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Is it possible to geometry to explain why only full rank matrices have inverses?

I understand the geometric explanation of inverses from this post (to undo the transformation of a matrix). And it is also clear that full rank matrices in a space are vectors that all go different directions (independent from each other). But I can't seem to combine these two geometric explanations together to explain why only full rank matrices have inverses.

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  • $\begingroup$ A non-full-rank matrix has notrivial kernel, i.e., there exists some nonzero $v$ with $Av=0$. So you'd need $A^{-1}0=v$, which is absurd $\endgroup$ – Hagen von Eitzen Sep 30 '16 at 21:23
  • $\begingroup$ Matrices are called invertible, not inversable (this is hard to write, my computer immediately wants to convert this back to invertible). $\endgroup$ – Dietrich Burde Sep 30 '16 at 21:33
  • $\begingroup$ @HagenvonEitzen Your explanation makes sense! $\endgroup$ – susanz Oct 1 '16 at 13:35
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To elaborate on the first comment, if you multiply a full rank matrix with any vector (i.e. take linear combination of the columns of full rank matrix) it will not collapse to origin, unless the vector you multiply is zero vector. So, you should be able to apply some inverse transformation to bring every transformed vector to it's initial form. Khan Academy has an excellent series of short videos on geometric interpretation of linear transformations.

https://www.khanacademy.org/math/linear-algebra/eola-topic

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