91影视

The kernel of linear transformationis defined as follows:

The kernel of a linear transformation$T\left(\stackrel{鈫�}{x}\right)=A\stackrel{鈫�}{x}$ from${\mathrm{鈩�}}^m$ to${\mathrm{鈩�}}^n$ consists of all zeros of the transformation, i.e., the solutions of the equations $T\left(\stackrel{鈫�}{x}\right)=A\stackrel{鈫�}{x}=0$.

It is denoted by ker$\left(T\right)$or ker$\left(A\right)$.

Given the matrix role="math" localid="1664189844252" $A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]$.

Let us first compute the product $A^2,A^3$.

$$\begin{gathered} A^2=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right] \\ 鈬�A^2=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right] \end{gathered}$$

role="math" localid="1664190000379" $$\begin{gathered} A^3=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right] \\ {A}^3=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right] \end{gathered}$$

The image of$A$ consists of all vectors of the form:

role="math" localid="1664190450046" $$\begin{gathered} A\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right] \\ =x_1\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]+x_2\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]+x_3\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right] \end{gathered}$$

Thus, all the linear combinations of the column vectors of matrix$A$

role="math" localid="1664190569882" ${\stackrel{鈫�}{蠀}}_1=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right],{\stackrel{鈫�}{蠀}}_2=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],{\stackrel{鈫�}{蠀}}_3=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$,

.

Hence, the image of$A$ is the plane spanned by these three vectors.

The image of $A^2$consists of all vectors of the form:

role="math" localid="1664190729655" $$\begin{gathered} A^2\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right] \\ =x_1\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]+x_2\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]+x_3\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \end{gathered}$$

Thus, all the linear combinations of the column vectors of matrix $A^2$,

${\stackrel{鈫�}{蠀}}_1=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right],{\stackrel{鈫�}{蠀}}_2=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right],{\stackrel{鈫�}{蠀}}_3=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$.

Hence, the image of$A^2$ is the plane spanned by these three vectors.

The image of $A^3$consists of all vectors of the form:

$$\begin{gathered} A^3\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right] \\ =x_1\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]+x_2\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]+x_3\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$

Thus, all the linear combinations of the column vectors of matrix $A^3$,

${\stackrel{鈫�}{蠀}}_1=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right],{\stackrel{鈫�}{蠀}}_2=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right],{\stackrel{鈫�}{蠀}}_3=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$.

Hence, the image of$A^3$ is the plane spanned by these three vectors.

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

This gives

$$\begin{gathered} A\stackrel{鈫�}{x}=0 \\ 鈬�\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=0 \\ 鈬�\left[\begin{array}{c}{x}_2\\ x_3\\ 0\end{array}\right]=0 \\ 鈬�x_2=x_3=0 \end{gathered}$$

Thus, the kernel of$A$is, $\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$.

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

This gives

$$\begin{gathered} A^2\stackrel{鈫�}{x}=0 \\ 鈬�\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=0 \\ 鈬�\left[\begin{array}{c}{x}_3\\ 0\\ 0\end{array}\right]=0 \\ 鈬�x_3=0 \end{gathered}$$

Thus, the kernel of$A^2$is,$\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]=\left\{\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]\right\}$ .

Consider the matrix: role="math" localid="1664191989275" $A^3=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

This gives:

$$\begin{gathered} A^3\stackrel{鈫�}{x}=0 \\ 鈬�\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=0 \\ 鈬�\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]=0 \\ 鈬�x_1=x_2=x_3=0 \end{gathered}$$

Thus, the kernel of $A^3$is,$\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left\{\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]\right\}$.