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 December 19th, 2015, 08:56 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, 04:17 PM #2 Global Moderator   Joined: May 2007 Posts: 6,685 Thanks: 661 What is troubling you about P? Is it the constant outside? Thanks from mhhojati
 December 19th, 2015, 07:45 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, 06:54 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 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

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