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$\begin{bmatrix}0.95 & 0.03 \\ 0.05 & 0.97\\ \end{bmatrix} $

I found the eigen values to be 1 and .92

However, I forgot how to find that the eigenvectors are multiples of <3,5> and <1,-1>

please help me remember.

  • 1
    Hint: An eigenvector $v=(c_1,c_2)$ corresponding to eigenvalue $\lambda$ satisfies $Av=\lambda v.$ There should be some relation between $c_1$ and $c_2$ depending on what $\lambda$ you choose. – 2017-02-01

3 Answers 3

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The eigenvectors are the solutions of the equation

$$ Ax = \lambda x \Leftrightarrow (\lambda I - A) X = 0 $$

Plug in the lambdas you found. For $\lambda = 1$:

$$ \left[ \begin{array}{c c} \lambda - 0.95 & -0.03 \\ -0.05 & \lambda - 0.97 \end{array} \right] x = \left[ \begin{array}{c c} 0.05 & -0.03 \\ -0.05 & 0.03 \end{array} \right] x = 0 $$

The first row gives you $0.05 x_1 - 0.03 x_2 = 0 \Rightarrow x_1 = \frac{3}{5} x_2$ which is a multiple of $(3, 5)$. Try it yourself for $\lambda = 0.92$.

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You have your eigenvalues $\lambda_1=1$ and $\lambda_2=0.92$.

You have to therefore solve the following systems: $$A\mathbf{x}=\lambda_1\mathbf{x}$$ $$\begin{pmatrix} 0.95 & 0.03 \\ 0.05 & 0.97 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=1\begin{pmatrix} x \\ y \end{pmatrix} \tag{1}$$ And: $$A\mathbf{x}=\lambda_2\mathbf{x}$$ $$\begin{pmatrix} 0.95 & 0.03 \\ 0.05 & 0.97 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=0.92\begin{pmatrix} x \\ y \end{pmatrix} \tag{2}$$ You should only need to solve 1 part of the system of equations for each (The system's equations are redundant):

$$\begin{cases} 0.95x + 0.03y=x \\ 0.05x+0.97y=y \end{cases} \tag{1*}$$ $$\begin{cases} 0.95x + 0.03y=0.92x \\ 0.05x+0.97y=0.92y \end{cases} \tag{2*}$$

If you can't go further, feel free to let us know.

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I can't tell you what you don't remember.

But you have found eigenvalues $1, 0.92$ and eigenvectors $(3,5), (1,-1)$ for those eigenvalues.

$\begin{bmatrix} 0.95&0.3\\0.5&0.97\end{bmatrix} \begin{bmatrix} 3\\5\end{bmatrix} = \begin{bmatrix} 3\\5\end{bmatrix}$ and

$\begin{bmatrix} 0.95&0.3\\0.5&0.97\end{bmatrix} \begin{bmatrix} 1\\-1\end{bmatrix} = \begin{bmatrix} 0.92\\-0.92\end{bmatrix}$

What else are you looking for?

$\begin{bmatrix} 0.95&0.3\\0.5&0.97\end{bmatrix} = \frac {1}{8}\begin{bmatrix} 3&1\\5&-1\end{bmatrix} \begin{bmatrix} 1&\\&0.92\end{bmatrix}\begin{bmatrix} 1&1\\5&-3\end{bmatrix}$