mikeraj 0 Posted October 27, 2016 For the attached matrix equation above, can someone guide me on the steps to solve it? Highly appreciate any guidance ! matrix equation.pdf Share this post Link to post Share on other sites

Prometheus 341 Posted October 27, 2016 Hi mikeraj. What have you tried so far? Which bit in particular are you stuck on? Share this post Link to post Share on other sites

mikeraj 0 Posted October 27, 2016 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] Share this post Link to post Share on other sites

timo 539 Posted October 27, 2016 (edited) 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 October 27, 2016 by timo Share this post Link to post Share on other sites

mikeraj 0 Posted October 27, 2016 My question here is not about eigenvalue or eigenvector. It is specific to the matrix equation I attached to my original post Share this post Link to post Share on other sites

HallsofIvy 50 Posted October 28, 2016 (edited) The equation is [math]\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}[/math].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 [math]\begin{bmatrix} y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/math] 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 [math]\begin{bmatrix}-2z \\ z\end{bmatrix}[/math] will satisfy the matrix equation for z any number. You could also write that as [math]z\begin{bmatrix}-2 \\ 1 \end{bmatrix}[/math] or "any multiple of [math]\begin{bmatrix}-2 \\ 1 \end{bmatrix}[/math]". Edited October 28, 2016 by HallsofIvy Share this post Link to post Share on other sites

mikeraj 0 Posted October 28, 2016 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. example.pdf Share this post Link to post Share on other sites

timo 539 Posted October 28, 2016 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. Share this post Link to post Share on other sites

HallsofIvy 50 Posted October 29, 2016 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. Share this post Link to post Share on other sites

mikeraj 0 Posted October 31, 2016 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. Share this post Link to post Share on other sites