MHB Find components of map with respect to basis

[evinda]

Hello! (Wave)

Let linear map $f: \mathbb{R}^3 \to \mathbb{R}^2$, $B$ basis (unknown) of $\mathbb{R}^3$ and $c=[(1,2),(3,4)]$ basis of $\mathbb{R}^2$. We are given the information that $cf_s=\begin{pmatrix}
1 & 0 & 1\\
2 & 1 & 0
\end{pmatrix}$. Let $v \in \mathbb{R}^3$, of which the coordinates as for $B$ are $(1,1,1)$. If $f(v)=(x_1, x_2)$, what does $x_2-x_1$ equal to?

I haven't really understood how we can find $x_1$ and $x_2$. Could you give me a hint? (Thinking)

 

[I like Serena]

Hello! (Wave)

Let linear map $f: \mathbb{R}^3 \to \mathbb{R}^2$, $B$ basis (unknown) of $\mathbb{R}^3$ and $c=[(1,2),(3,4)]$ basis of $\mathbb{R}^2$. We are given the information that $cf_s=\begin{pmatrix}
1 & 0 & 1\\
2 & 1 & 0
\end{pmatrix}$. Let $v \in \mathbb{R}^3$, of which the coordinates as for $B$ are $(1,1,1)$. If $f(v)=(x_1, x_2)$, what does $x_2-x_1$ equal to?

I haven't really understood how we can find $x_1$ and $x_2$. Could you give me a hint? (Thinking)

Hey evinda!

What does $cf_s$ represent? (Wondering)

 

[evinda]

Hey evinda!

What does $cf_s$ represent? (Wondering)

It isn't stated... (Speechless) Is it maybe the matrix with which we change the basis? (Thinking)

 

[I like Serena]

It isn't stated... (Speechless) Is it maybe the matrix with which we change the basis? (Thinking)

With a different notation we have:
\begin{tikzpicture}
%preamble \usepackage{amsfonts}
\usetikzlibrary{cd}
\node {
\begin{tikzcd}[sep=8em,font=\Large]
\mathbb{R}^3 \arrow{r}{f_{E\to E}} & \mathbb{R}^2 \\
\mathbb{R}^3 \arrow{r}{f_{B\to C}}
\arrow{u}{(\mathbf b_1 \mathbf b_2 \mathbf b_3)}
\arrow{ur}[swap]{c\circ f_{B\to C}}{f_{B\to E}}
& \mathbb{R}^2 \arrow{u}[swap]{c=(\mathbf c_1 \mathbf c_2 )}
\end{tikzcd}
};
\end{tikzpicture}

I'm assuming that $f$ is $f_{B\to C}$. Is it? (Wondering)

Can it be that $cf_s$ is $c\circ f_{B\to C}$? Or is it one of the other arrows with $f$?
If so, can we find $(x_1,x_2)$ then? (Wondering)

 

[evinda]

With a different notation we have:
\begin{tikzpicture}
%preamble \usepackage{amsfonts}
\usetikzlibrary{cd}
\node {
\begin{tikzcd}[sep=8em,font=\Large]
\mathbb{R}^3 \arrow{r}{f_{E\to E}} & \mathbb{R}^2 \\
\mathbb{R}^3 \arrow{r}{f_{B\to C}}
\arrow{u}{(\mathbf b_1 \mathbf b_2 \mathbf b_3)}
\arrow{ur}[swap]{c\circ f_{B\to C}}{f_{B\to E}}
& \mathbb{R}^2 \arrow{u}[swap]{c=(\mathbf c_1 \mathbf c_2 )}
\end{tikzcd}
};
\end{tikzpicture}

I'm assuming that $f$ is $f_{B\to C}$. Is it? (Wondering)

Can it be that $cf_s$ is $c\circ f_{B\to C}$? Or is it one of the other arrows with $f$?
If so, can we find $(x_1,x_2)$ then? (Wondering)

If we assume that it is like that, how do we compute $c\circ f_{B\to C}$ ? Do we multiply $C$ by $f(v)=(x_1, x_2)$ ? Or how else do we do it? (Thinking)

 

[I like Serena]

If we assume that it is like that, how do we compute $c\circ f_{B\to C}$ ? Do we multiply $C$ by $f(v)=(x_1, x_2)$ ? Or how else do we do it? (Thinking)

If we have:
$$(c\circ f_{B\to C})(v) =\begin{pmatrix}
1 & 0 & 1\\
2 & 1 & 0
\end{pmatrix} \begin{pmatrix}1 \\ 1 \\ 1 \end{pmatrix}$$
and we have that this must be equal to:
$$c\begin{pmatrix}x_1 \\ x_2\end{pmatrix} = \begin{pmatrix} 1 & 3\\2&4\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}$$
Can we solve it then? (Wondering)

 

[evinda]

If we have:
$$(c\circ f_{B\to C})(v) =\begin{pmatrix}
1 & 0 & 1\\
2 & 1 & 0
\end{pmatrix} \begin{pmatrix}1 \\ 1 \\ 1 \end{pmatrix}$$
and we have that this must be equal to:
$$c\begin{pmatrix}x_1 \\ x_2\end{pmatrix} = \begin{pmatrix} 1 & 3\\2&4\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}$$
Can we solve it then? (Wondering)

We would get that $x_1=x_2=\frac{1}{2}$.

But could you explain to me how we get this equality? (Worried)

 

[I like Serena]

We would get that $x_1=x_2=\frac{1}{2}$.

But could you explain to me how we get this equality? (Worried)

We assume that:

• $f=f_{B\to C}$
• $cf_s=c\circ f_{B\to C}$
• $v=(1,1,1)$ with respect to the basis B
• $f(v)=(x_1,x_2)$, where $(x_1,x_2)$ is with respect to the basis C.
• $c$ is the coordinate transformation from C to standard coordinates E. It is given by: $c=(\mathbf c_1 \quad \mathbf c_2)$.

Then we have that;
$$(c\circ f)(v)=c(f(v))=c(x_1,x_2)$$
don't we? (Wondering)

 

[evinda]

We assume that:

• $f=f_{B\to C}$
• $cf_s=c\circ f_{B\to C}$
• $v=(1,1,1)$ with respect to the basis B
• $f(v)=(x_1,x_2)$, where $(x_1,x_2)$ is with respect to the basis C.
• $c$ is the coordinate transformation from C to standard coordinates E. It is given by: $c=(\mathbf c_1 \quad \mathbf c_2)$.

Then we have that;
$$(c\circ f)(v)=c(f(v))=c(x_1,x_2)$$
don't we? (Wondering)

Yes, I got it so far... (Nod)

 

[I like Serena]

Yes, I got it so far... (Nod)

Okay... what is left? (Wondering)

 

[evinda]

I understand. Using the same logic, I tried to solve the following.

Let linear map $f: \mathbb{R}^2 \to \mathbb{R}^3$, $B$ basis (unknown) of $\mathbb{R}^2$ and $C=\{ (3,-1,1), (-1,0,2), (-1,1,0)\}$ basis of $\mathbb{R}^3$. We are given that $cf_B=\begin{pmatrix}
-1 & -1\\
1 & -2\\
0 & 1
\end{pmatrix}$. Let $v \in \mathbb{R}^2$, the coordinates of which as for $B$ are $(2,-1)$. If $f(v)=(x_1,x_2,x_3)$, then to what is $3x_1+2x_2+x_3$ equal to?

I have thought the following:

$cf_B(v)=\begin{pmatrix}
-1 & -1\\
1 & -2\\
0 & 1
\end{pmatrix}\begin{pmatrix}
2\\
-1
\end{pmatrix}=\begin{pmatrix}
-1\\
4\\
-1
\end{pmatrix}$

and this should be equal to

$c\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=\begin{pmatrix}
3x_1-x_2-x_3\\
-x_1+x_3\\
x_1+2x_2
\end{pmatrix}$.

Thus $x_1=1, x_2=-1, x_3=5$ and $3x_1+2x_2+x_3=6$.

But this isn't a possible answer. Am I doing something wrong? (Thinking)

 

[I like Serena]

I understand. Using the same logic, I tried to solve the following.

Let linear map $f: \mathbb{R}^2 \to \mathbb{R}^3$, $B$ basis (unknown) of $\mathbb{R}^2$ and $C=\{ (3,-1,1), (-1,0,2), (-1,1,0)\}$ basis of $\mathbb{R}^3$. We are given that $cf_B=\begin{pmatrix}
-1 & -1\\
1 & -2\\
0 & 1
\end{pmatrix}$. Let $v \in \mathbb{R}^2$, the coordinates of which as for $B$ are $(2,-1)$. If $f(v)=(x_1,x_2,x_3)$, then to what is $3x_1+2x_2+x_3$ equal to?

I have thought the following:

$cf_B(v)=\begin{pmatrix}
-1 & -1\\
1 & -2\\
0 & 1
\end{pmatrix}\begin{pmatrix}
2\\
-1
\end{pmatrix}=\begin{pmatrix}
-1\\
4\\
-1
\end{pmatrix}$

and this should be equal to

$c\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=\begin{pmatrix}
3x_1-x_2-x_3\\
-x_1+x_3\\
x_1+2x_2
\end{pmatrix}$.

Thus $x_1=1, x_2=-1, x_3=5$ and $3x_1+2x_2+x_3=6$.

But this isn't a possible answer. Am I doing something wrong?

I don't see anything wrong.
However, we didn't properly establish what $cf_B$ and $f$ are exactly.
Can it be that it should be written as ${}^Cf_{B}$? Or perhaps ${}_Cf_{B}$?
If I'm not mistaken those are notations that are sometimes used to indicate $f_{B\to C}$.
And perhaps $f$ is then intended to represent $f_{B\to E}$ in this case.

If so then we have:
$$c(f_{B\to C}(v)) = f_{B\to E}(v) = (x_1,x_2,x_3)$$

Would that result in one of the possible answers?
What are the possible answers? (Wondering)

 

[evinda]

I don't see anything wrong.
However, we didn't properly establish what $cf_B$ and $f$ are exactly.
Can it be that it should be written as ${}^Cf_{B}$? Or perhaps ${}_Cf_{B}$?
If I'm not mistaken those are notations that are sometimes used to indicate $f_{B\to C}$.
And perhaps $f$ is then intended to represent $f_{B\to E}$ in this case.

If so then we have:
$$c(f_{B\to C}(v)) = f_{B\to E}(v) = (x_1,x_2,x_3)$$

Would that result in one of the possible answers?
What are the possible answers? (Wondering)

The possible answers are $45$, $-15$, $35$, $10$, $-11$, $-10$... (Thinking)

 

[I like Serena]

The possible answers are $45$, $-15$, $35$, $10$, $-11$, $-10$... (Thinking)

Since all the answers are numbers and do not contain unknown symbols, I actually see no other possible interpretation.
And indeed, with my latest interpretation I get $-11$. (Thinking)

 

[evinda]

Since all the answers are numbers and do not contain unknown symbols, I actually see no other possible interpretation.
And indeed, with my latest interpretation I get $-11$. (Thinking)

You mean that we have $cf_B(v)=(x_1,x_2,x_3)$ ? (Thinking)
I tried this, but the dimensions do not agree... (Worried)

 

[I like Serena]

You mean that we have $cf_B(v)=(x_1,x_2,x_3)$ ? (Thinking)
I tried this, but the dimensions do not agree... (Worried)

I meant:
$$c({}^Cf_B(v))=f(v)=(x_1,x_2,x_3) \quad\Rightarrow\quad
\begin{bmatrix}3&-1&-1\\-1&0&1\\1&2&0\end{bmatrix}
\begin{bmatrix}-1&-1\\1&-2\\0&1\end{bmatrix}
\begin{bmatrix}2\\-1\end{bmatrix}
=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}
$$
Those dimensions agree don't they? (Wondering)

The corresponding diagram is:
\begin{tikzpicture}
%preamble \usepackage{amsfonts}
\usetikzlibrary{cd}
\node {
\begin{tikzcd}[sep=8em,font=\Large]
\mathbb{R}^2 & \mathbb{R}^3 \\
B:\mathbb{R}^2 \arrow{r}{{}^Cf_{B}}
\arrow{u}{(\mathbf b_1 \mathbf b_2)}
\arrow{ur}{f=c\ \circ\ {}^Cf_B}
& C:\mathbb{R}^3 \arrow{u}[swap]{c=(\mathbf c_1 \mathbf c_2 \mathbf c_3)}
\end{tikzcd}
};
\end{tikzpicture}