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 December 19th, 2015, 07:56 AM #1 Member   Joined: Jan 2012 Posts: 57 Thanks: 0 p matrix $\displaystyle K=\begin{bmatrix} 1.5 & -0.5 \\ -0.5 & 1.5 \end{bmatrix}$ A is Eigenvalues matrix of K $\displaystyle A=\begin{bmatrix} 1 &0 \\ 0&2 \end{bmatrix}$ but what about P I have not seen this type of matrix before $\displaystyle P=\frac{1}{\sqrt{2}}\begin{bmatrix} -1 &-1 \\ -1&1 \end{bmatrix}$ December 19th, 2015, 03:17 PM #2 Global Moderator   Joined: May 2007 Posts: 6,770 Thanks: 700 What is troubling you about P? Is it the constant outside? Thanks from mhhojati December 19th, 2015, 06:45 PM #3 Member   Joined: Jan 2012 Posts: 57 Thanks: 0 sorry I should not ask like this I thought that matrix P is a kind of special matrix I am unaware of that it is about part d of the answer of this problem I don't know how they came up with matrix P A is Eigenvalues matrix of Kx $\displaystyle A=k_{x}=\begin{bmatrix} 1 &0 \\ 0&2 \end{bmatrix} P=\frac{1}{\sqrt{2}}\begin{bmatrix} -1 &-1 \\ -1&1 \end{bmatrix}$ $\displaystyle P^{T}=\frac{1}{\sqrt{2}}\begin{bmatrix} -1 &-1 \\ -1&1 \end{bmatrix} k_{y}=\begin{bmatrix} 1 &0 \\ 0&2 \end{bmatrix} m_{y}=\frac{1}{\sqrt{2}}\begin{bmatrix} -1 \\ -1 \end{bmatrix}$ December 20th, 2015, 05:54 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 If an n by n matrix, A, has n distinct eigenvalues, then it is "diagonalizable"- there exist a matrix, P, such that $\displaystyle P^{-1}AP= D$ where D is a diagonal matrix having the eigenvalues on the diagonal. I think that is the "P" you are talking about. P is simply the matrix having the eigenvectors as columns. Here, $\displaystyle A= \begin{bmatrix}3/2 & -1/2 \\ -1/2 & 3/2\end{bmatrix}$. Its "characteristic equation" is $\displaystyle \left|\begin{array}3/2- \lambda & -1/2 \\ -1/2 & 3/2- \lambda\end{array}\right|= (3/2- \lambda)^2- 1/4= 0$ which has roots $\displaystyle \lambda= 3/2+ 1/2= 2$ and $\displaystyle \lambda= 3/2- 1/2= 1$. So, yes, the diagonal matrix, D, is $\displaystyle \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}$. The eigenvectors corresponding to eigenvalue 1 must satisfy $\displaystyle \begin{bmatrix}3/2 & -1/2 \\ -1/2 & 3/2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}(3/2)x- (1/2)y \\ (-1/2)x+ (3/2)y\end{bmatrix}= \begin{bmatrix}x \\ y \end{bmatrix}$ which gives us the two equations (3/2)x- (1/2)y= x and (-1/2)x+ (3/2)y= y. Those are the same as (1/2)x- (1/2)y= 0 and (-1/2)x+ (1/2)y= 0 both of which reduce to y= x. (The fact that there is no single solution is due to the fact that $\displaystyle \lambda$ is an eigenvalue.) Any multiple of $\displaystyle \begin{bmatrix}1 \\ 1\end{bmatrix}$ is an eigenvector corresponding to eigenvalue 1. In particular, since the length of $\displaystyle \begin{bmatrix}1 \\ 1 \end{bmatrix}$ is $\displaystyle \sqrt{2}$, dividing by $\displaystyle \sqrt{2}$ gives $\displaystyle \begin{bmatrix}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{bmatrix}$, a unit eigenvectorl Similarly, the eigenvectors corresponding to eigenvalue 2 must satisfy $\displaystyle \begin{bmatrix}3/2 & -1/2 \\ -1/2 & 3/2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix}(3/2)x- (1/2)y \\ (-1/2)x+ (3/2)y\end{bmatrix}= \begin{bmatrix} 2x \\ 2y\end{bmatrix}$ which gives us the two equations (3/2)x- (1/2)y= 2x an (-1/2)x+ (3/2)y= 2y. Those are the same a (-1/2)x- (1/2)y= 0 and (-1/2)x- (1/2)y= 0 both of which reduce to y= -x. Any multiple of $\displaystyle \begin{bmatrix}1 \\ -1 \end{bmatrix}$ is an eigenvector corresponding to eigenvalue 2. Since that has length 2, we can divide by $\displaystyle \sqrt{2}$ to get the unit eigenvector $\displaystyle \begin{bmatrix}\frac{\sqrt{2}}{2} \\ -\frac{\sqrt{2}}{2}\end{bmatrix}$ Using those as columns, we have $\displaystyle P= \begin{bmatrix}\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{bmatrix}$. Thanks from mhhojati Tags matrix Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post silviatodorof Linear Algebra 2 March 22nd, 2015 05:28 AM Achira Linear Algebra 0 September 25th, 2014 08:21 PM Ter Linear Algebra 4 March 20th, 2013 06:43 PM

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