Find components of map with respect to basis

In summary: 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
  • #1
evinda
Gold Member
MHB
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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)
 
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  • #2
evinda said:
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)
 
  • #3
I like Serena said:
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)
 
  • #4
evinda said:
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)
 
  • #5
I like Serena said:
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)
 
  • #6
evinda said:
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)
 
  • #7
I like Serena said:
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)
 
  • #8
evinda said:
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)
 
  • #9
I like Serena said:
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)
 
  • #10
evinda said:
Yes, I got it so far... (Nod)

Okay... what is left? (Wondering)
 
  • #11
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)
 
  • #12
evinda said:
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)
 
  • #13
I like Serena said:
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)
 
  • #14
evinda said:
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)
 
  • #15
I like Serena said:
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)
 
  • #16
evinda said:
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}
 

FAQ: Find components of map with respect to basis

1. What is the basis of a map?

The basis of a map refers to a set of vectors that can be used to express all other vectors in the map. It is essentially the set of building blocks that make up the map.

2. How do you find the components of a map with respect to a basis?

To find the components of a map with respect to a basis, you can use the coordinate transformation matrix. This matrix can be used to convert the coordinates of a vector in the original basis to its coordinates in the new basis.

3. Can you have more than one basis for a map?

Yes, it is possible to have multiple bases for a map. In fact, a map can have an infinite number of bases, as long as they all satisfy certain criteria.

4. How do you determine the components of a vector in a specific basis?

To determine the components of a vector in a specific basis, you can use the dot product formula. This involves taking the dot product of the vector with each basis vector, and then dividing by the magnitude of the basis vector.

5. Why is finding the components of a map with respect to a basis important?

Finding the components of a map with respect to a basis is important because it allows us to express vectors in different coordinate systems. This is particularly useful in fields such as physics and engineering, where different coordinate systems are often used to simplify calculations and problem-solving.

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