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