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When learning Linear Algebra, I learned that rotation matrices don't have real eigenvalues or eigenvectors, because their characteristic polynomial does not factor over $\mathbb{R}$. Over the complex numbers it does, so an $n$-dimensional rotation matrix has $n$ linearly independent eigenvectors.

Somehow I concluded that over the complex numbers, all $n\times n$-matrices have $n$ linearly independent eigenvectors. But the following shear matrix does not:

$$ S = \left(\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}\right) $$

This matrix has two eigenvalues $\lambda_{1,2} = 1$, and the eigenvector $(1,0)$. So the question arises, what is the other eigenvector? Mathematica says $(0,0)$ is the other eigenvector, but I find that a cheap answer, since $(0,0)$ is trivially an eigenvector of every matrix.

So, my first question is, does one really consider $(0,0)$ an eigenvector of $S$?

Second question: Is there a condition for an $n\times n$-matrix to have $n$ linearly independent eigenvectors? Maybe the symmetry of the matrix? Maybe that the eigenvalues are nondegenerate?

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No, $(0,0)$ is not an eigenvector. The matrix $S$ you wrote down has only one eigenvector. It has one eigenvalue, $1$, which has a algebraic multiplicity (the number of times it appears as a root of the characteristic polynomial) of $2$ and a geometric multiplicity (the dimension of its eigenspace) of $1$.

In general, a $n\times n$ matrix does not need to have $n$ linearly independent eigenvectors. For a detailed explanation, I suggest you read about the Jordan canonical form of a matrix, and connected to that, algebraic and geometric multiplicity of eigenvalues, invariant subspaces and characteristic/minimal polynomials. The entire theory behind this is not very complicated (it's taught usually at the end of an introductory linear algebra class), but it does take a couple of lessons to fully cover.

However, there are cases when we can be sure that the matrix has $n$ eigenvectors (in other words, we say that the matrix is diagonalizable). One simple condition that assures this is if the matrix is symmetric.

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  • $\begingroup$ "In general, a $n×n$ matrix does not have $n$ linearly independent eigenvectors" - should it be "does not necessarily" (as opposed to the OP "over the complex numbers, all $n×n$-matrices have $n$ linearly independent eigenvectors")? $\endgroup$ Commented Oct 3, 2023 at 13:25
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The general criterion is this:

An $n\times n$ matrix with coefficients in a field $K$ is diagonalisable over $K$ if and only if its minimal polynomial splits as a product of distinct linear factors.

By the Hamilton-Cayley theorem, a matrix $A$ is a root of its characteristic polynomial, hence $\{p(x)\in K[x]\mid p(A)=0\}$ is a non-zero ideal in $K[x]$. A monic generator of this ideal is the minimal polynomial of $A$. It is a divisor of the characteristic polynomial.

One proves the minimal polynomial and the characteristic polynomial have the same roots.

Example:

If $A=\begin{bmatrix}0& 1&0\\0&0&0\\0&0&0\end{bmatrix}$, the characteristic polynomial of $A$ is $-x^3$, but its minimal polynomial is $x^2$.

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    $\begingroup$ Terrible answer! If he knew what a minimal polynomial is he would never ask that question in the first place. At least give some links or explanations. Even though it is the most correct answer as it is the only necessary and sufficient condition to my knowledge. $\endgroup$ Commented Nov 10, 2015 at 18:47
  • $\begingroup$ @Bernard, just read about the Minimal Polynomial. You say "...splits as a product of distinct linear factors", does that mean that the minimal polynomial must not contain the same factor twice? $\endgroup$ Commented Nov 10, 2015 at 21:55
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    $\begingroup$ Yes, it should have only simple roots. But the characteristic polynomial can have roots with order of multiplicity $>1$. $\endgroup$ Commented Nov 10, 2015 at 21:58

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