Question

Consider a matrix $\mathit{A}\mathbf{=}\left[\begin{array}{cc}0.36& 0.48\\ 0.48& 0.64\end{array}\right]$.

1. Describe ker (role="math" localid="1664194291935" $\mathit{A}$) and im ($\mathit{A}$) geometrically.
2. Find ${\mathit{A}}^{\mathbf{2}}$. If $\mathit{\upsilon }$is in the image of $\mathit{A}$, what can you say about $\mathit{A}\overrightarrow{\mathbf{\upsilon }}$?
3. Describe the transformation $\mathit{T}\left(\overrightarrow{x}\right)\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$geometrically.

Short answer

1. $\ker \left(A\right)$represents a line in ${\mathrm{ℝ}}^2$, which is an $im\left(A\right)$.
2. $A^2=\left[\begin{array}{cc}0.36& 0.48\\ 0.48& 0.64\end{array}\right]$and $A\overrightarrow{\upsilon }=t·\left[\begin{array}{c}6\\ 8\end{array}\right], t\in \mathrm{ℝ}$.
3. The linear transformation$T\left(\overrightarrow{x}\right)=A\overrightarrow{x}$is the translation of the vector$\overrightarrow{x}$on the line spanned by the vector .

Step-by-step solution

Define kernel of linear transformation

Thekernel of linear transformationis defined as follows:

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

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

Let$A$ be the matrix given by,

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

(a) Describe the kernel of A and image of A geometrically

Let $\overrightarrow{x}\in \ker \left(A\right)$.

$\therefore A\overrightarrow{x}=0$.

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 $R_2\to R_2-\frac{8}{6}{R}_1$, we get,

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

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

(b) Find A2 and discuss about Aυ→

Let us first calculate $A^2$.

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

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

Thus, $A=A^n, \forall n\in \mathrm{ℝ}$.

Therefore, if $\overrightarrow{\upsilon }\in im\left(A\right)$, then $A\overrightarrow{\upsilon }\in im\left(A\right)$.

Consider$A\overrightarrow{\upsilon }=\overrightarrow{\upsilon }$ since from part (a),

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

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

Therefore, $\overrightarrow{\upsilon }\in im\left(A\right)\iff \overrightarrow{\upsilon }=t \left[\begin{array}{c}6\\ 8\end{array}\right], t\in \mathrm{ℝ}$.

Thus,

role="math" localid="1664196146461" $$\begin{gathered} A\overrightarrow{\upsilon }=\left[\begin{array}{cc}0.36& 0.48\\ 0.48& 0.64\end{array}\right]·t·\left[\begin{array}{c}6\\ 8\end{array}\right] \\ \Rightarrow A\overrightarrow{\upsilon }=t·\left[\begin{array}{cc}0.36& 0.48\\ 0.48& 0.64\end{array}\right]\left[\begin{array}{c}6\\ 8\end{array}\right] \\ A\overrightarrow{\upsilon }=t·\left[\begin{array}{c}6\\ 8\end{array}\right], t\in \mathrm{ℝ} \end{gathered}$$

(c) Describe the transformation T(x→)=Ax→ geometrically

From the part (b), we have,

$A\overrightarrow{\upsilon }=t·\left[\begin{array}{c}6\\ 8\end{array}\right]$.

That is, $A\overrightarrow{x}=t·\left[\begin{array}{c}6\\ 8\end{array}\right]$.

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