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Thekernel 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$\left(A\right)$.

Let$A$ be the matrix given by,

$A=\left[\begin{array}{cc}0.36& 0.48\\ 0.48& 0.64\end{array}\right]$.

Let $\stackrel{鈫�}{x}鈭�\ker \left(A\right)$.

$鈭�A\stackrel{鈫�}{x}=0$.

role="math" localid="1664195128910" $$\begin{gathered} \left[\begin{array}{cc}0.36& 0.48\\ 0.48& 0.64\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ ~\left[\begin{array}{c}0.36x_1+0.48x_2\\ 0.48x_1+0.64x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \end{gathered}$$

By elementary transformation $R_2鈫�R_2-\frac{8}{6}{R}_1$, we get,

$$\begin{gathered} ~\left[\begin{array}{c}0.36x_1+0.48x_2\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ 鈭�0.36x_1+0.48x_2=0 \end{gathered}$$

This shows that geometrically $\ker \left(A\right)$, represents a line in ${\mathrm{鈩�}}^2$, which is an $\mathrm{im}\left(\mathrm{A}\right)$.

Let us first calculate $A^2$.

$$\begin{gathered} A^2=\left[\begin{array}{cc}0.36& 0.48\\ 0.48& 0.64\end{array}\right]\left[\begin{array}{cc}0.36& 0.48\\ 0.48& 0.64\end{array}\right] \\ 鈬�A^2=\left[\begin{array}{cc}0.36& 0.48\\ 0.48& 0.64\end{array}\right] \end{gathered}$$

It is observed that $A=A^2$.

Thus, $A=A^n,鈥�鈥�鈥�鈥�鈭赌n鈭�\mathrm{鈩�}$.

Therefore, if $\stackrel{鈫�}{蠀}鈭�im\left(A\right)$, then $A\stackrel{鈫�}{蠀}鈭�im\left(A\right)$.

Consider$A\stackrel{鈫�}{蠀}=\stackrel{鈫�}{蠀}$ since from part (a),

$\mathrm{im}\left(\mathrm{A}\right)=$span$\left\{\left[\begin{array}{c}0.36\\ 0.48\end{array}\right]\right\}$

$鈬�\mathrm{im}\left(\mathrm{A}\right)=$span$\left\{\left[\begin{array}{c}6\\ 8\end{array}\right]\right\}$

Therefore, $\stackrel{鈫�}{蠀}鈭�im\left(A\right)鈬�\stackrel{鈫�}{蠀}=t鈥�\left[\begin{array}{c}6\\ 8\end{array}\right],鈥�鈥�鈥�t鈭�\mathrm{鈩�}$.

Thus,

role="math" localid="1664196146461" $$\begin{gathered} A\stackrel{鈫�}{蠀}=\left[\begin{array}{cc}0.36& 0.48\\ 0.48& 0.64\end{array}\right]路t路\left[\begin{array}{c}6\\ 8\end{array}\right] \\ 鈬�A\stackrel{鈫�}{蠀}=t路\left[\begin{array}{cc}0.36& 0.48\\ 0.48& 0.64\end{array}\right]\left[\begin{array}{c}6\\ 8\end{array}\right] \\ A\stackrel{鈫�}{蠀}=t路\left[\begin{array}{c}6\\ 8\end{array}\right],鈥�鈥�t鈭�\mathrm{鈩�} \end{gathered}$$

From the part (b), we have,

$A\stackrel{鈫�}{蠀}=t路\left[\begin{array}{c}6\\ 8\end{array}\right]$.

That is, $A\stackrel{鈫�}{x}=t路\left[\begin{array}{c}6\\ 8\end{array}\right]$.

Hence, the linear transformation$T\left(\stackrel{鈫�}{x}\right)=A\stackrel{鈫�}{x}$ is the translation of the vector$\stackrel{鈫�}{x}$ on the line spanned by the vector $\left[\begin{array}{c}6\\ 8\end{array}\right]$.