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Lat say matrix:

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

(assume $\beta \ne 0$).

This 2x2 matrix has degree 2, inverse (the same matrix replacing $-\beta$ with $\beta$), determinant=1, ... . It seems very "normal".

Characteristic polynomial is $(1-\lambda)^2=0$, eigenvalue $\lambda=1$ with multiplicity two and one eigenvector=(0,1).

However, I can not find the second independent eigenvector:

$$\begin{pmatrix} 0 & 0\\ -\beta & 0 \end{pmatrix}\begin{pmatrix} a\\ b \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix} \rightarrow a=0$$

Even I've tried to find it testing usual possibilities $(1,0)$, $(\beta,1)$, ... . No luck.

pasaba por aqui
  • 865
  • You have already figured out your answer. I would also like to point out that an eigenvalue's geometric multiplicity (i.e. the max number of linearly independent eigenvectors corresponding to the eigenvalue) cannot exceed its algebraic multiplicity (i.e. its multiplicity as a root of the characteristic polynomial). Thus, if an $n\times n$ real matrix $A$ has $n$ distinct eigenvalues, it will have an eigenbasis for $\mathbb{R}^n$. However, the inverse need not be true: if $A$ has repeated eigenvalues, it need not have an eigenbasis. – Golden_Ratio Jan 18 '22 at 20:43

1 Answers1

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You will not be able to find another independent eigenvector cause there isn't. The dimension of the eigenspace $E(1)$ is only one - not every matrix must have a basis consisting of eigenvectors, only diagonalizable matrices do, and for $\beta \ne 0$ this matrix is not diagonalizable.

Lukas
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    A matrix $A$ is called diagonalizable iff there is an invertible matrix $P$ such that $P^{-1}AP$ is a diagonal matrix. This is equivalent to saying "A matrix $A$ is diagonalizable iff there is a basis of eigenvectors of $A$". The equivalence of these two statements should be a standard linear algebra result - roughly speaking: $P^{-1}AP$ is modeling a base change - if there is a base change such that the matrix of $A$ with respect to the new base is a diagonal matrix then this base is the base of eigenvectors. – Lukas Jan 18 '22 at 20:15