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December 19th, 2015, 07:56 AM   #1
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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
$\displaystyle P=\frac{1}{\sqrt{2}}\begin{bmatrix}
-1 &-1 \\
-1&1
\end{bmatrix}$
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December 19th, 2015, 03:17 PM   #2
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What is troubling you about P? Is it the constant outside?
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December 19th, 2015, 06:45 PM   #3
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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}
$
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December 20th, 2015, 05:54 AM   #4
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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 $\displaystyle P= \begin{bmatrix}\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{bmatrix}$.
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