Finding the Set of Permutatable Matrices with Algebra

  • Thread starter Thread starter esmeco
  • Start date Start date
  • Tags Tags
    Matrices
AI Thread Summary
To find the set of permutable matrices with respect to a matrix A, one must identify matrices X that satisfy the condition AX = XA. The discussion clarifies that a matrix is permutable if all its rows and columns are identical, leading to the conclusion that 2x2 permutable matrices take the form [a, a; a, a]. The only permutable matrix that commutes with any matrix A is the zero matrix [0, 0; 0, 0]. The terminology confusion between "exchangeable" and "permutable" is acknowledged, emphasizing the need for clarity in definitions. Understanding these concepts is crucial for solving the homework problem effectively.
esmeco
Messages
144
Reaction score
0
I have this question as homework from my Algebra class:
A square matrix X is called exchangeable with A if AX=XA.Determine the set of permutable matrices with
matrix.jpg
.

My question is,how do I find that set?I know that a matrix to be permutable all rows and columns must be the same and that a square matrix is composed by the same number of rows and columns.
Thanks in advance for the help!
 
Physics news on Phys.org
?? You defined "exchangeable" with A and then asked for "permutable" with A?? Then you defined "permutable" matrix without any reference to a matrix A?? What am I missing?


If you want to find all matrices that are "exchangeable" with A (standard terminology: "that commute with A"), then look at
\left[\begin{array}{cc}a & b \\ c & d \end{array}\right]\left[\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right]= \left[\begin{array}{cc}1 & 1 \\0 & 1\end{array}\right]\left[\begin{array}{cc}a & b \\c & d\end{array}\right]

If I understand your definition of "permutable" correctly: "all rows and columns must be the same", then all 2 by 2 permutable matrices are of the form
\left[\begin{array}{cc}a & a \\ a & a\end{array}\right]
and the only "permutable" matrix that is "exchangeable" with A is
\left[\begin{array}{cc}0 & 0 \\ 0 & 0 \end {array} \right]
 
...

Sorry for the mistypeing!:redface: When I said "exchengeable I meant to say permutable,so it would be like:

"A square matrix X is called permutable with A if AX=XA..."
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top