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I have a matrix $$B= \begin{bmatrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{bmatrix}$$

with two eigenvalues $\lambda=-1$ with eigenvector u=$(0,-1,1)$ and the double eigenvalue $\lambda=1$ to which corresponds the eigenspace $S_2$ given by $(x,y,z)\in\mathbb{R^3}:y-z=0$

My question is:

How can I find a basis of the eigenspace $S_2$?

In general, is a basis of a degenerate eigenspace given by its eigenvectors that are equal in number to the degeneracy and are linearly independent?

Could you please show me how linear independence is required?

I would do this:

One eigenstate can be $(2,1,1)$ and the other eigenstate $(0,1,1)$, now what about the independence?

and generally speaking, how do you find the basis of an eigenspace?

Salmon
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    If you know how to compute a kernel then just apply the definition : $S_2$ is the kernel of $B-I_3$ – Lelouch Jan 28 '22 at 22:59
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    The proposed duplicate target concerns how to find the basis for a subspace defined by a first degree homogeneous polynomial condition on components. With a little effort that technique could be applied here, but it would be worth explaining how. Here we need (for eigenvalue $1$) a basis for solutions to $Bv = v$ Alternatively use elimination to find a basis for solutions to $(B-I)v = 0$ as @Lelouch suggests. – hardmath Jan 29 '22 at 19:44

1 Answers1

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$y-z=0$ means $y=z$ and $x$ is free since we have no condition on it, therefore the vectors inside $S_2$ have shape $(x,y,y)=x(1,0,0)+y(0,1,1)$. Since $(1,0,0)$ and $(0,1,1)$ are independent then these two vectors are your basis.

Lelouch
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  • is $(-2,1,1)$ and $(1,1,1)$ also a basis? – Salmon Jan 29 '22 at 11:04
  • @Salmone If they both are in the eigenspace and are linearly independent, yes. Can you verify that ? – Lelouch Jan 29 '22 at 13:25
  • Yes, they both are in the eigenspace and linearly independent. So the general rule is: let $n$ be the degeneracy of the eigenspace, if I want to find a basis I need $n$ eigenvectors inside that eigenspace which are linearly independent, right? So the basis are infinite? – Salmon Jan 29 '22 at 21:22