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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.

maple
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  • See: http://math.stackexchange.com/questions/37398/what-is-the-geometric-interpretation-of-the-transpose – Casteels Sep 30 '13 at 14:53
  • See also: http://math.stackexchange.com/questions/484844/intuition-behind-definition-of-transpose-map – Christian Blatter Sep 30 '13 at 15:19
  • This is critical to understand why the Rank and Determinant of $A^T$ are the same as $A$. Singular Value Decomposition is a good start. – Kuo Mar 15 '24 at 14:33

3 Answers3

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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 $u$.

Berci
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    thank you. But how to use this conclusion? Is there any geometry explaination? – maple Jan 10 '13 at 11:58
  • Real case: $\langle Av, w \rangle = (Av)^\top w = v^\top A^\top w = \langle v, A^\top w \rangle$ Complex case: $\langle Av, w \rangle = (Av)^\top \overline{w} = v^\top A^\top \overline{w} = v^\top \overline{A^* w} = \langle v, A^* w \rangle$ – dwolfeu Mar 18 '24 at 08:49
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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).

widawen
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  • That'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. – Patrick Da Silva Apr 16 '15 at 11:47
  • Yes, but in this case transpose is equal to inverse, I agree .. it is a special case. – widawen Apr 16 '15 at 11:54
  • I'm doing the comment because I myself do not have much intuition about the transpose... the answer didn't help me! – Patrick Da Silva Apr 16 '15 at 14:31
  • Maybe 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. – widawen Apr 17 '15 at 09:35
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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.

Chris Bonnell
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