Question

Find an orthonormal basis of the kernel of the matrix $\mathbf{A}\mathbf{=}\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 3& 4\end{array}\right]$.

Short answer

The solution is $\left\{\overrightarrow{v_1}\left[\begin{array}{c}1\\ 2\\ 1\\ 0\end{array}\right],\overrightarrow{v_2}=\left[\begin{array}{c}2\\ -3\\ 0\\ 1\end{array}\right]\right\}$.

Step-by-step solution

`Explanation of the solution

Consider the matrix A as follows:

$\mathrm{A}=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 3& 4\end{array}\right]$

Let’s find out the kernel of the matrix A.

$K=\left\{\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]\in R^4:\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 3& 4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\right\}$

Now apply for solve the above linear system as follows:

$\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 3& 4\end{array}\right]\stackrel{{\mathrm{R}}_2\to {\mathrm{R}}_3-{\mathrm{R}}_1}{\to }\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& 2& 3& 4\end{array}\right]\stackrel{{\mathrm{R}}_2\to {\mathrm{R}}_3-{\mathrm{R}}_1}{\to }\left[\begin{array}{cccc}1& 0& -1& -2\\ 0& 1& 2& 3\end{array}\right]$

So, the kernel is as follows:

$$\begin{gathered} \mathrm{K}=\left\{\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\\ {\mathrm{x}}_4\end{array}\right]\in {\mathrm{R}}^4:{\mathrm{x}}_1-{\mathrm{x}}_2-2{\mathrm{x}}_4=0,{\mathrm{x}}_2+2{\mathrm{x}}_3+3{\mathrm{x}}_4=0\right\} \\ \mathrm{K}=\left\{\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\\ {\mathrm{x}}_4\end{array}\right]\in {\mathrm{R}}^4:{\mathrm{x}}_1-{\mathrm{x}}_2-2{\mathrm{x}}_4,{\mathrm{x}}_2=-2{\mathrm{x}}_3-3{\mathrm{x}}_4\right\} \\ \mathrm{K}=\left\{\left[\begin{array}{c}{\mathrm{x}}_3+2{\mathrm{x}}_4\\ -2{\mathrm{x}}_3-3{\mathrm{x}}_4\\ {\mathrm{x}}_3\\ {\mathrm{x}}_4\end{array}\right]\in {\mathrm{R}}^4:{\mathrm{x}}_3+{\mathrm{x}}_4\in \mathrm{R}\right\} \\ \mathrm{K}=\left\{{\mathrm{x}}_3\left[\begin{array}{c}1\\ -2\\ 1\\ 0\end{array}\right]+\left[\begin{array}{c}1\\ -2\\ 1\\ 0\end{array}\right]:{\mathrm{x}}_3,{\mathrm{x}}_4\in \mathrm{R}\right\} \\ \mathrm{K}=\mathrm{span}\left\{\left[\begin{array}{c}1\\ -2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}2\\ -3\\ 0\\ 1\end{array}\right]\right\} \end{gathered}$$

Thus, the basis for K is $\left\{{\overrightarrow{v}}_1=\left[\begin{array}{c}1\\ -2\\ 1\\ 0\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}2\\ -3\\ 0\\ 1\end{array}\right]\right\}$.