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It is said that a matrix $A$ is said to be invertible if there exists a matrix $B$ such that $AB=I=BA$, but if we only know $AB=I$, can we be sure that $BA=I$?

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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Shashwat
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  • If $A$ is square, then yes, always (see linked duplicate). If $A$ is not square, then no, never (look at the maximum ranks). – Arthur Jul 17 '18 at 07:35

2 Answers2

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If $A$ is invertible then $$AB=I\implies ABA=IA=A\implies A^{-1}ABA=A^{-1}A\implies BA=I.$$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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    True if $A^{-1}$ exists. But what about the case $A=\pmatrix{1&0&0\cr0&1&0\cr}$, $B=\pmatrix{1&0\cr0&1\cr0&0\cr}$? – David Jul 17 '18 at 07:33
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If your matrix are square, then is true. A proof is here. The proof doesn't suppose $A^{-1}$ exists. If not, for example $$ A = \begin{pmatrix} 2 & 0 & 1\\ 0 & 0 & 2 \end{pmatrix}\qquad B = \frac{1}{4}\begin{pmatrix} 2 & -1\\ 0 & 0 \\ 0 & 2 \end{pmatrix} $$ Then $$ AB= I_2 \qquad BA = \begin{pmatrix}1& 0 &0\\ 0 & 0 &0 \\ 0& 0 & 1\end{pmatrix} $$

Rafael Gonzalez Lopez
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