Assume, that $A$ and $B$ are matrices. Which of the following statements are true and which are false? Note that such an assertion is proved to be untrue by giving a counterexample, namely concrete matrices.

  1. If $AB$ = $B$ and $B$ $≠$ $O$, then $A$ $=$ $I$

  2. If $AB$ $=$ $I$, then $A$ is the inverse of $B$

This has been driving me nuts... any help on how to proceed this. I just cannot seem to find the correct concrete examples like they're asking.

  • $\begingroup$ (2) is the definition of what it means for $A$ to be the inverse of $B$ or vice versa for $B$ to be the inverse of $A$. As for (1), you should immediately be suspicious. As a hint towards a counterexample... diagonal matrices are particularly easy to multiply. $\begin{bmatrix}a_1&0\\0&a_2\end{bmatrix}\begin{bmatrix}b_1&0\\0&b_2\end{bmatrix}=\begin{bmatrix}a_1b_1&0\\0&a_2b_2\end{bmatrix}$ $\endgroup$ – JMoravitz Nov 8 at 16:25
  • $\begingroup$ @JMoravitz (2) is not the definition. See this post $\endgroup$ – Omnomnomnom Nov 8 at 16:32
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    $\begingroup$ Are your matrices in $M_n(K)$, i.e., square matrices? $\endgroup$ – Dietrich Burde Nov 8 at 16:37
  • $\begingroup$ @Omnomnomnom I don't see where in the page you are referring to, but I clearly remember for square matrices $A$ and $B$ having the inverse defined as such in the textbooks from which I was taught. Granted, if your complaint is how $A$ and $B$ aren't said to be square in this case, then yes, I admit then that $A$ shouldn't have received the name "inverse" of $B$ but instead received the name "left-inverse" of $B$ and that counterexamples would exist, for example $A=[1~0~0~0]$ and $B=[1~0~0~0]^T$, resulting in $AB=[1]$. $\endgroup$ – JMoravitz Nov 8 at 16:37
  • $\begingroup$ @JMoravitz The conventional definition is that "$B$ is the inverse of $A$ if and only if $BA = AB = I$". I have not yet seen and would be surprised to find a textbook definition that only requires either $BA = I$ or $AB = I$ (even if $A$ and $B$ are said to be square). $\endgroup$ – Omnomnomnom Nov 8 at 16:40

You said it correctly. One should look for "concrete matrices". Take $$ A=\begin{pmatrix} 0 & 0 \cr 0 & 1\end{pmatrix},\; B=\begin{pmatrix} 0 & 0 \cr 1 & 0\end{pmatrix} $$ Then $AB=B$ and $B\neq 0$, $A\neq I$.


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