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I know that when the the geometric multiplicity and algebraic multiplicity of a n by n matrix are not equal, n independent eigenvectors can't be found, hence the matrix is not diagonalizable. And I have read some good explanations of this phenomen, like this: Algebraic and geometric multiplicities and this: Repeated eigenvalues: How to check if eigenvectors are linearly independent or not?.

But it still seems to be quite abstract, could someone explain what does this situation eventually mean and what properties does such a matrix have?

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    $\begingroup$ The most satisfying answer is that Jordan canonical form gives a generalization of diagonalization when the geometric multiplicities fall short. You get a decomposition into invariant subspaces of dimension $>1$. $\endgroup$ – Ted Shifrin Feb 27 '18 at 18:50
  • $\begingroup$ At the root level, and I'm not sure this is quite the explanation you're seeking, but if your geometric multiplicity doesn't add up to your algebraic, then your eigenspace (space spanned by the eigenvectors) doesn't span the space of the column vectors of the matrix. You don't have enough information to do what you need to do to diagonalize. $\endgroup$ – Adrian Keister Feb 27 '18 at 19:22
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There is a geometric interpretation that I find helpful.

Consider for example the matrix $\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}$, and think of it as a linear transformation on the plane $\mathbb{R}^2$, namely a linear function $T :\mathbb{R}^2 \to \mathbb{R}^2$ given by the formula $$T(x,y) = (x,y) \begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix} = (x+5y,y) $$ The eigenvalue $\lambda=1$ has algebraic multiplicity 2 but there is only 1 linearly independent eigenvector, namely $(1,0)$ (or anything parallel to it). You can visualize the linear transformation $T$ as follows:

  • The transformation $T$ fixes each point on the $x$-axis, $T(x,0)=(x,0)$ (this, geometrically, is what it means to say that the $x$-axis is the eigenspace generated by the eigenvector $(1,0)$ with eigenvalue $1$).
  • For each horizontal line $L$, given by an equation $y=b$, the transformation $T$ takes each point on that line to another point on that line, with the effect of translating the entire line by a horizontal amount equal to $5b$: $$T(x,b) = (x+5b,b) = (x,b) + (5b,0) $$

This kind of behavior is sometimes described by saying that $T$ is a "shearing transformation" or a "shear mapping".

Shearing maps can be used to give a general description of the behavior when the algebraic multiplicity of an eigenvalue is higher than its geometric multiplicity: the matrix can be decomposed as a product of a "pure shearing transformation" (like the example I just gave) composed with a diagonalizable matrix.

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    $\begingroup$ Also, a "first order" generalized eigenvector of a pure shearing transformation is one that is sent to itself plus an actual eigenvector. So for a pure shear like this, it's just some vector perpendicular to the shearing direction. $\endgroup$ – Ian Feb 27 '18 at 19:15
  • $\begingroup$ I've read it again, great answer! Hopefully I've understand it correctly: does it mean that such matrix transfers a n-dimensional vector in less than n dimensions(less then n positions of the coordinate are changed)? $\endgroup$ – Xichu Feb 28 '18 at 7:48
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    $\begingroup$ For that matrix your observation is correct, less than $n$ positions of the coordinate are changed, although that's a very special situation that occurs only when an eigenvalue $1$ occurs having geometric multiplicity less than its algebraic multiplicity. $\endgroup$ – Lee Mosher Feb 28 '18 at 13:19
  • $\begingroup$ got it, thanks. $\endgroup$ – Xichu Feb 28 '18 at 16:53

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