# Geometrically, what does it mean for a matrix to be degenerate? (i.e. have non-distinct eigenvalues)

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?

• 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. – Rahul Apr 22 at 16:43
• @Rahul Thanks, I think that's more specifically what I meant! Is there anything notable about defective matrices when viewed geometrically? – questionmark Apr 22 at 16:51
• 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. – Rahul Apr 22 at 17:05
• You might find my answer here to be helpful – Omnomnomnom Apr 22 at 17:16
• 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? – questionmark Apr 22 at 17:51