Question

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

Short answer

The solution is the vectors of the orthonormal basis of the matrix A is$\overrightarrow{u_1}=\left(\frac{\sqrt{2}}{2},0,0,\frac{\sqrt{2}}{2}\right)$ and ${\overrightarrow{\mathrm{u}}}_2=\left(0,\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0\right)$.

Step-by-step solution

`Explanation of the solution

Consider the matrix A as follows:

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

Suppose that $\overrightarrow{\mathrm{x}}=({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4)\in \ker \left(\mathrm{A}\right)$.

Therefore $A\overrightarrow{x}=\overrightarrow{0}$. Let’s find the form of all the vectors that belongs to the kernel of A.

$$\begin{gathered} \ker \left(\mathrm{A}\right)=\left\{\overrightarrow{\mathrm{x}}\in {\mathrm{R}}^4:\mathrm{A}\overrightarrow{\mathrm{x}}=\overrightarrow{0}\right\} \\ =\left\{{\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4\right\}:\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -1& -1& 1\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\\ {\mathrm{x}}_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ =\left\{\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4\right):\left[\begin{array}{c}{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4\\ {\mathrm{x}}_1-{\mathrm{x}}_2-{\mathrm{x}}_3-{\mathrm{x}}_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\right\} \\ =\left\{\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4\right):{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=0\stackrel{,}{\mathrm{U}}{\mathrm{x}}_1-{\mathrm{x}}_2-{\mathrm{x}}_3-{\mathrm{x}}_4=0\right\} \end{gathered}$$

Further solve above expression,

$$\begin{gathered} \ker \left(\mathrm{A}\right)=\left\{\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4\right):{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=0\stackrel{,}{\mathrm{U}}2{\mathrm{x}}_1+2{\mathrm{x}}_4=0\right\} \\ =\left\{\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4\right):{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=0\wedge {\mathrm{x}}_4=-\mathrm{x}\right\} \\ =\left\{\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4\right):{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3-{\mathrm{x}}_1=0\wedge {\mathrm{x}}_4=-{\mathrm{x}}_1\right\} \\ =\left\{\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4\right):{\mathrm{x}}_2+{\mathrm{x}}_3=0\wedge {\mathrm{x}}_4=-{\mathrm{x}}_1\right\} \end{gathered}$$

Further solve above expression,

role="math" localid="1660108293755" $$\begin{gathered} \ker \left(\mathrm{A}\right)=\left\{\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4\right):{\mathrm{x}}_3=-{\mathrm{x}}_2\wedge {\mathrm{x}}_4=-{\mathrm{x}}_1\right\} \\ =\left\{\left({\mathrm{x}}_1,{\mathrm{x}}_2,-{\mathrm{x}}_2,-{\mathrm{x}}_1\right),{\mathrm{x}}_1,{\mathrm{x}}_2\in \mathrm{R}\right\} \\ =\left\{{\mathrm{x}}_1\left(1,0,0,-1\right)+{\mathrm{x}}_2\left(0,1,,-1,0\right),{\mathrm{x}}_1,{\mathrm{x}}_2\in \mathrm{R}\right\} \\ =\mathrm{span}\left\{\left(1,0,0,-1\right),\left(0,1,,-1,0\right)\right\} \end{gathered}$$

Since, the vectors that span ker(A) are linearly independent. They form to basis.

Thus,${\overrightarrow{\mathrm{V}}}_1=(1,0,0,-1)$and${\overrightarrow{\mathrm{V}}}_2=(0,1,-1,0)$form a basis for ker(A).

Now, let’s find an orthonormal basis to applying the Gram-Schmidt process on the vectors of the basis of ker(A) as follows:

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{1}{||{\overrightarrow{v}}_1||}{\overrightarrow{v}}_1 \\ =\frac{1}{\sqrt{1^2+0^2+0^2+{(-1)}^2}}(1,0,0,-1) \\ =\frac{1}{\sqrt{2}}(1,0,0,-1) \\ {\overrightarrow{u}}_1=\left(\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2}\right) \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} {\overrightarrow{v}}_2^{\perp }={\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1{\overrightarrow{v}}_1\right){\overrightarrow{u}}_1 \\ =\left(0,1,-1,0\right)-\left(\frac{\sqrt{2}}{2}.0+0.1+0.\left(-1\right)+\left(\frac{\sqrt{2}}{2}\right).0\right)\left(\frac{\sqrt{2}}{2},0,0,-\frac{\sqrt{2}}{2}\right) \\ =\left(0,1,-1,0\right)-0\left(\frac{\sqrt{2}}{2},0,0,-\frac{\sqrt{2}}{2}\right) \\ {\overrightarrow{v}}_2^{\perp }=\left(0,1,-1,0\right) \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{1}{||{{\overrightarrow{v}}^{\perp }}_1||}{\overrightarrow{v}}_2^{\perp } \\ =\frac{1}{\sqrt{0^2+1^2+{(-1)}^2+0^2}}(0,1,-1,0) \\ =\frac{1}{\sqrt{2}}(0,1,-1,0) \\ {\overrightarrow{u}}_2=\left(0,\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},0\right) \end{gathered}$$

Therefore, vectors ${\overrightarrow{u}}_1=\left(\frac{\sqrt{2}}{2},0,0,-\frac{\sqrt{2}}{2}\right)$and ${\overrightarrow{u}}_2=\left(0,\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},0\right)$ form an orthonormal basis from ker(A).