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Find the eigenvalues and their eigenspaces:

\begin{bmatrix} 2 & -3 & 1 \\ 1 & -2 & 1 \\ 1 & -3 & 2 \\ \end{bmatrix}

The three eigenvalues I get are $\lambda_1=0$, $\lambda_1=1$, and $\lambda_2=1$.

Does this mean that that since two lambda's have the same value, I only have two eigenvalues $\lambda_1=0$, $\lambda_1=1$? Or is there still 3: $\lambda_1=0$, $\lambda_1=1$, and $\lambda_2=1$

Because the answer I computed were:

from $\lambda_1=0$ one answer,

\begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix}

and from $\lambda_2=1$ I got two answers:

\begin{bmatrix} 3\\ 1\\ 0\\ \end{bmatrix}

\begin{bmatrix} -1\\ 0\\ 1\\ \end{bmatrix}

Is this three together correct? Also, how would I write this as an eigenspace?

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Your result is correct. The matrix have an eigenvalue $\lambda =0$ of algebraic multiplicity $1$ and another eigenvalue $\lambda =1$ of algebraic multiplicity $2$. The fact that for for this last eigenvalue you find two distinct eigenvectors means that its geometric multiplicity is also $2$.


this means that the eigenspace of $\lambda=1$ has dimension $2$ and can be expressed as the set of vectors of the form: $$ \begin{bmatrix} 3y-z\\y\\z \end{bmatrix}= y\begin{bmatrix} 3\\1\\0 \end{bmatrix}+z\begin{bmatrix} -1\\0\\1 \end{bmatrix} $$

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  • $\begingroup$ Thank you! How would I write these results as an eigenspace though? $\endgroup$ – BurstFlame May 29 '17 at 20:04
  • $\begingroup$ @BurstFlame You can write that the eigenspace of $\lambda=1$ is: $$E_1=<\begin{bmatrix} 3\\ 1\\ 0\\ \end{bmatrix},\begin{bmatrix} -1\\ 0\\ 1\\ \end{bmatrix}>$$ where $<\cdot>$ means "span of". $\endgroup$ – John Doe May 29 '17 at 20:15
  • $\begingroup$ I've added to may answer. I hope it's useful :) $\endgroup$ – Emilio Novati May 29 '17 at 20:17
  • $\begingroup$ Thank you! Very helpful $\endgroup$ – BurstFlame May 29 '17 at 20:19

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