I am working on a problem and I think that knowing if there is some sort of pattern to a transpose graphically to its normal matrix could potentially be useful. Is there any reflection, rotation, etc that can be seen by looking at u and v where $A$x=u and $A^{T}$x=v? In other words is there some $C$ where $C$u=v. A three dimensional understanding of this should suffice to get me unstuck, but more general cant hurt of course.
Thanks for any help

Matt Staab
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  • Well, if $A$ is an orthonormal matrix, then $A^{-1} = A^T$. All pure rotation matrices are orthonormal, and their transpose is the inverse rotation. In the general case, you need to look at the row and column vectors of $A$. Transpose is the operation that turns the row vectors into column vectors, and vice versa. Instead of looking at the transpose itself, look at how it functions (as part of some formula where transpose is necessary). In the orthonormal matrix case, the row vectors represent the inverse operation compared to the one represented by the column vectors, and vice versa. – Nominal Animal Jun 14 '17 at 17:51
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    you should see the answer [here](https://math.stackexchange.com/questions/37398/what-is-the-geometric-interpretation-of-the-transpose?rq=1), I think it would help. – Andres Mejia Jun 14 '17 at 18:26

1 Answers1


A handy visualization comes from polar decomposition. In particular, every transformation $A$ can be decomposed into a product $$ A = UP $$ where $U$ is orthogonal and $P$ is (symmetric and) positive definite. In other words: $P$ is a dilation along orthogonal axes, and $U$ is a rotation and/or reflection. Every transformation can be broken down into a "stretch/squish" according to $P$, followed by a rotation/reflection according to $U$.

With that in mind, we have $$ A^T = (UP)^T = PU^T = PU^{-1} $$ So, if $A$ consists of a stretch/squish followed by a flip, then $A^T$ corresponds to first reversing the flip, then applying the same stretch/squish.

Ben Grossmann
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