Question

Consider a 2 × 2 matrix A with ${\mathbf{A}}^{\mathbf{2}}\mathbf{=}\mathbf{A}$.

a. If$\overrightarrow{\mathbf{w}}$ is in the image of A, what is the relationship between$\overrightarrow{\mathbf{w}}$ and A $\overrightarrow{\mathbf{w}}$?

b. What can you say about A if rank(A) = 2? What if rank(A) = 0?

c. If rank(A) = 1, show that the linear transformation$\mathit{T}\left(\overrightarrow{x}\right)\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$ is the projection onto im(A) along ker(A). See Exercise 2.2.33.

Short answer

a)$A\overrightarrow{v}=\overrightarrow{w}$

b) If rank(A)= 2, then the columns of A are linearly independent.$\mathrm{If}\mathrm{rank}\left(\mathrm{A}\right)=0,\mathrm{then}\mathrm{A}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$.

Step-by-step solution

Step 1: Objective

Consider a 2 × 2 matrix A with $A^2=A.a$)If $\overrightarrow{w}$is in the image of A, the objective is to determine the relationship between $\overrightarrow{w}$ and A$\overrightarrow{w}$.

Proving w→=Aw→.

Since the vector $\overrightarrow{w}$is in the image of A, there exists a vector $\overrightarrow{v}$in $R^2$such that

$$\begin{gathered} \mathrm{A}\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{w}} \\ \mathrm{A}.\mathrm{A}\overrightarrow{\mathrm{v}}=\mathrm{A}\overrightarrow{\mathrm{w}}\left[\mathrm{Multiplying}\mathrm{both}\mathrm{sideswith}\mathrm{A}\right] \\ {\mathrm{A}}^2\overrightarrow{\mathrm{v}}=\mathrm{A}\overrightarrow{\mathrm{w}} \\ \mathrm{A}\overrightarrow{\mathrm{v}}=\mathrm{A}\overrightarrow{\mathrm{w}}\left[\mathrm{Since}{\mathrm{A}}^2=\mathrm{A}\right] \\ \mathrm{It}\mathrm{is}\mathrm{concluded}\mathrm{that}\mathrm{A}\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{w}} \end{gathered}$$

If rank(A) = 2

If rank (A) = 2, then the columns of A are linearly independent.

$$\begin{gathered} \mathrm{The}\mathrm{matrix}\mathrm{is}\mathrm{A}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \mathrm{If}\mathrm{rank}\left(\mathrm{A}\right)=0,\mathrm{then}\mathrm{A}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]. \end{gathered}$$

Proving (c)

Let rank(A)=1. Consider two nonparallel lines ${\mathrm{L}}_1$and L₂ in R². A vector $\overrightarrow{v}$in R² can be expressed uniquely as $\overrightarrow{v}={\overrightarrow{v}}_1+\overrightarrow{v_2}$, where ${\overrightarrow{v}}_1$is on ${\mathrm{L}}_1$and ${\overrightarrow{v}}_1$is on ${\mathrm{L}}_2$The transformation $\mathrm{T}\left(\overrightarrow{\mathrm{v}}\right)=\overrightarrow{{\mathrm{v}}_1}$is called the projection onto ${\mathrm{L}}_1$along L₂. The objective is to

show that the linear transformation$\mathrm{T}\left(\overrightarrow{\mathrm{x}}\right)=\mathrm{A}\overrightarrow{\mathrm{x}}$is the projection onto im(A) along ker(A).

Step 5:Proving (c)

Consider the transformation $\mathrm{T}\left(\overrightarrow{\mathrm{x}}\right)=\mathrm{A}\overrightarrow{\mathrm{x}}$. The vector can be written as $\overrightarrow{x}=A\overrightarrow{x}+\left(\overrightarrow{x}-A\overrightarrow{x}\right)$.

localid="1660731745293" $$\begin{gathered} \mathrm{Here}A\overrightarrow{x}\mathrm{is}\mathrm{the}\mathrm{column}\mathrm{space}\mathrm{of}\mathrm{A}.\mathrm{Since}\mathrm{the}\mathrm{im}\left(A\right)=\mathrm{column}\mathrm{space}\mathrm{of}\mathrm{A}.\mathrm{A}\overrightarrow{\mathrm{x}}\in \mathrm{im}\left(A\right). \\ \mathrm{Consider}\overrightarrow{\mathrm{x}}-\mathrm{A}\overrightarrow{\mathrm{x}}.\mathrm{Verify}\mathrm{that}\overrightarrow{\mathrm{x}}-\mathrm{A}\overrightarrow{\mathrm{x}}\in \ker (\mathrm{A}). \\ \mathrm{A}(\overrightarrow{\mathrm{x}}-\mathrm{A}\overrightarrow{\mathrm{x}})=\mathrm{A}\overrightarrow{\mathrm{x}}-{\mathrm{A}}^2\overrightarrow{\mathrm{x}} \\ =\mathrm{A}\overrightarrow{\mathrm{x}}-\mathrm{A}\overrightarrow{\mathrm{x}}[\mathrm{since}{\mathrm{A}}^2=\mathrm{A}] \\ =\overrightarrow{0} \\ \ker (A)\mathrm{consists}\mathrm{of}\mathrm{all}\mathrm{vectors}\overrightarrow{\mathrm{y}}\mathrm{such}\mathrm{that}\mathrm{A}\overrightarrow{\mathrm{y}}=0. \\ \mathrm{So},\overrightarrow{\mathrm{x}}-\mathrm{A}\overrightarrow{\mathrm{x}}\mathrm{is}{\mathrm{inR}}^2\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\overrightarrow{\mathrm{x}}-\mathrm{A}\overrightarrow{\mathrm{x}}\mathrm{is}\mathrm{in}\mathrm{the}\ker \left(\mathrm{A}\right). \\ \mathrm{Hence},\mathrm{every}\mathrm{vector}\overrightarrow{\mathrm{x}}\mathrm{in}{\mathrm{R}}^2\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\overrightarrow{\mathrm{x}}=\mathrm{A}\overrightarrow{\mathrm{x}}+\overrightarrow{\mathrm{y}}\mathrm{where}\overrightarrow{\mathrm{y}}=\overrightarrow{\mathrm{x}}-\mathrm{A}\overrightarrow{\mathrm{x}}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{linear}\mathrm{transformation}\mathrm{T}\left(\overrightarrow{\mathrm{x}}\right)=\mathrm{A}\overrightarrow{\mathrm{x}}\mathrm{is}\mathrm{the}\mathrm{projection}\mathrm{onto}\mathrm{im}\left(\mathrm{A}\right)\mathrm{along}\ker \left(\mathrm{A}\right). \end{gathered}$$