2
$\begingroup$

Suppose that we have a $p$ dimensional square matrix $A$ whose rank is less than $p$. We know that such a matrix cannot have an inverse and there are several different ways to prove that the $A$ does not have an inverse.

However, I am struggling to obtain an intuition behind why the inverse does not exist. I considered the following ideas to generate an intuition but failed to do so.

  1. The matrix $A$ can be viewed as a set of transformations such as scaling, translation, rotation etc. Thus, when we apply $A$ to a vector in $p$ dimensions it always maps the $p$ dimensional vector to a vector in a subspace spanned by $A$ if it is less than full rank. Lack of an inverse implies that we cannot reverse the transformations. Why not?

  2. The columns of $A$ span only a subspace of $R^p$ if it is less than full rank. Thus, a transformation such as $A y$ takes a vector from $R^p$ to a vector that always belongs to that subspace. The above viewpoint did not help in obtaining an intuition either.

Is there a way to obtain an intuition as to why a rank deficient matrix does not have an inverse?

$\endgroup$
  • 2
    $\begingroup$ I think point 2 will give you the intuition if you consider the action on a basis. (The image of a basis is a basis of the image.) $\endgroup$ – saulspatz Jan 17 at 14:57
  • $\begingroup$ To elaborate on the above comment, if $A$ does not have full rank, then a nontrivial subspace gets sent to the zero vector. What happens if we try to invert that? In other words, how can we find the inverse image of the zero vector if lots of vectors are mapped to it (the answer: we can't!). $\endgroup$ – OldGodzilla Jan 17 at 15:00
  • $\begingroup$ It hasn't an inverse but it has a pseudo-inverse, a very useful notion. $\endgroup$ – Jean Marie Jan 25 at 15:44
2
$\begingroup$

"Invertible" when talking about linear transformations means "reversible". In other words, a linear transformation (and the corresponding matrix in a given basis) is invertible iff it is possible, given an output, to figure out exactly what the input was.

A rank deficient linear transformation will collapse at least one dimension, meaning each output could be the result of any of a number of different inputs. Specifically, it will have a non-trivial kernel, so multiple different inputs result in the output $\vec 0$.

$\endgroup$
1
$\begingroup$

The row reduced echelon form of your matrix will have some rows of zero at the bottom , which is not invertible.

An invertible matrix when reduced to its row reduced echelon form becomes the identity matrix.

$\endgroup$
1
$\begingroup$

This stems from the fact that a mapping $f:V\to V$ cannot possess any left inverse if it is not injective and it cannot possess any right inverse if it is not surjective:

  • if $f$ is not injective, i.e. if $f(u)=f(v)$ for some $u\ne v$, then $f$ cannot have any left inverse, otherwise we would have $u=(f^{-1}\circ f)(u)=f^{-1}(f(u))=f^{-1}(f(v))=(f^{-1}\circ f)(v)=v$, which is a contradiction;
  • if $f$ is not surjective, i.e. if there is some member $w$ of $V$ that lies outside $f(V)$, then $f$ cannot possess any right inverse, otherwise we would have $w=f(f^{-1}(w))\in f(V)$, which is a contradiction.

Now, if a square matrix $A$ is rank deficient, its columns are linearly dependent. Therefore $Au=0$ for some nonzero vector $u$. In other words, $A$ maps both $u$ and $0$ to $0$. Hence $A$ has not any left inverse, because the mapping $x\mapsto Ax$ is not injective. (If you invert back, what should $A^{-1}0$ be? $u$ or $0$?)

Also, as $A$ is rank deficient, its column space $A$ is a proper subspace of the ambient space. Hence $A$ has not any right inverse, because the mapping $x\mapsto Ax$ is not surjective. (If $w$ lies outside the column space of $A$ and it has an inverse image, then $w$ itself is the image of its own inverse image, hence $w$ also lies inside the column space of $A$. How paradoxical!)

$\endgroup$
0
$\begingroup$

The way i learned is was like this: You should see the Det function as sending a matrix (of dim=n),to the oriented n-dim volume it's column vector's span. if this is zero, this will be a n-1 dim hyper-surface. Thus losing surjectivity into its image, we know injectivity and surjectivity are equivalent for finite dim square matrices. So it can't be invertible.

$\endgroup$
  • $\begingroup$ All functions are surjective onto their image. Onto their codomains, on the other hand... $\endgroup$ – Arthur Jan 17 at 15:17
  • $\begingroup$ yes ofcourse, i meant onto it's n dim vector space $\endgroup$ – Aylon Pinto Jan 17 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.