For the matrix A below, find a nonzero vector in nul A and a nonzero vector in col A.

$$A=\left[\begin{array}{cccc}1& -2& 5& 6\\ 5& 1& -10& 15\\ 1& -2& 8& 4\end{array}\right]$$

This question aims to find the null space which represents the set of all solutions to the homogeneous equation and column space which represents the range of a given vector.

The concepts that we need to solve this question are null space, column space, homogeneous equation of vectors, and linear transformations. The null space of a vector is written as $NulA$ is a set of all possible solutions to the homogeneous equation $Ax=0$. The column space of a vector is written as $ColA$ is the set of all possible linear combinations or range of the given matrix.

Expert Anwer

The homogeneous equation is given as:

$$AX=0$$

The matrix $A$ is given in the question and $X$ is a column vector with $4$ unknown variables. We can assume matrix $X$ to be:

$$X=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]$$

Using row operations on matrix $A$ to reduce the matrix to echelon form.

$$R_2\to R_2-5R_1,R_3\to R_3-R_1$$

$$A=\left[\begin{array}{cccc}1& -2& 5& 6\\ 0& 1& -35& -15\\ 0& 0& 3& -2\end{array}\right]$$

$$R_2\to R_2/11,R_1\to R_1+2R_2$$

$$A=\left[\begin{array}{cccc}1& 0& -15/11& 36/11\\ 0& 1& -35/11& -15/11\\ 0& 0& 3& -2\end{array}\right]$$

$$R_3\to R_3/3,R_1\to R_1+15R_2/11$$

$$A=\left[\begin{array}{cccc}1& 0& 0& 26/11\\ 0& 1& -35/11& -15/11\\ 0& 0& 1& -2/3\end{array}\right]$$

$$R_1\to R_1–35R_3/11$$

$$A=\left[\begin{array}{cccc}1& 0& 0& 26/11\\ 0& 1& 0& -115/33\\ 0& 0& 1& -2/3\end{array}\right]$$

The matrix $A$ contains $2$ pivot columns and $2$ free columns. Substituting the values in homogeneous equation, we get:

$$A=\left[\begin{array}{cccc}1& 0& 0& 26/11\\ 0& 1& 0& -115/33\\ 0& 0& 1& -2/3\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right]$$

Solving for unknown variables, we get:

$$x_1+{\displaystyle \frac{26}{11}}{x}_4=0⟶x_1=-{\displaystyle \frac{26}{11}}$$

$$x_2-{\displaystyle \frac{115}{33}}{x}_4=0⟶x_2={\displaystyle \frac{115}{33}}$$

$$x_3-{\displaystyle \frac{2}{3}}{x}_4=0⟶x_3={\displaystyle \frac{2}{3}}$$

The parametric solution is given as:

$$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}-{\displaystyle \frac{26}{11}}{x}_4\\ {\displaystyle \frac{115}{33}}{x}_4\\ {\displaystyle \frac{2}{3}}{x}_4\\ x_4\end{array}\right]$$

$$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}-{\displaystyle \frac{26}{11}}\\ {\displaystyle \frac{115}{33}}\\ {\displaystyle \frac{2}{3}}\\ 1\end{array}\right]x_4$$

Numerical Result

The nonzero vector in $NulA$ is:

$$\left\{\begin{array}{c}\left[\begin{array}{c}-{\displaystyle \frac{26}{11}}\\ {\displaystyle \frac{115}{33}}\\ {\displaystyle \frac{2}{3}}\\ 1\end{array}\right]\end{array}\right\}$$

The pivot columns in the echelon form of matrix $A$ points to $ColA$, which are given as:

$$\left\{\begin{array}{c}\left[\begin{array}{c}1\\ 5\\ 1\end{array}\right],\left[\begin{array}{c}-2\\ 1\\ -2\end{array}\right],\left[\begin{array}{c}5\\ -10\\ 8\end{array}\right]\end{array}\right\}$$

Example

Find the column space of the given matrix below:

$$\left[\begin{array}{cc}-3& 2\\ -5& -9\end{array}\right]$$

The echelon form of the given matrix found to be:

$$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$

The $Col$ space of the given matrix is given as:

$$\left\{\begin{array}{c}\left[\begin{array}{c}-3\\ -5\end{array}\right],\left[\begin{array}{c}2\\ -9\end{array}\right]\end{array}\right\}$$

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