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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.

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marked as duplicate by Marc van Leeuwen linear-algebra Apr 16 at 11:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

See:… –  Casteels Sep 30 '13 at 14:53

3 Answers 3

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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thank you. But how to use this conclusion? Is there any geometry explaination? –  maple Jan 10 '13 at 11:58

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.

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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).

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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 at 11:47
Yes, but in this case transpose is equal to inverse, I agree .. it is a special case. –  widawen Apr 16 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 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 at 9:35

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