Finding limitations for matrix

[nuuskur]

Homework Statement

Let [itex]A\in Mat_{3,4}(K)[/itex]. Find all matrices X such that [itex]\forall X| A\cdot X = A'[/itex], where A' is the same as A with 2nd and 4th column swapped.

Homework Equations

The Attempt at a Solution

First we determine the size of matrix X. By definition the first factor must have as many columns as the second has rows and the end product is [itex]A_{m,n}\cdot B_{n,p} = C_{m,p}[/itex]. X must be 4 x 4. Let [itex]a'_{i,j}\in A', a_{i,j}\in A[/itex]

Haven't been able to (in my opinion) come up with anything sensible. Essentially, I have noticed that if we denote rows of A and columns of X as vectors, such that [itex]a_1 = (a_{1,1},a_{1,2},a_{1,3},a_{1,4})[/itex](by row up to a₃) and [itex]x_1 = (x_{1,1},x_{2,1},x_{3,1},x_{4,1})[/itex](by column up to x₄)
Then the dot products equal to the corresponding element of A except when we multiply by either x₂ or x₄. Problem is, where do I find the conditions for matrix X?
I can also write out the dot products individually, but then I have 3 equations and 4 variables..

I have concluded that:
[tex]
\begin{cases}
a'_{i,j} = a_{i,j}= \sum_{k=1}^4 a_{i,k}\cdot x_{k,j},& j\in\{1,3\}\\
a'_{i,2} = \sum_{k=1}^4 a_{i,k}\cdot x_{k, 2} = a_{i,4}\\
a'_{i,4} = \sum_{k=1}^4 a_{i,k}\cdot x_{k, 4} = a_{i,2}

\end{cases}

[/tex]
THIS LOOKS UGLY! I can't think of any other way to show conditions for all x in X :s
This definitely is not acceptable, suggestions?

(I want to avoid writing out individual x - what if the dimensions of the matrix are countible? Would take a long time to write them out xD)

 

[Svein]

If X were the identity matrix it would look like this: [itex]
\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{matrix}[/itex]. The second row controls the second column in the product, and the fourth row controls the fourth column.

 

[Ray Vickson]

Homework Statement

Let [itex]A\in Mat_{3,4}(K)[/itex]. Find all matrices X such that [itex]\forall X| A\cdot X = A'[/itex], where A' is the same as A with 2nd and 4th column swapped.

Homework Equations

The Attempt at a Solution

First we determine the size of matrix X. By definition the first factor must have as many columns as the second has rows and the end product is [itex]A_{m,n}\cdot B_{n,p} = C_{m,p}[/itex]. X must be 4 x 4. Let [itex]a'_{i,j}\in A', a_{i,j}\in A[/itex]

Haven't been able to (in my opinion) come up with anything sensible. Essentially, I have noticed that if we denote rows of A and columns of X as vectors, such that [itex]a_1 = (a_{1,1},a_{1,2},a_{1,3},a_{1,4})[/itex](by row up to a₃) and [itex]x_1 = (x_{1,1},x_{2,1},x_{3,1},x_{4,1})[/itex](by column up to x₄)
Then the dot products equal to the corresponding element of A except when we multiply by either x₂ or x₄. Problem is, where do I find the conditions for matrix X?
I can also write out the dot products individually, but then I have 3 equations and 4 variables..

I have concluded that:
[tex]
\begin{cases}
a'_{i,j} = a_{i,j}= \sum_{k=1}^4 a_{i,k}\cdot x_{k,j},& j\in\{1,3\}\\
a'_{i,2} = \sum_{k=1}^4 a_{i,k}\cdot x_{k, 2} = a_{i,4}\\
a'_{i,4} = \sum_{k=1}^4 a_{i,k}\cdot x_{k, 4} = a_{i,2}

\end{cases}

[/tex]
THIS LOOKS UGLY! I can't think of any other way to show conditions for all x in X :s
This definitely is not acceptable, suggestions?

(I want to avoid writing out individual x - what if the dimensions of the matrix are countible? Would take a long time to write them out xD)

Google 'matrix column operations'; for example, see
http://stattrek.com/matrix-algebra/elementary-operations.aspx
or
https://unapologetic.wordpress.com/2009/08/27/elementary-row-and-column-operations/

These show explicitly how to find the matrix X. (They leave unanswered the question of whether X is unique---they just show how to find one possible X.)

 

[nuuskur]

Okay, if I approach this deductively, then
[itex]a'_{11} = a_{11}\cdot x_{11} + a_{12}\cdot x_{21} + a_{13}\cdot x_{31} + a_{14}\cdot x_{41} = a_{11} [/itex] if x₁₁ = 1, then the other three summands would sum to 0 and the resulting matrix is very similar to the identity matrix, BUT
x₁₁ does not have to be 1 in which case everything breaks.

Intuitively I am quite sure that the X is the identity matrix with 2, 4 column swapped - but this is all deduction based math. I am not satisfied.

 

[Ray Vickson]

Okay, if I approach this deductively, then
[itex]a'_{11} = a_{11}\cdot x_{11} + a_{12}\cdot x_{21} + a_{13}\cdot x_{31} + a_{14}\cdot x_{41} = a_{11} [/itex] if x₁₁ = 1, then the other three summands would sum to 0 and the resulting matrix is very similar to the identity matrix, BUT
x₁₁ does not have to be 1 in which case everything breaks.

Intuitively I am quite sure that the X is the identity matrix with 2, 4 column swapped - but this is all deduction based math. I am not satisfied.

What is unsatisfactory about it? What is wrong with building on past knowledge developed by others?