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My matrix is

$$A=\begin{bmatrix}0 & -1\\-1 & 0\end{bmatrix}$$

I used $$det(\lambda I-A)=0$$

And found my two eigenvalues, $1$ and $-1$.

If I from here try to use $\lambda = -1$ it results in the equations

$$-v_1+v_2=0$$ $$v_1-v_2=0$$

Am I then correct in that I can from this point just state that $$v_1=v_2$$ and therefore the eigenvector can be expressed as $$\begin{bmatrix}v_1 \\v_1\end{bmatrix}$$ where I can just pick a value for $v_1$?

Also, how would you express this answer concisely (if my work is correct that is)?

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  • $\begingroup$ Yes, that is correct. It is sometimes conventional to normalize the eigenvector, which means that the representative eigenvector would be $\langle 1/\sqrt{2}, 1/\sqrt{2} \rangle.$ Also, you can check to see if that eigenvector works by multiplying it by the original matrix and seeing if it only changes the vector by scalar multiplication (in this case the scalar should be the eigenvalue). $\endgroup$ – Jbag1212 Feb 15 at 0:42
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    $\begingroup$ Remember that there’s no such thing as “the” eigenvector. Any nonzero scalar multiple of an eigenvector is also one. $\endgroup$ – amd Feb 15 at 0:44
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It's correct. You could express the entire eigenspace as $E_{-1}=\{t\cdot\begin{pmatrix}1\\1\end{pmatrix}\mid t\in \Bbb R\}$.

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You know that $(A-I)(A+I)=(A+I)(A-I)=0$. So the column vectors of $A+I$ are eigenvectors with eigenvalue $-1$. And the column vectors of $A-I$ are eigenvectors with eigenvalue $1$.

The column space of $A-I$ is one-dimensional: $$ A+I = \left[\begin{array}{cc}-1 & -1 \\ -1 & -1\end{array}\right]. $$ The column space of $A+I$ is one-dimensional: $$ A-I = \left[\begin{array}{cc}1 & -1 \\ -1 & 1 \end{array}\right]. $$

This is another way to get at what you are doing.

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