Finding the Matrix Associated with a Linear Mapping on a Real Matrix

In summary: Many, many thanks for your quick response. I understand what you said so far in your response, but unfortunately I am still don't know how to proceed. Let's say I would like using the simplest bases, which are$$\begin{bmatrix}1 &0\\0 &0\end{bmatrix}, \begin{bmatrix}0 &1\\0 &0\end{bmatrix},\begin{bmatrix}0 &0\\1 &0\end{bmatrix},\begin{bmatrix}0 &0\\0 &1\
  • #1
rputra
35
0
I am working on a two-by-two real matrix $M$, with a linear mapping $F$ that returns the sum of $M$ and its transpose. I need to find out the matrix that is associated with the mapping. To the best of my understanding:

$$
M + M^T =
\begin{bmatrix}
r &s\\
t &u
\end{bmatrix}
+
\begin{bmatrix}
r &t\\
s &u
\end{bmatrix}
=
\begin{bmatrix}
2r &s+t\\
s+t &2u
\end{bmatrix}
,$$
therefore I need to find $A$ of two-by-two of real entries, such that for any $M,$

$$
F(M) = AM =
\begin{bmatrix}
2r &s+t\\
s+t &2u
\end{bmatrix}
.$$

But I am lost on finding the $A,$ any hints or help would be very much appreciated. Thank you in advance for your time and effort.
 
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  • #2
Tarrant said:
I am working on a two-by-two real matrix $M$, with a linear mapping $F$ that returns the sum of $M$ and its transpose. I need to find out the matrix that is associated with the mapping. To the best of my understanding:

$$
M + M^T =
\begin{bmatrix}
r &s\\
t &u
\end{bmatrix}
+
\begin{bmatrix}
r &t\\
s &u
\end{bmatrix}
=
\begin{bmatrix}
2r &s+t\\
s+t &2u
\end{bmatrix}
,$$
therefore I need to find $A$ of two-by-two of real entries, such that for any $M,$

$$
F(M) = AM =
\begin{bmatrix}
2r &s+t\\
s+t &2u
\end{bmatrix}
.$$

But I am lost on finding the $A,$ any hints or help would be very much appreciated. Thank you in advance for your time and effort.
Such a matrix $A$ does not exist. Just take $a, b, c, d$ to be the entries of $A$. If you want $AM=M+M^t$ to be true for all $M$, you'll see $A$ cannot exist.
 
  • #3
caffeinemachine said:
Such a matrix $A$ does not exist. Just take $a, b, c, d$ to be the entries of $A$. If you want $AM=M+M^t$ to be true for all $M$, you'll see $A$ cannot exist.

Oh, that is strange! Let me double check and I will get back with you. Thanks again.
 
  • #4
caffeinemachine said:
Such a matrix $A$ does not exist. Just take $a, b, c, d$ to be the entries of $A$. If you want $AM=M+M^t$ to be true for all $M$, you'll see $A$ cannot exist.

Forget about what I wrote after "To the best of my understanding...," they are totally erroneous and that was what distracted you in the first response. It turned out the mapping should be $F: M_{2 \times 2}(\mathbb R) \rightarrow M_{2 \times 2}(\mathbb R)$, such that $F$ should be have domain and co-domain of dimension 4, simply because a $2 \times 2$ matrix has dimension of 4. (Does it mean that the matrix then will be $4 \times 4$?)

I hope this should clear you up and I am still waiting for you to take a second look. Thank you and am looking forward to hear from you again soon.

PS. Congratulation on your new badge!
 
Last edited:
  • #5
Tarrant said:
Forget about what I wrote after "To the best of my understanding...," they are totally erroneous and that was what distracted you in the first response. It turned out the mapping should be $F: M_{2 \times 2}(\mathbb R) \rightarrow M_{2 \times 2}(\mathbb R)$, such that $F$ should be have domain and co-domain of dimension 4, simply because a $2 \times 2$ matrix has dimension of 4. (Does it mean that the matrix then will be $4 \times 4$?)

I hope this should clear you up and I am still waiting for you to take a second look. Thank you and am looking forward to hear from you again soon.

PS. Congratulation on your new badge!

Hey thanks for the wishes.
___
We can think of $M_{2\times 2}(\mathbf R)$ is as $\mathbf R^4$ under the correspondence
$$\begin{bmatrix}r& s\\ t & u\end{bmatrix}\leftrightarrow (r, s, t, u)$$
So basically we have a linear map $T:\mathbf R^4\to \mathbf R^4$ sending $(r, s, t, u)$ to $(2r, s+t, s+t, 2u)$. What you want is to find the matrix of this linear map (with respect to some basis of your choice). Do you how to proceed from here?
 
  • #6
caffeinemachine said:
Hey thanks for the wishes.
___
We can think of $M_{2\times 2}(\mathbf R)$ is as $\mathbf R^4$ under the correspondence
$$\begin{bmatrix}r& s\\ t & u\end{bmatrix}\leftrightarrow (r, s, t, u)$$
So basically we have a linear map $T:\mathbf R^4\to \mathbf R^4$ sending $(r, s, t, u)$ to $(2r, s+t, s+t, 2u)$. What you want is to find the matrix of this linear map (with respect to some basis of your choice). Do you how to proceed from here?

Many, many thanks for your quick response. I understand what you said so far in your response, but unfortunately I am still don't know how to proceed. Let's say I would like using the simplest bases, which are
$$
\begin{bmatrix}
1 &0\\
0 &0
\end{bmatrix}
,
\begin{bmatrix}
0 &1\\
0 &0
\end{bmatrix}
,
\begin{bmatrix}
0 &0\\
1 &0
\end{bmatrix}
,
\begin{bmatrix}
0 &0\\
0 &1
\end{bmatrix}
.$$
I am still confused on getting the matrix associated with the mapping, this is especially true after I was told the matrix will be a $4 \times 4$ one, rightly or wrong. Again many, many thanks to all your patient helps so far.
 
  • #7
Tarrant said:
Many, many thanks for your quick response. I understand what you said so far in your response, but unfortunately I am still don't know how to proceed. Let's say I would like using the simplest bases, which are
$$
\begin{bmatrix}
1 &0\\
0 &0
\end{bmatrix}
,
\begin{bmatrix}
0 &1\\
0 &0
\end{bmatrix}
,
\begin{bmatrix}
0 &0\\
1 &0
\end{bmatrix}
,
\begin{bmatrix}
0 &0\\
0 &1
\end{bmatrix}
.$$
I am still confused on getting the matrix associated with the mapping, this is especially true after I was told the matrix will be a $4 \times 4$ one, rightly or wrong. Again many, many thanks to all your patient helps so far.

I think I get it now, the matrix associated with the $F$ is
$$\begin{bmatrix}
2 &0 &0 &0\\
1 &1 &0 &0\\
1 &1 &0 &0\\
0 &0 &0 &2
\end{bmatrix},$$

simple because

$$
F(r, s, t, u) =
\begin{bmatrix}
2 &0 &0 &0\\
1 &1 &0 &0\\
1 &1 &0 &0\\
0 &0 &0 &2\\
\end{bmatrix}
\begin{bmatrix}
r \\
s \\
t \\
u \\
\end{bmatrix}
=
(2r, s+t, s+t, 2u).
$$
Thank you though for all your help and time.
 
  • #8
Tarrant said:
I think I get it now, the matrix associated with the $F$ is
$$\begin{bmatrix}
2 &0 &0 &0\\
1 &1 &0 &0\\
1 &1 &0 &0\\
0 &0 &0 &2
\end{bmatrix},$$

simple because

$$
F(r, s, t, u) =
\begin{bmatrix}
2 &0 &0 &0\\
1 &1 &0 &0\\
1 &1 &0 &0\\
0 &0 &0 &2\\
\end{bmatrix}
\begin{bmatrix}
r \\
s \\
t \\
u \\
\end{bmatrix}
=
(2r, s+t, s+t, 2u).
$$
Thank you though for all your help and time.
You miscalculated.
The matrix is
$$\begin{bmatrix}
2 &0 &0 &0\\
0&1 &1 &0\\
0&1 &1 &0\\
0 &0 &0 &2
\end{bmatrix}$$
because
$$
\begin{bmatrix}
2 &0 &0 &0\\
0 &1 &1 &0\\
0 &1 &1 &0\\
0 &0 &0 &2\\
\end{bmatrix}
\begin{bmatrix}
r \\
s \\
t \\
u \\
\end{bmatrix}
=
(2r, s+t, s+t, 2u)
$$
 
  • #9
Yes, you are correct. I stand by to be corrected. One final question: I didn't see any link between the bases that I provided earlier with the matrix associated with the mapping. Could you please explain? Thanks again.
 
  • #10
Tarrant said:
Yes, you are correct. I stand by to be corrected. One final question: I didn't see any link between the bases that I provided earlier with the matrix associated with the mapping. Could you please explain? Thanks again.
Let

$$
e_1=
\begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix},
e_2=
\begin{bmatrix}
& 1\\
0 & 0
\end{bmatrix},
e_3=
\begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix},
e_4=
\begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix}
$$

What is $F(e_1)$? It is the matrix
$$
\begin{bmatrix}
2 & 0 \\
0 & 0
\end{bmatrix}
$$
Thus $F(e_1)=2e_1$. Similarly, $F(e_2)=e_2+e_3 = F(e_3)$, and $F(e_4)=2e_4$.

Now do you see how the matrix of $F$ with respect to the basis $\{e_1, e_2, e_3, e_4\}$ is got?
 

FAQ: Finding the Matrix Associated with a Linear Mapping on a Real Matrix

What is a linear mapping?

A linear mapping is a mathematical operation that takes in a vector and returns a transformed vector. In other words, it is a function that maps one vector space to another in a way that preserves the algebraic structure of the original vector space.

What is a matrix associated with a linear mapping?

A matrix associated with a linear mapping is a matrix representation of the transformation. It is a square matrix that can be used to perform the same transformation on a vector as the linear mapping function.

How do you find the matrix associated with a linear mapping?

To find the matrix associated with a linear mapping, you first need to determine the basis vectors for the domain and range of the mapping. Then, you can use the transformation of these basis vectors to construct the matrix by placing the transformed vectors as columns in the matrix.

What is the purpose of finding the matrix associated with a linear mapping?

Finding the matrix associated with a linear mapping allows us to perform the same transformation on any vector in the domain by simply multiplying it by the matrix. This makes it easier to apply the linear mapping to a large number of vectors and to perform calculations involving the mapping.

In what situations would you need to find the matrix associated with a linear mapping?

Finding the matrix associated with a linear mapping is necessary in many applications of linear algebra, such as computer graphics, image processing, and physics. It is also crucial in solving systems of linear equations and studying the properties of linear transformations.

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