Is a matrix invertible only when it is a square matrix?

What about a matrix of the order $m \cdot n$ with $m \gt n$ and such that it is row-equivalent to a row-reduced echelon matrix with more non-zero rows than columns?

What is the motivation behind the concepts of left and right inverse? Is it only useful when dealing with non-square matrix?

  • $\begingroup$ One question per post, or at least only directly related questions. Also you should make "guesses" for each of these questions, with "justification", so that we can help get to the root of your problem. $\endgroup$ – Zach466920 Jun 23 '15 at 1:43
  • $\begingroup$ There is such a thing as a generalized inverse for non-square matrices en.wikipedia.org/wiki/Generalized_inverse $\endgroup$ – Alex S Jun 23 '15 at 1:46
  • $\begingroup$ If you do have a real (or complex) square matrix $A$ that has a left inverse $B$, then $BA=I$. Then neither the determinant of $A$ nor the determinant of $B$ can be zero, so both are invertible. Thus, $B^{-1}BA=B^{-1}$, and $AB=B^{-1}B=I$. Thus, $B=A^{-1}$. So if you have a left or right inverse for a square matrix, it is both. $\endgroup$ – Alex S Jun 23 '15 at 1:51

Let $A$ be a full rank $m\times n$ matrix. By full rank we mean $\DeclareMathOperator{rank}{rank}\rank(A)=\min\{m,n\}$.

  • If $m<n$, then $A$ has a right inverse given by $$ A^{-1}_{\text{right}}=A^\top(AA^\top)^{-1} $$

  • If $m>n$, then $A$ has a left inverse given by $$ A^{-1}_{\text{left}}=(A^\top A)^{-1} A^\top $$

Of course, a right inverse is useful for solving an equation of the form $XA=Y$ and a left inverse is useful for solving an equation of the form $AX=Y$.


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