Give a linear map that satisfies given properties

[mathmari]

Hey!

Let $v_1:\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}, \ \ v_2:\begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}\in \mathbb{R}^3$.

1. Let $w=\begin{pmatrix}1 \\ 0 \\2\end{pmatrix}\in \mathbb{R}^3$. If possible, give a linear map $\phi:\mathbb{R}^3\rightarrow \mathbb{R}^2$ such that $\phi (v_1)=\begin{pmatrix}1 \\ 0\end{pmatrix}, \ \ \phi (v_2)=\begin{pmatrix}0 \\ 1\end{pmatrix}, \ \ \phi (w)=\begin{pmatrix}0 \\ 0\end{pmatrix}$.
2. Let $w'=\begin{pmatrix}0 \\ 1 \\0\end{pmatrix}\in \mathbb{R}^3$. If possible, give a linear map $\phi:\mathbb{R}^3\rightarrow \mathbb{R}^2$ such that $\phi (v_1)=\begin{pmatrix}1 \\ 0\end{pmatrix}, \ \ \phi (v_2)=\begin{pmatrix}0 \\ 1\end{pmatrix}, \ \ \phi (w')=\begin{pmatrix}0 \\ 0\end{pmatrix}$.

I have done the following:

1. The three vectors $v_1,v_2, w$ are linearly independent and so they form a basis of $\mathbb{R}^3$.
   
   We consider an arbitrary vector in $\mathbb{R}^3$ and we write it as a linear combination of the vectors of the basis.
   
   \begin{equation*}\begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}=c_1\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}+c_2\begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}+c_3\begin{pmatrix}1 \\ 0 \\2\end{pmatrix}=\begin{pmatrix}1 & 1 & 1\\ 1 & 0 & 0 \\ 1 & 1 & 2\end{pmatrix}\begin{pmatrix}c_1\\ c_2 \\ c_3\end{pmatrix}\end{equation*}
   
   We multiply by the inverse matrix and we get
   \begin{equation*}\begin{pmatrix}c_1\\ c_2 \\ c_3\end{pmatrix}=P^{-1}\begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}=\begin{pmatrix}0 & 1 & 0\\ 2 & -1 & -1 \\ -1 & 0 & 1\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}=\begin{pmatrix}x_2 \\ 2x_1-x_2-x_3\\ -x_1+x_3\end{pmatrix}\end{equation*}
   
   So
   \begin{equation*}\begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}=x_2\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}+(2x_1-x_2-x_3)\begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}+( -x_1+x_3)\begin{pmatrix}1 \\ 0 \\2\end{pmatrix}\end{equation*}
   
   We calculate $\phi \begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}$ using the linearity of $\phi$:
   \begin{align*}\phi \begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}&=\phi \left (x_2\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}+(2x_1-x_2-x_3)\begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}+( -x_1+x_3)\begin{pmatrix}1 \\ 0 \\2\end{pmatrix}\right )\\ & =x_2\cdot \phi \begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}+(2x_1-x_2-x_3)\cdot \phi \begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}+( -x_1+x_3)\cdot \phi \begin{pmatrix}1 \\ 0 \\2\end{pmatrix}\\ & =x_2 \begin{pmatrix}1 \\ 0\end{pmatrix}+(2x_1-x_2-x_3) \begin{pmatrix}0\\ 1\end{pmatrix}+( -x_1+x_3) \begin{pmatrix}0 \\ 0 \end{pmatrix}\\ & = \begin{pmatrix}x_2 \\ 2x_1-x_2-x_3 \end{pmatrix}\end{align*}
2. The vectors $v_1, v_2, w'$ are linearly dependent, since \begin{equation*}\begin{pmatrix}0 \\ 1\\ 0\end{pmatrix}=\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}-\begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}\Rightarrow w'=v_1-v_2\end{equation*}
   
   We apply $\phi$ at this equation and we get: \begin{equation*}\phi (w')=\phi (v_1-v_2)=\phi (v_1)-\phi (v_2)\Rightarrow \begin{pmatrix}0 \\ 0\end{pmatrix}=\begin{pmatrix}1 \\ 0\end{pmatrix}-\begin{pmatrix}0 \\ 1\end{pmatrix}\Rightarrow \begin{pmatrix}0 \\ 0\end{pmatrix}=\begin{pmatrix}1 \\ -1\end{pmatrix}\end{equation*}
   
   Therefore there is no linear map $\phi:\mathbb{R}^3\rightarrow \mathbb{R}^2$ such that$\phi (v_1)=\begin{pmatrix}1 \\ 0\end{pmatrix}$, $\phi (v_2)=\begin{pmatrix}0 \\ 1\end{pmatrix}$ und $\phi (w')=\begin{pmatrix}0 \\ 0\end{pmatrix}$.

Is everything correct? (Wondering)

 

[I like Serena]

Hey mathmari!

Looks correct to me. (Nod)

Btw, here is how we can do 1. in a more straight forward fashion.
Let $\phi(x)=Ax$, then:
$$A\begin{pmatrix}v_1&v_2&w\end{pmatrix}=\begin{pmatrix}\phi(v_1)&\phi(v_2)&\phi(w)\end{pmatrix}
\implies A\begin{pmatrix}1&1&1\\1&0&0\\1&1&2\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}
\implies A=\begin{pmatrix}1&1&1\\1&0&0\\1&1&2\end{pmatrix}^{-1}\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}$$
(Nerd)

 

[mathmari]

Looks correct to me. (Nod)

Btw, here is how we can do 1. in a more straight forward fashion.
Let $\phi(x)=Ax$, then:
$$A\begin{pmatrix}v_1&v_2&w\end{pmatrix}=\begin{pmatrix}\phi(v_1)&\phi(v_2)&\phi(w)\end{pmatrix}
\implies A\begin{pmatrix}1&1&1\\1&0&0\\1&1&2\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}
\implies A=\begin{pmatrix}1&1&1\\1&0&0\\1&1&2\end{pmatrix}^{-1}\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}$$
(Nerd)

Ahh ok! Thanks a lot! (Star)

 

[HallsofIvy]

An alternative method: Since A maps R3 to R2 it can be represented by a matrix with three columns and two rows. Write it as
$A= \begin{pmatrix}a & b & c \\ d & e & f\end{pmatrix}$
A maps $\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}$ to $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ so $
\begin{pmatrix}a & b & c \\ d & e & f\end{pmatrix}
\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}= \begin{pmatrix}a+ b+ c \\ d+ e+ f\end{pmatrix}= \begin{pmatrix}1 \\ 0 \end{pmatrix}
$
so we have the equations a+ b+ c= 1 and d+ e+ f= 0.

A maps $\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix}$ to $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ so $\begin{pmatrix}a & b & c \\ d & e & f\end{pmatrix}
\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix}= \begin{pmatrix}a+ c \\ d+ f\end{pmatrix}= \begin{pmatrix}0 \\ 1 \end{pmatrix}
$
so we have the equations a+ c= 0 and d+ f= 1.

A maps $\begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}$ to $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ so $\begin{pmatrix}a & b & c \\ d & e & f\end{pmatrix}
\begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}= \begin{pmatrix}a+ 2c \\ d+ 2f\end{pmatrix}= \begin{pmatrix}0 \\ 0 \end{pmatrix}
$
so we have the equations a+ 2c= 0 and d+ 2f= 0.

We have six equations, a+ b+ c= 1, d+ e+ f= 0, a+ c= 0, d+ f= 1, a+ 2c= 0, and d+ 2f= 0, to solve for a, b, c, d, e, and f.


From a+ c= 0 and a+ 2c= 0, c= 0. From d+ f= 1 and d+ 2f= 0, f= -1. Then a+ 0= 0 so a= 0 and d- 1= 1 so d= 2. a+ b+ c= 0+ b+ 0= 1 so b= 1. d+ e+ f= 2+ e- 1= 0 so e= -1.
$A= \begin{pmatrix}0 & 1 & 0 \\ 2 & -1 & -1 \end{pmatrix}$.