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Find the eigenvectors and eigenvalues of
$$\mathbf{C}= \left[\begin{array}{r r r} 4&6&2\\6&0&3\\2&3&-1 \end{array}\right]$$

I found the the eigenvalues and are equal to $8.33, -6.63$ and $1.30.$ When I solve $(\lambda \mathbf{I} - \mathbf{C})\mathbf{E}=0$ I get the trivial solution as an answer while I shouldn't. What am I doing wrong?

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  • $\begingroup$ Of course taking $E$ trivial, that is, $E = 0$, gives $0 = 0$. Can you find an other solution for each eigenvalue? Also, there may be a typo in your matrix, the one shown here has eigenvalues $9.43{\dots}$, $-1.57{\dots}$, and $-4.85{\dots}$. $\endgroup$
    – Eric Towers
    Feb 25, 2021 at 8:25
  • $\begingroup$ Where are those $ 8.33, -6.63$ and $1.30$ decimal values coming from? How are you solving the equation to find the eigen vectors? $\endgroup$
    – mathcounterexamples.net
    Feb 25, 2021 at 8:25
  • $\begingroup$ I solved the characteristic equation $\lambda^3-3\lambda^2-53\lambda+72=0$, Matlab gives your answers, but when you do it manually you get different answers than Matlab $\endgroup$
    – Diana
    Feb 25, 2021 at 8:28
  • $\begingroup$ The characteristic equation is not the one you have written. You have the sign of the constant coefficient flipped: "$\dots - 72$". $\endgroup$
    – Eric Towers
    Feb 25, 2021 at 8:31
  • $\begingroup$ $(\lambda - 4) \begin{array}{|r r |} \lambda &3\\ 3&\lambda +1\end{array}- 6\begin{array}{|r r |} 6&3\\2&\lambda +1\end{array} +2\begin{array}{|r r |} 6&\lambda\\2&3\end{array}=(\lambda - 4) \big[\lambda (\lambda+1)-9\big]-6\big[ 6(\lambda+1)-6)\big]+2\big[18-2\lambda \big]$ $\endgroup$
    – Diana
    Feb 25, 2021 at 8:34

1 Answer 1

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We are given the matrix

$$\mathbf{C}= \left[\begin{array}{r r r} 4&6&2\\6&0&3\\2&3&-1 \end{array}\right]$$

Find the characteristic polynomial using $|\lambda I - C|=0$ or $|C - \lambda I|=0$, just be consistent

$$|\lambda I - C| = \left|\begin{array}{r r r} \lambda-4&-6&-2\\-6&\lambda&-3\\-2&-3&\lambda+1\end{array}\right|$$

Use the Laplace expansion to write

$$(\lambda - 4)~ \begin{array}{|r r |} \lambda &-3\\ -3&\lambda +1\end{array}+ 6~\begin{array}{|r r |} -6&-3\\-2&\lambda +1\end{array} -2~\begin{array}{|r r |} -6&\lambda\\-2&-3\end{array}=\lambda^3 - 3 \lambda^2-53 \lambda - 72$$

The roots are

$$\lambda_{1,2,3} = 9.430018403873884, -1.5715091919701965, -4.858509211903689$$

For the eigenvectors, we solve $[\lambda I-C]v_i = 0$ using RREF, so for $9.430018403873884$, we have

$$\left( \begin{array}{ccc} 1 & 0 & -2.424177326987695 \\ 0 & 1 & -1.860554583299498 \\ 0 & 0 & 0 \\ \end{array} \right)v_1 = 0$$

If you want to see steps, you can use this RREF Calculator

So, we have a free choice of $c$ and will choose $c = 1$, so $$v_1 = (2.424177326987695, 1.860554583299498, 1)$$

Repeat this process for the other two eigenvalues and find

$$v_2 = (-0.545323960202366, 0.17304624281151228, 1)\\v_3 = (1.176702188770226, -2.070637863148047, 1)$$

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