I'm trying to understand matrix operations as geometric transformations. For example, in the $2$x$2$ case, the matrix $$ \begin{bmatrix} 2 & 1 \\ 0 & 2 \\ \end{bmatrix} $$ produces a shear. However, this is also an upper triangular matrix, so the eigenvalues are on the diagonal, and are not distinct, i.e. both in this case are $2$. I understand what a shear looks like geometrically, but more generally, how do I "spot" a degenerate matrix transformation? What, geometrically, would it mean for a matrix transformation to be degenerate?

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    $\begingroup$ Note that your matrix has a more interesting property than simply having duplicate eigenvalues: it is also defective, that is, it does not have a complete set of eigenvectors. $\endgroup$ – Rahul Apr 22 at 16:43
  • $\begingroup$ @Rahul Thanks, I think that's more specifically what I meant! Is there anything notable about defective matrices when viewed geometrically? $\endgroup$ – questionmark Apr 22 at 16:51
  • $\begingroup$ I guess the definition is geometric enough: the eigenvectors do not span $\mathbb R^n$. Equivalently, there exists a lower-dimensional subspace (in your example, the $x$-axis) outside of which there are no eigenvectors. $\endgroup$ – Rahul Apr 22 at 17:05
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    $\begingroup$ You might find my answer here to be helpful $\endgroup$ – Omnomnomnom Apr 22 at 17:16
  • $\begingroup$ I guess I mean maybe in general then - by just having a matrix and observing its transformation, can you "guess" whether its eigenvectors span or do not span ℝ**𝑛? $\endgroup$ – questionmark Apr 22 at 17:51

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