# Give a linear map that satisfies given properties

• MHB
• mathmari
In summary, two possible linear maps are presented, one that satisfies the given conditions and one that does not. The first linear map, $\phi:\mathbb{R}^3\rightarrow \mathbb{R}^2$, maps the vectors $v_1$ and $v_2$ to the standard basis vectors in $\mathbb{R}^2$ and maps $w$ to the zero vector. The second linear map, also denoted by $\phi$, cannot be constructed as the vectors $v_1$, $v_2$, and $w'$ are linearly dependent. The alternative method of solving for the linear map using a matrix representation is also shown.
mathmari
Gold Member
MHB
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)

Last edited by a moderator:
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)

Klaas van Aarsen said:
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)

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.
$A= \begin{pmatrix}0 & 1 & 0 \\ 2 & -1 & -1 \end{pmatrix}$.

## 1. What is a linear map?

A linear map, also known as a linear transformation, is a mathematical function that maps one vector space to another while preserving the operations of vector addition and scalar multiplication.

## 2. How do you give a linear map that satisfies given properties?

To give a linear map that satisfies given properties, you must first determine the properties that the map must satisfy. Then, using the properties, you can construct a matrix representation of the linear map and verify that it satisfies the given properties.

## 3. What are some common properties that a linear map can satisfy?

Some common properties that a linear map can satisfy include preserving vector addition and scalar multiplication, preserving linear combinations, and preserving the zero vector.

## 4. Can a linear map satisfy multiple properties at once?

Yes, a linear map can satisfy multiple properties at once. In fact, many linear maps are defined by a combination of properties that they must satisfy.

## 5. How are linear maps used in science?

Linear maps are used in science to model and analyze various physical, biological, and social phenomena. They are particularly useful in fields such as physics, engineering, and economics, where they can be used to represent and solve complex systems and equations.

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