I got -2, 1, 3 as my eigenvalues but when I put the equation into wolframalpha it shows 5, 3, and -2 as the exponents over the e so I might have...

Wednesday, January 1, 2014

I got -2, 1, 3 as my eigenvalues but when I put the equation into wolframalpha it shows 5, 3, and -2 as the exponents over the e so I might have...

Hello!

The system has the form $\displaystyle{x}'={A}{x},$ where $\displaystyle{A}$ is the matrix with the given constant coefficients.

If $\displaystyle\lambda$ is an eigenvalue of $\displaystyle{A}$ and $\displaystyle{r}$ is the corresponding eigenvector, then by definition $\displaystyle{A}{r}=\lambda{r}.$ So if we take $\displaystyle{x}{\left({t}\right)}={{e}}^{{\lambda{t}}}\cdot{r},$ then $\displaystyle{x}'{\left({t}\right)}=\lambda\cdot{{e}}^{{\lambda{t}}}\cdot{r},$ so such $\displaystyle{x}$ is the solution of the system $\displaystyle{x}'={A}{x}.$

Actually, if there are 3 different real eigenvalues $\displaystyle\lambda_{{1}},$ $\displaystyle\lambda_{{2}}$ and $\displaystyle\lambda_{{3}}$ and their corresponding eigenvectors are $\displaystyle{r}_{{1}},$ $\displaystyle{r}_{{2}}$ and $\displaystyle{r}_{{3}},$ then the general solution for $\displaystyle{x}'={A}{x}$ is

$\displaystyle{x}{\left({t}\right)}={C}_{{1}}\cdot{{e}}^{{\lambda_{{1}}{t}}}\cdot{r}_{{1}}+{C}_{{2}}\cdot{{e}}^{{\lambda_{{2}}{t}}}\cdot{r}_{{2}}+{C}_{{3}}\cdot{{e}}^{{\lambda_{{3}}{t}}}\cdot{r}_{{3}}$

for any constants $\displaystyle{C}_{{1}},$ $\displaystyle{C}_{{2}},$ $\displaystyle{C}_{{3}}.$

So we have to find not only eigenvalues but also eigenvectors. I agree with you about eigenvalues, they are $\displaystyle\lambda_{{1}}={1},$ $\displaystyle\lambda_{{2}}=-{2}$ and $\displaystyle\lambda_{{3}}={3}.$ To find an eigenvector corresponding to $\displaystyle\lambda$ , we have to solve $\displaystyle{\left({A}-\lambda\cdot{I}\right)}{r}={0}.$ In our case r's may be found in the form $\displaystyle{\left({1},{a},{b}\right)}.$

For example, for $\displaystyle\lambda_{{1}}={1}$

$\displaystyle{\left({A}-\lambda_{{1}}\cdot{I}\right)}{r}={\left({A}-{I}\right)}\cdot{\left({1},{a},{b}\right)}={\left(-{a}+{4}{b},{3}+{a}-{b},{2}+{a}-{2}{b}\right)}={0}.$

Solving this simple linear system for $\displaystyle{a}$ and $\displaystyle{b}$ we obtain $\displaystyle{a}=-{4}$ and $\displaystyle{b}=-{1},$ so the eigenvector is $\displaystyle{r}_{{1}}={\left({1},-{4},-{1}\right)}.$

(I don't know how to draw matrices here, so vectors are written horizontal).

The same way we find $\displaystyle{r}_{{2}}={\left({1},-{1},-{1}\right)}$ and $\displaystyle{r}_{{3}}={\left({1},{2},{1}\right)}.$ This gives us the general solution of the original system of equations.

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Wednesday, January 1, 2014