I know the transpose is to swap the columns and rows of a matrix. And $A^T$$A$ is a symmetric matrix which elements are the inner product of each column of $A$. But I didn't understand the intuition of transpose. Suppose $A_{m \times n}$, and A transform a vector from $\Bbb R^n$ to $\Bbb R^m$. But $A^T$ transform a vector from $\Bbb R^m$ to $\Bbb R^n$. What's the relationship between them? Could anyone please explain the relationship between $A^T$,$A$,the inner product and symmetric matrix. I think there would be a intuition explaination.
What's the intuition of the transpose of a matrix?
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0See: http://math.stackexchange.com/questions/37398/what-is-the-geometric-interpretation-of-the-transpose – 2013-09-30
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0See also: http://math.stackexchange.com/questions/484844/intuition-behind-definition-of-transpose-map – 2013-09-30
3 Answers
Well, $A^T$ is the adjoint matrix of $A$ with respect to the ordinary inner products, i.e. $A^T$ is the only linear mapping $B$ such that $$\langle Av,w\rangle = \langle v,Bw\rangle$$ for all $v\in\Bbb R^n$ and $w\in\Bbb R^m$. You can easily see it if you verify it on the standard bases, noting that $\langle u,e_i\rangle$ gives the $i$th coordinate of $u$.
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2thank you. But how to use this conclusion? Is there any geometry explaination? – 2013-01-10
One aspect of this to consider is that the transpose lets you do the same thing in different ways. The regular matrix gets multiplied by a column on the right to give your answer as a column. The transpose gets multiplied by a row on the left to give a row as the answer.
Is one better than the other? Not really, they're equivalent.
Good geometric ilustration of transpose is if we take as linear operator rotation matrix R. In this case is easy to see that < Rv, w> = < v, $R^T$ w> as we have 2 opportunities to change an angle between vector v and w to the same value. One opportunity is to rotate v what Rv operation does, the second one to rotate w in reverse direction what $R^T$ does ( transpose of R is inverse of R).
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0That's the geometric intuition of an inverse matrix, not of the transpose ; you took the case where $R^{\top} = R^{-1}$, which is called an orthogonal matrix. – 2015-04-16
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0Yes, but in this case transpose is equal to inverse, I agree .. it is a special case. – 2015-04-16
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0I'm doing the comment because I myself do not have much intuition about the transpose... the answer didn't help me! – 2015-04-16
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0Maybe some additional intuition about transpose can be taken from decomposition of any matrix $A$ into symmetric and skew-symmetric part. In this case transpose gives the same symmetric part and skew-symmetric with minus sign. – 2015-04-17