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I'm looking for a way to determine linearly independent eigenvectors if an eigenvalue has a multiplicity of e.g. $2$. I've looked for this online but cannot really seem to find a satisfying answer to the question. Given is a matrix A:

$$ A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix}$$

I know an exact formula to calculate the eigenvalues of this matrix since a previous exercise learned me that if we have a matrix:

$$ M = \begin{pmatrix} a^2 + t & ab & ac \\ ab & b^2 + t & bc \\ ac & bc & c^2 + t\end{pmatrix} $$

The determinant of this matrix is $det(M) = t^2(t + a^2 + b^2 + c^2)$. When creating the characteristic polynomial for $A$, I find the following values for the variables $a, b, c$ and $t$:

$$ \begin{cases} a = 1 \\ b = 2 \\ c = 3 \\ t = -\lambda \end{cases} $$

After calculating the eigenvalues using this trick, I find them to be $\lambda_1 = 14$ and $\lambda_2 = 0$ (with multiplicity $\mu = 2$). I can find the eigenvector for $\lambda_1$, but when I try and find the eigenvectors for $\lambda_2$, I never get the same results as the solution provides, which are two linearly independent vectors:

$$\vec{x_1} = (2, -1, 0)^T$$ and $$\vec{x_2} = (3, 0, -1)^T$$

Can someone explain me how to find these eigenvectors?

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  • $\begingroup$ What vectors do you get? Note that for eigenvalues of higher multiplicity, the answer is not necesarily unique for example $(1,1,-1)$ and $(1,-2,2)$ is also a good answer. $\endgroup$
    – N. S.
    Jun 3, 2020 at 19:57

2 Answers 2

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You want to find a basis of the null space of $A - \lambda I$. Gaussian elimination is the standard way of doing this.

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  • $\begingroup$ Yes, I did that and I found the following result: $$\begin{pmatrix} 1 & 2 & 3 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$$ But I don't see how I should choose my free variables $c_1$ and $c_2$ to obtain the result now. $\endgroup$ Jun 3, 2020 at 20:15
  • $\begingroup$ The free variables are $x_2$ and $x_3$. If you take $x_2=-1$ and $x_3 = 0$ you get $x_1 = 2$, thus $[2,-1,0]^T$. If $x_2 = 0$ and $x_3=-1$ you get $x_1 = 3$, thus $[3,0,-1]^T$. More natural might be $-1$ instead of $1$, but apparently the author of the solutions preferred the first entry positive instead of negative. $\endgroup$ Jun 3, 2020 at 20:43
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You just have to provide a basis of $\ker(A)$. Therefore, calculate a row-reduced echelon form of $A$:

$$\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$

You now see that the kernel is two-dimensional. Hence, you need two linearly independent vectors which fulfill the equation $x + 2y + 3z = 0$. Suitable choices would be $x_1 = (2, -1, 0)^\intercal$ and $x_2 = (3, 0, -1)^\intercal$ as in your solutions, but there are several possibilities: You could for example also use $(6, -3, 0)^\intercal$ and $(9, 3, -5)^\intercal$ etc.

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