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Given the matrix

$$\begin{pmatrix} 3 & 0& 0 \\ -3& 4& 9 \\ 0 & 0& 3 \end{pmatrix}$$

you get eigenvalues $3$ (twice) and $4$. However, when solving for the eigenvector of $3$, you end up with

$$\begin{pmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

How do you solve for the eigenvectors with the free variables?

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As Eigenvectors are not unique, this will happen every time you want to calculate the eigenvectors.

What you can do is in order to calculate this underdetermined system is to assume that $x_1 = s$, $x_2 = t$ and calculate $x_3$ depending on these two parameters. If you do this, you will end up with a vector depending on these two parameters, or if you separate the parameters, an eigenvector which is a linear combination of two linearly independent vectors where $s,t$ are the coefficients: $v= sv_1 + sv_2$, i.e. your Eigenspace is $span \{v_1,v_2\}$

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Your caracteristic equation is $(3-\lambda)^2(4-\lambda)$ so we need to find two vectors such that

$\begin{bmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{bmatrix}v = 0$

Fortunately, you have the freedom do to that.

I think the easiest way to do these is to plug one of your entries equal to 0, and then plug a different entry equal to 0

$\begin{bmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}v_1\\v_2\\0\end{bmatrix} = -3u_1 + u_2 = 0$

$u_2 = 3u_1$

$u = \begin{bmatrix}1\\3\\0\end{bmatrix}$

then

$\begin{bmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}v_1\\0\\v_3\end{bmatrix} = -3v_1 + 9v_3 = 0$

$v_1 = 3v_3$

$v = \begin{bmatrix}3\\0\\1\end{bmatrix}$

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Solve the linear system as usual. Intuitively, when you get a row of zeros, you know you have more than one solution. In this case, you have more than one eigenvector for the corresponding eigenvalue.

$$\begin{cases} -3x_1 +x_2+9x_3 = 0 \\ x_2 = t \\ x_3 = s \end{cases}$$ $$\implies x_1 = \frac{1}{3}t + 3s$$

Thus a basis for the eigenspace corresponding to the eigenvalue of $3$ is $$B_{\lambda_3} = \Bigg \{ t\begin{bmatrix} \frac{1}{3} \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} \Bigg \}$$

This means all eigenvectors can be written as a linear combination of the vectors in this basis.

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  • $\begingroup$ Thanks for your response, a few questions though. Why did you set x2 equal to t, doesn't x2 correspond to -3x1 + x2 + 9x3, and x1 x3 can be s and t. Also, your answer is close, but it doesn't quite match what this calculator is giving me: mathportal.org/calculators/matrices-calculators/… $\endgroup$ – Jane Doe Dec 3 '15 at 8:04
  • $\begingroup$ You forget that all zero rows must be below non-zero rows. Hence, by looking at the pivots of the matrix, you see that you have a pivot in the first row which means $x_1$ is the independent variable and $x_2$ and $x_3$ are free variables. Also, note that a basis for a space is not unique. $\endgroup$ – Kevin Zakka Dec 3 '15 at 8:09
  • $\begingroup$ Oh right, and yes I know that the basis isn't unique, however I think that your answer is still different. I could be wrong, did you plug in the numbers to check? $\endgroup$ – Jane Doe Dec 3 '15 at 8:13

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