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

[rputra]

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.

 

[caffeinemachine]

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.

 

[rputra]

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.

 

[rputra]

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!

 

[caffeinemachine]

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?

 

[rputra]

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.

 

[rputra]

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.

 

[caffeinemachine]

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
$$
\begin{bmatrix}
2 &0 &0 &0\\
0 &1 &1 &0\\
0 &1 &1 &0\\
0 &0 &0 &2\\
\end{bmatrix}
\begin{bmatrix}
r \\
s \\
t \\
u \\
\end{bmatrix}
=
(2r, s+t, s+t, 2u)
$$

 

[rputra]

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.

 

[caffeinemachine]

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?