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If we have a matrix A with some eigenvectors $ev_1, ev_2, ...$

and eigenvectors are independent but not orthogonal, we can represent

$k(ev_1)+v=ev_2$

where v is a vector orthoganal to $ev_1$, and k is a constant.

then we can do

$$A(k(ev_1)+v)=A(ev_2)$$ $$ =\lambda_1(k)(ev_1)+Av =\lambda_2(ev_2)=\lambda_2(k(ev_1)+v)=\lambda_2(k(ev_1))+\lambda_2(v)$$ thus $$\lambda_1(k)(ev_1)+Av=\lambda_2(k(ev_1))+\lambda_2(v)$$ reorder refactor $$( \lambda_1 -\lambda_2)(k)ev_1=\lambda_2v-Av$$

since LHS is a $\lambda_1$ ev, then RHS is also a $\lambda_1$ ev. If v is an ev then there is a clear degeneracy because it lies on the same plane as ev_1 and ev_2.

If vs is not an ev, then that implies that A cannot take v off the plane of v and ev_1 (and ev_2 by that regard. ) Does this imply degeneracy?

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    $\begingroup$ The eigevectors are not necessarily orthogonal. They are if the matrix is symmetric. $\endgroup$
    – Doug M
    Jun 27, 2017 at 0:32
  • $\begingroup$ Eigenvectors of different eigenvalues are not necessarily orthogonal w/r the “standard” inner product. However, in this situation you can decompose $A$ into a sum of projections $P_\lambda$ onto the eigenspaces such that $P_\lambda\mathbf v=0$ when $\mathbf v$ is an eigenvector of a different eigenvalue. $\endgroup$
    – amd
    Jun 27, 2017 at 1:28

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Consider $A=\begin{bmatrix} 1&1\\ 0&2 \end{bmatrix}$. Clearly $Ae_{1}=e_{1}$ and $A(e_{1}+e_{2})=2(e_{1}+e_{2}).$ We can see that here $v=e_{2},$ and $(2I-A)v=\begin{bmatrix} 1&-1\\ 0&0\end{bmatrix}v=-e_{1}=(1-2)e_{1},$ as you noted above, and indeed, $A$ does not move $v$ out of the $e_{1}/e_{2}$-plane (or the $e_{1}/(e_{1}+e_{2})$-plane), but $A$ is clearly not degenerate ($Ax=0$ implies $2x_{2}=0,$ so $x_{2}=0,$ and $x_{1}+x_{2}=0,$ so $x_{1}=0$).

It's worth pointing out that since matrix multiplication is linear, any hyperplane will be mapped to a hyperplane: $A(\alpha_{1} x_{1}+\alpha_{2}x_{2}+\cdots+\alpha_{n}x_{n})=\alpha_{1}Ax_{1}+\cdots+\alpha_{n}Ax_{n},$ so $A$ takes $\mathrm{span}(\{x_{1},\ldots,x_{n}\})$ to $\mathrm{span}(\{Ax_{1},\ldots,Ax_{n}\}).$ A matrix is degenerate when it maps some hyperplanes to hyperplanes of strictly lower dimension. Indeed, if we consider $A=\begin{bmatrix} 1&1\\ 0&0\end{bmatrix},$ then $Ae_{1}=e_{1},$ but $A(e_{1}-e_{2})=0,$ so $A$ maps the $e_{1}/e_{2}$-plane to $\mathrm{span}(\{e_{1}\}),$ which has strictly lower dimension.

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No.

Example: \begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} has eigenvector $(1, 1)$ with eigenvalue $2$, and $(1, 0)$ with eigenvalue $1$.

As Doug M notes, if the matrix is symmetric, then the eignevectors for distinct eigenvalues are orthogonal. (For the same eigenvalue, they need not be orthogonal; for example, the $2 \times 2$ identity has all nonzero vectors as eigenvectors for $1$.)

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    $\begingroup$ Great minds think alike, at least for examples. The result to which you are referring is usually called the spectral theorem for symmetric matrices, but Sylvester's law of inertia refers to something else. The statement of that result may be found here. $\endgroup$ Jun 27, 2017 at 1:30
  • $\begingroup$ I'll edit -- thanks. I'd somehow gotten it in my head -- probably from my linear algebra course in about 1971 -- that Sylvester's law had to do with diagonalizability rather than with the invariance of the signature of a quadratic form. Live and learn, I guess, albeit slowly. :( $\endgroup$ Jun 27, 2017 at 9:47

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