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I am following Gilbert Strang's Linear Algebra course (link).

In lecture 33, he mentions that a rectangular matrix can’t have a two-sided inverse because either that matrix or its transpose has a nonzero null space. I know the condition for the null space for a square matrix to be invertible (null space must contain only the zero vector). However, I am not able to figure out how the null space comes in when we are talking about left and right inverses.

Referring to the example given in the course lecture: We take an $m\times n$ matrix $A$ with $m>n$. $A$ has full column rank. We can compute the left inverse as $A_{left}^{-1} = (A^TA)^{-1}A^T$ using the invertibility of $A^TA$. He mentions that the right inverse will not exist. I think this has to do something with $N(A^T)$ being nontrivial, but I am not able to figure out the exact relation.

Could someone please explain the relation between invertibility for rectangular matrices and its relation between the two null spaces?

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    $\begingroup$ If $xA=0$ and $AB=I$ then $xAB=(xA)B=0B=0$ but also $xAB=x(AB)=xI=x$, contradiction. $\endgroup$ Jul 2, 2021 at 7:08
  • $\begingroup$ @GerryMyerson I'd say that is actually an answer rather than a comment! $\endgroup$ Jul 2, 2021 at 7:10
  • $\begingroup$ @math, chances are the question is a duplicate. But feel free to write it up as an answer. $\endgroup$ Jul 2, 2021 at 7:12
  • $\begingroup$ @GerryMyerson, the closest question I found to mine was math.stackexchange.com/questions/2225575/…. However, it asks only about the right inverse and does not have any mention of the relation with null spaces, so I thought of asking a new question $\endgroup$
    – temp_user
    Jul 2, 2021 at 7:15
  • $\begingroup$ If a matrix has a two sided inverse then it induces a linear trasformation between vectorspaces $\mathbb R^a\to \mathbb R^b$ that is an isomorphisms, i.e. that has a two sided inverse. Conclude that $a=b$. $\endgroup$
    – MphLee
    Jul 2, 2021 at 7:30

1 Answer 1

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If $π‘₯𝐴=0$ and $𝐴𝐡=𝐼$ then $π‘₯𝐴𝐡=(π‘₯𝐴)𝐡=0𝐡=0$ but also $π‘₯𝐴𝐡=π‘₯(𝐴𝐡)=π‘₯𝐼=π‘₯$

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