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I'm having trouble finding the eigenvectors for an eigenvalue. For example, I've finally found my two eigenvalues, 0 (multiplicity of 1) and 2 (multiplicity of 2). So naturally, I start to solve for the eigenvectors associated with lambda=2. After elementary row operations, I get: $$ \begin{matrix} 1 & -4/3 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} $$

This is where I start to mess up. I let x+z=(4/3)y and from this, one of the eigenvectors I get is [2,3,1]. This isn't correct, as the book tells me the two eigenvectors for lambda = 2 are [4,3,0] and [-2,0,1].

I don't understand, what am I doing wrong and what can I fix?

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  • $\begingroup$ Your pivot is in the first column; the other two variables are free. $\endgroup$ Commented Mar 28, 2018 at 1:15
  • $\begingroup$ Note that $(2,3,1)$ is a linear combination of $(4,3,0)$ and $(-2,0,1)$. $\endgroup$ Commented Mar 28, 2018 at 1:31
  • $\begingroup$ You might find this useful for understanding how to read a basis for the null space from the rref matrix. $\endgroup$ Commented Mar 28, 2018 at 7:17

2 Answers 2

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The only problem with what you've written is a typo ($x + 2z = (4/3)y$); all the work is correct. The $\lambda = 2$ eigenspace is two-dimensional, and your book is giving a basis. But of course this basis is not unique -- in fact you can see that your eigenvector is the sum of the two given in the answer. So just find another linearly independent eigenvector (for example, you could pick either of the ones from the book) and you will be done.

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When you get only one equation with three 'variables' (in this case x, y, z) you should let $x=4/3y-2z$, and then in this space any vector is of the form

$$\begin{pmatrix}4/3y-2z\\y\\z\end{pmatrix}=y\begin{pmatrix}4/3\\1\\0\end{pmatrix}+z\begin{pmatrix}-2\\0\\1\end{pmatrix}$$ where $y, z$ are real numbers, therefore, this is a basis for the eigenspace, i.e. this eigenspace has dimension 2

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