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How to solve this matrix equation?


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#1 mikeraj

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Posted 27 October 2016 - 09:15 AM

For the attached matrix equation above, can someone guide me on the steps to solve it?

 

Highly appreciate any guidance !

 

 

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#2 Prometheus

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Posted 27 October 2016 - 09:30 AM

Hi mikeraj. What have you tried so far? Which bit in particular are you stuck on?


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#3 mikeraj

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Posted 27 October 2016 - 09:46 AM

I am reading a textbook on eigenvalue/eigenvector and this question is from there.

 

Normally, I would use the method where the inverse of the matrix on the LHS is used to multiply the matrix on the RHS. However there are two issues here when trying to use this method. Firstly the RHS matrix is singular and the determinant cannot be found. Secondly any matrix multiplied with the zero matrix on the RHS will be zero anyway. The answer given in the book for [y z] is [2 -1]


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#4 timo

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Posted 27 October 2016 - 05:25 PM

If an eigenvector has an associated eigenvalue of zero, then all multiples of this eigenvector solve the equation => find the eigenvectors and their eigenvalues.


Edited by timo, 27 October 2016 - 05:26 PM.

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#5 mikeraj

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Posted 27 October 2016 - 11:19 PM

My question here is not about eigenvalue or eigenvector.

 

It is specific to the matrix equation I attached to my original post


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#6 HallsofIvy

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Posted 28 October 2016 - 12:37 PM

The equation is \begin{bmatrix}1 & 2 \\ 2 & 4 \end{bmatrix}\begin{bmatrix}y \\ z\end{bmatrix}= \begin{bmatrix}y+ 2z \\ 2y+ 4z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}.
The determinant of the matrix is 0 which means it does not have an inverse. That, in turn, means that there is no unique solution to this equation. Either there is no solution or there are an infinite number of solutions. Obviously \begin{bmatrix} y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix} is a solution so there must be an infinite number of solutions. To find them, I would write this matrix equation as the system of equations y+ 2z= 0 and 2y+ 4z= 0. Dividing the second equation by 2 gives y+ 2z= 0, the same as the first equation. Any y and z that satisfy y+ 2z= 0, which is the same as y= -2z, will satisfy both equations and so \begin{bmatrix}-2z \\ z\end{bmatrix} will satisfy the matrix equation for z any number. You could also write that as z\begin{bmatrix}-2 \\ 1 \end{bmatrix} or "any multiple of \begin{bmatrix}-2 \\ 1 \end{bmatrix}".


Edited by HallsofIvy, 28 October 2016 - 12:41 PM.

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#7 mikeraj

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Posted 28 October 2016 - 01:34 PM

Hi Hallsoflvy, thanks for your inputs! If you take a look at the attachment, the solution given for [y z] is [2 -1]. Sorry as I am not familiar with latex, I typed the matrix in row form for convenience.

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#8 timo

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Posted 28 October 2016 - 02:06 PM

My question here is not about eigenvalue or eigenvector. It is specific to the matrix equation I attached to my original post

So is my reply. I even explicitly told you how to solve for the [x y] vector, and why that approach gives you the solution.


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#9 HallsofIvy

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Posted 29 October 2016 - 01:18 AM



Hi Hallsoflvy, thanks for your inputs! If you take a look at the attachment, the solution given for [y z] is [2 -1]. Sorry as I am not familiar with latex, I typed the matrix in row form for convenience.

 

  I said​ that any multiple of (-2, 1) is a solution.  Clearly (2, -1) is a solution because it is -1 times (-2, 1).  In fact, we could say that all solutions are multiples of (2, -1). 

 

  However, it is wrong to say that (2, -1) is the solution.  It is a solution- one of an infinite number of solutions.


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#10 mikeraj

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Posted 31 October 2016 - 02:20 AM

Hi HallsofIvy, thanks again. I understand it now. The solution is a basis vector of the eigenspace, which represents the multiples (infinite number) of the basis vector.


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