Timeline for Is a matrix $A$ with an eigenvalue of $0$ invertible?
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| Apr 17, 2014 at 1:24 | comment | added | Amitesh Datta | Dear @Ricky, thanks for your comment! It was implicit that we were referring to square matrices with entries in a field throughout the above discussion. I agree that injective self-maps need not be bijective in general! Infinite-dimensional math is hard :) | |
| Apr 17, 2014 at 0:09 | comment | added | user57159 | The 1-by-1 matrix over $\mathbb{R}\lbrack x\rbrack$ whose entry is $x$, and the $\mathbb{N}$-by-$\mathbb{N}$ matrix with ones on the subdiagonal $\hspace{.45 in}$ and zeros everywhere else, are both non-invertible matrices that do not have 0 as an eigenvalue. $\hspace{.66 in}$ | |
| Apr 16, 2014 at 8:50 | comment | added | Amitesh Datta | Dear @Marc, the hypothesis of the question is that $A$ is a square matrix. I guess you could argue that "for square matrices $A$, invertibility is equivalent to injectivity (as a linear operator)" is non-trivial but that doesn't invalidate my answer. Anyway, when I'm used to thinking about something for a long time, my brain identifies logically (but non-trivially) equivalent statements as literally the same :) | |
| Apr 16, 2014 at 8:10 | comment | added | Marc van Leeuwen | But this answer is not quite correct. While the implications in the question can be made into equivalences to show that "$A$ has $0$ as eigenvalue" is equivalent to "there exists a nonzero $v$ with $Av=0$" (in fact I would say that is what the definition of eigenvalue amounts to), the latter means "(the linear map with matrix) $A$ is not injective" which is not the same a "$A$ is not invertible". It is true that for square matrices the linear map is injective if and only if $A$ is invertible, but this is a theorem whose proof is a bit more complicated. | |
| Apr 16, 2014 at 3:38 | history | answered | Amitesh Datta | CC BY-SA 3.0 |