91影视

$x_1-x_2+2x_3+4x_4=0$

Let V be a subspace ${\mathrm{鈩�}}^4$such that

localid="1659931383607" $$\begin{gathered} V=\left\{\left(x_1,x_2,x_3,x_4\right)鈭�{\mathrm{鈩�}}^4;x_1-x_2+2x_3+4x_4=0\right\} \\ 鈬�V=\left\{\left(x_1-2x_3-4x_4,x_2,x_3,x_4\right)鈭�{\mathrm{鈩�}}^4;x_1=x_2-2x_3-4x_4\right\} \\ 鈬�V=\left\{x_2\left[\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\right]+x_3\left[\begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}\right]+x_4\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\} \end{gathered}$$$鈬�BasisofV=\left\{\left[\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\}$.

Since the basis of the subspace V= $\left\{\left[\begin{array}{c}\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\end{array}\right],\left[\begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\}$, thus the dimension of the subspace V is three.

Also, the vectors in the basis of V are linearly independent so, we have

$lm\left(T\right)=V\mathrm{and}Ker\left(T\right)=\left\{\right\}$

Thus if T is a linear transformation from${\mathrm{鈩�}}^4$to ${\mathrm{鈩�}}^4$such that $T\left(\stackrel{鈫�}{x}\right)=A\stackrel{鈫�}{x}$then

$A{=}^{\left[\begin{array}{ccc}\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}& \begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}& \begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\end{array}\right]}$

The basis of the subspace V of ${\mathrm{鈩�}}^4$defined by the equation$x_1-x_2+2x_3+4x_4=0$is localid="1659870465922" $\left\{\left[\begin{array}{c}\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\end{array}\right],\left[\begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\}$and the matrix of the linear transformation T from ${\mathrm{鈩�}}^3$to ${\mathrm{鈩�}}^4$is

localid="1659870529535" $A{=}^{\left[\begin{array}{ccc}\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}& \begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}& \begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\end{array}\right]}$.