Question

Find an orthonormal basis of the image of the matrix $\mathbf{A}\mathbf{=}\left[\begin{array}{ccc}1& 2& 1\\ 2& 1& 1\\ 2& -2& 0\end{array}\right]$.

Short answer

The solution is ${\overrightarrow{u}}_1=\left(\frac{1}{3},\frac{2}{3},\frac{2}{3}\right)$and${\overrightarrow{u}}_2=\left(\frac{2}{3},\frac{1}{3},\frac{-2}{3}\right)$

Step-by-step solution

`Explanation of the solution

Consider the matrix A as follows:

$\mathrm{A}=\left[\begin{array}{ccc}1& 2& 1\\ 2& 1& 1\\ 2& -2& 0\end{array}\right]$

Now, going to transform the matrix A to its row echelon form.

The columns of its row echelon form which contain pivots are actually the columns of the matrix A that form the basis for its image.

Therefore, let’s find the row echelon form of A as follows:

$$\begin{gathered} A=\left[\begin{array}{ccc}1& 2& 1\\ 2& 1& 1\\ 2& -2& 0\end{array}\right]\stackrel{R_2-2R_1\to R_2}{\to }\left[\begin{array}{ccc}1& 2& 1\\ 0& -3& -1\\ 2& -2& 0\end{array}\right]\stackrel{R_3-2R_1\to R_3}{\to }\left[\begin{array}{ccc}1& 2& 1\\ 0& -3& -1\\ 2& -6& -2\end{array}\right] \\ \stackrel{R_3-2R_2\to R_3}{\to }\left[\begin{array}{ccc}1& 2& 1\\ 0& -3& -1\\ 0& 0& 0\end{array}\right]\stackrel{\left(\frac{-1}{3}\right)R_3\to R_3}{\to }\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& \frac{1}{3}\\ 0& 0& 0\end{array}\right] \end{gathered}$$

Since, the first two columns of the row echelon form of A contain pivots that the first and the second column of A form the basis for Im(A).

Let’s denote these columns by${\overrightarrow{\mathrm{a}}}_1$and$\overrightarrow{{\mathrm{a}}_2}$respectively.

If these columns are already orthogonal then find an orthonormal basis for Im(A) by normalizing them, but if they are not going to have to use the Gram-Schmidt algorithm to find one as follows.

$$\begin{gathered} {\overrightarrow{a}}_1.\overrightarrow{a_2}={\left(1,2,2\right)}^{\perp }.\left(2,1,-2\right) \\ =2+2-4 \\ {\overrightarrow{a}}_1.\overrightarrow{a_2}=0 \end{gathered}$$

Since, the first two column vectors of A are orthogonal and going to normalize them as follows.

$$\begin{gathered} {\overrightarrow{\mathrm{u}}}_1=\frac{1}{||\overrightarrow{{\mathrm{a}}_1}||}{\overrightarrow{\mathrm{a}}}_1 \\ =\frac{1}{\sqrt{1^2+2^2+2^2}}\left(1,2,2,\right) \\ =\frac{1}{\sqrt{9}}\left(1,2,2\right) \\ {\overrightarrow{\mathrm{u}}}_1=\left(\frac{1}{3},\frac{2}{3},\frac{2}{3}\right) \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{1}{||\overrightarrow{a_2}||}{\overrightarrow{a}}_2 \\ =\frac{1}{\sqrt{2^2+1^2+{\left(-2\right)}^2}}\left(2,1,-2\right) \\ =\frac{1}{\sqrt{9}}\left(2,1,-2\right) \\ =\frac{1}{3}\left(2,1,-2\right) \\ {\overrightarrow{u}}_2=\left(\frac{2}{3},\frac{1}{3},\frac{-2}{3}\right) \end{gathered}$$

Thus,localid="1660104194486" ${\overrightarrow{\mathrm{u}}}_1=\left(\frac{1}{3},\frac{2}{3},\frac{2}{3}\right)$andlocalid="1660104218801" ${\overrightarrow{\mathrm{u}}}_2=\left(\frac{2}{3},\frac{1}{3},\frac{-2}{3}\right)$form an orthonormal basis for Im(A).