & 0 & 3
\end{bmatrix}
\begin{bmatrix}
2 \\ -1 \\ 0
\end{bmatrix}
=2+0+0=2,
\\
\mathbf{u}_{1}^{T}\mathbf{u}_{1}
&=
\mathbf{w}_{1}^{T}\mathbf{w}_{1}
=
\begin{bmatrix}
1 & 0 & 3
\end{bmatrix}
\begin{bmatrix}
1 \\ 0 \\ 3
\end{bmatrix}
=1+0+9=10.
\end{align*}


Next, we define
\[
\mathbf{u}_{2}
=
\mathbf{w}_{2}
-\dfrac{\mathbf{u}_{1}^{T}\mathbf{w}_{2}}
{\mathbf{u}_{1}^{T}\mathbf{u}_{1}}
\mathbf{u}_{1}
=
\begin{bmatrix}
2 \\ -1 \\ 0
\end{bmatrix}
-\dfrac{2}{10}
\begin{bmatrix}
1 \\ 0 \\ 3
\end{bmatrix}
=
\begin{bmatrix}
10/5 \\ -1 \\ 0
\end{bmatrix}

\begin{bmatrix}
1/5 \\ 0 \\ 3/5
\end{bmatrix}
=
\begin{bmatrix}
9/5 \\ -1 \\ -3/5
\end{bmatrix}
.
\] By Gram-Schmidt orthogonalization, $\{\mathbf{u}_{1},\mathbf{u}_{2}\}$ is an orthogonal basis for the span of the vectors $\mathbf{w}_{1}$ and $\mathbf{w}_{2}$.

Remark

Note that since scalar multiplication by a nonzero number does not change the orthogonality of vectors and the new vectors still form a basis, we could have used $5\mathbf{u}_2$, instead of $\mathbf{u}_2$ to avoid a fraction in our computation.
We have
\[5\mathbf{u}_2=\begin{bmatrix}
10 \\
-5 \\
0
\end{bmatrix}-\begin{bmatrix}
1 \\
0 \\
3
\end{bmatrix}=\begin{bmatrix}
9 \\
-5 \\
-3
\end{bmatrix},\] and $\{\mathbf{u}_1, 5\mathbf{u}_2\}$ is an orthogonal basis for the span.


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