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let us suppose that we have some operator $T$ which acts as a transpose of given matrix in other word

$T(A)=A^T$

question is what is $T^{-1}?$

generally from this answer Transpose of inverse vs inverse of transpose its clear that inverse of this matrix $A^T$ is transpose of inverse of this matrix, but when we are talking about operators, can should i define inverse of operator?i am following like this way

let $B$ be inverse operator or

$B(A)=A^{-1}$ and $T$ be transpose operator, or $T(A)=A^T$

then question comes

what is equal $B(T)$ ? from the above facts we got that

$B(T(A))=T(B(A))$

on following video

https://www.youtube.com/watch?v=-X04WJoTDBc

students said that $T^2=I$ or $T^{-1}=T$ or inverse of transpose operator is itself transpose operator but how did he got? of course

$A^{-1} A=I$

but from all those things, i can't get point what does mean inverse of operator, can you make please clear all those stuff?thanks in advance

EDITED

about $T^2=I$ that means that if we apply operator twise

$T(T(A))=A$ so we get original matrix, does it means that $T^2=I$ ?

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    $\begingroup$ Transposition of matrices is an example of an involution, i.e. an operation that is its own inverse. $\qquad$ $\endgroup$ – Michael Hardy Sep 6 '16 at 17:34
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By the inverse of $T$, you mean the map that "undoes" the matrix transpose. And that's accomplished by transposing again! So $T^{-1} = T$. You're right that $T^2 = I$, which means that $T$ is an involution (and yes, $T^2(A)$ is the same thing as $T(T(A))$.) Finally, the inverse of $T$ doesn't have anything to do with inverting its input argument $A$. After all, transposition is defined for all square matrices, not just the invertible ones.

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  • $\begingroup$ aa i got operation that undo transpose of transposed matrix itself transpose $\endgroup$ – dato datuashvili Sep 6 '16 at 17:21

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