91影视

The given matrix has a reduced row echelon form, we have that the dimension of the image is 2. Therefore, from the nullity theorem follows that the dimension of the kernel is equal to$\mathbf{3}\mathbf{=}\mathbf{5}\mathbf{-}\mathbf{dim}\mathbf{\left(}\mathbf{im}\mathbf{\right)}\mathbf{=}\mathbf{dim}\mathbf{\left(}\mathbf{ker}\mathbf{\right)}\mathbf{.}$ .

Column vectors of the given matrix are

$\stackrel{鈫�}{V_1}=\left[\begin{array}{c}1\\ 0\end{array}\right],\stackrel{鈫�}{V_2}=\left[\begin{array}{c}2\\ 0\end{array}\right],\stackrel{鈫�}{V_3}=\left[\begin{array}{c}0\\ 1\end{array}\right],\stackrel{鈫�}{V_4}=\left[\begin{array}{c}3\\ 4\end{array}\right],\stackrel{鈫�}{V_5}=\left[\begin{array}{c}5\\ 6\end{array}\right]$

It is not hard to conclude that $\stackrel{鈫�}{{\mathrm{V}}_2},\stackrel{鈫�}{{\mathrm{V}}_4}\mathrm{and}\stackrel{鈫�}{{\mathrm{V}}_5}$are redundant vectors, because

$\begin{array}{l}\stackrel{鈫�}{V_2}=2\stackrel{鈫�}{V_1}\\ \stackrel{鈫�}{V_4}=\frac{3}{2}\stackrel{鈫�}{V_2}+4\stackrel{鈫�}{V_3}\\ \stackrel{鈫�}{V_5}=\frac{5}{3}\stackrel{鈫�}{V_4}+\frac{2}{3}\stackrel{鈫�}{V_4}\end{array}$

i.e.

$\begin{array}{l}0=2\stackrel{鈫�}{V_1}-\stackrel{鈫�}{V_2}\\ 0=\frac{3}{2}\stackrel{鈫�}{V_2}+4\stackrel{鈫�}{V_3}-\stackrel{鈫�}{V_4}\\ 0=\frac{5}{3}\stackrel{鈫�}{V_4}+\frac{2}{3}\stackrel{鈫�}{V_4}-\stackrel{鈫�}{V_5}\end{array}$

This relations gives us that the vectors

role="math" localid="1659422557127" $\left[\begin{array}{c}2\\ -1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 4\\ -1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 5/3\\ -2/3\\ -1\end{array}\right]$

Are in ker of the given matrix. Since the dimension of the is 3,

To prove that those three vectors form a basis, we need to show that rank of matrix is 3 .

$\left[\begin{array}{c}2\\ -1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 4\\ -1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 5/3\\ -2/3\\ -1\end{array}\right]$

Finding reduced row echelon form of the given matrix by using the row operations.

Add the first row divided by 2 into second row i.e.$R_2鈫�\frac{1}{2}{R}_1+R_2$,R

localid="1659422791460" $\left[\begin{array}{c}2\\ 0\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 4\\ -1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 5/3\\ -2/3\\ -1\end{array}\right]$

$R_3鈫�\frac{5}{3}{R}_5+R_3$

localid="1659423694909" $\left[\begin{array}{c}2\\ 0\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 4\\ -1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 0\\ -2/3\\ -1\end{array}\right]$

$R_3鈫�R_5-\frac{2}{3}{R}_3$and $R_1=\frac{1}{2}{R}_1$

localid="1659423918380" $\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\begin{array}{c}0\\ \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 4\\ -1\\ 0\end{array}\begin{array}{c}0\\ 0\\ 0\\ 0\\ -1\end{array}\right]$

$R_2鈫�\frac{3}{2}{R}_{2}and$$R鈫�-1R_5$

localid="1659424013075" $\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\begin{array}{c}0\\ 1\\ 4\\ -1\\ 0\end{array}\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right]$

$R_2鈫�R_2-4R_3$and $R_4鈫�R_2+R_4$

localid="1659424576791" $\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right]$

Swap fifth and third row to obtain rref

$rref=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right]$

Those three vectors are linearly independent, then from the fact that follows $dim\left(\mathrm{ke}r\right)=3$that they form a basis of ker .

The vectors form a basis for ker .

$\left[\begin{array}{c}2\\ -1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 4\\ -1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 5/3\\ -2/3\\ -1\end{array}\right]$