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Does $AA^T=I$ imply that $A^TA=I$?

The wiki article defines the orthogonal group as:

$$o(n,\Bbb C) = \{ A\in M_n(\Bbb C): AA^T=A^TA=I \}$$

My book writes:

$$o(n,\Bbb C) = \{ A\in M_n(\Bbb C): AA^T=I \}$$

I couldn't show it just by manipulationg:

$$AA^T=I\implies AA^TA=A\implies A^TAA^T=A^T\implies A^T=A^T$$

and so on, never helped. Thanks. I haven't done much linear algebra

Frosty cake
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1 Answers1

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A general property in $GL_n(K)$ asserts that if $B$ is a right inverse for $A$ then $B$ is also a left inverse for $A$. In other words if $AB=I_n$ for some matrices then necessarily $BA=I_n$.

Clément Guérin
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  • This is a very misleading answer, because it is only true for square matrices. Non-square matrices are not elements of the group, and, for them, it is not generally true that $A^\top A = I \implies AA^\top = I$, – SRobertJames Jun 01 '23 at 20:51
  • @SRobertJames, OP's question is about orthogonal matrices and, in particular, square matrices. The underlying question is really about why "$A^T$ is a right inverse" implies that "$A^T$ is an inverse". – Clément Guérin Sep 20 '23 at 05:08