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This question already has an answer here:

I don't understand the motivation of the transpose (or better yet, I haven't even seen one). It feels like just something pulled out of a hat. Thinking about it makes it seem like a product of being able to write a matrix using either columns or rows 'first'. E.g., when we 'reflect down the diagonal' we are really keeping all the information of our old matrix, just changing rows to columns.

This is much like how $f \text{ from} \{1,2,...,m-1,m\} \times \{1,2,...,n\} \text{ to some field } \mathbf{F}$ is an $m \times n$ matrix,
● if we switch the order of the product, so that we would define: $g \text{ is a map from } \{1,2,...,n\} \times \{1,2,...,m - 1, m\} \text{ to } \mathbf{F}$
● but allow $f(i,j) = g(j,i)$.
Then we somehow get a 'natural' map from the space of $m \times n$ matrices to $n \times m$ matrices.

Is that what transpose is - a 'natural' map for those spaces? What do I even mean by natural here (serious question, I'm not being mysterious)? If I had to guess, it is a linear bijective map? Is that close enough?

Moving away from this, does the transpose have any useful application in Euclidean geometry (other than orthogonal matrices being defined in terms of transposes?).

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marked as duplicate by Marc van Leeuwen linear-algebra Apr 16 '15 at 11:58

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.

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The transpose is closely related to dual spaces. A linear transformation $T:V\to W$ gives rise to a linear transformation $T^*:W^*\to V^*$ of the dual spaces. The corresponding matrix is the transpose of the original one, when you consider dual bases.

See http://en.wikipedia.org/wiki/Dual_space#Transpose_of_a_linear_map.

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The transpose can be thought of as a generalization, or perhaps linearization, of the transpose of a binary relation (defined by $x R^T y \Leftrightarrow y R x$; for example, "is the parent of" is the transpose of "is the child of"). Indeed, it is possible to represent a relation between two sets $A, B$ as an $|A|$-by-$|B|$ matrix of $0$s and $1$s, and then the transpose of this matrix is the matrix of the transpose relation. If one thinks of a linear transformation between inner product spaces as a "linear relation," then its transpose is the "transpose linear relation" (which always exists, unlike the inverse).

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  • $\begingroup$ But the transpose of a linear map, interpreted as a binary relation, is not the usual linear transpose, it's the inverse map (or rather, the inverse one-to-many relation). Why do you say the transpose of a relation corresponds to the transpose of a linear map? $\endgroup$ – Jack M May 15 '18 at 23:02
  • $\begingroup$ @Jack: this is exactly what is being explained in the second sentence of this answer. $\endgroup$ – Qiaochu Yuan May 15 '18 at 23:08
  • $\begingroup$ I guess. But the matrix representation of a binary relation is not really related to the matrix representation of a linear map in any way other than that it happens to be a two dimensional grid (ie, if you take the matrix representation of linear map as a binary relation, you get an uncountably infinite grid of ones and zeros, not the usual matrix of the map), so I'm not sure if there's really an interesting conceptual connection here. Certainly not one that's obvious to me. $\endgroup$ – Jack M May 15 '18 at 23:13
  • $\begingroup$ @Jack: if you interpret the entries of the matrix as living in the two-element Boolean algebra, then matrix multiplication becomes composition of relations. $\endgroup$ – Qiaochu Yuan May 15 '18 at 23:25
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In $\mathbb{R}^n$ with the standard inner product, $\langle x,y\rangle=\sum_i x_iy_i$, and for $A:\mathbb{R}^n\to\mathbb{R}^n$ linear, we have $\langle Ax,y\rangle=\langle x,A^ty\rangle$. for instance, an orthogonal transformation $A$ preserves the inner product so that $\langle x,y\rangle=\langle Ax,Ay\rangle=\langle x,A^tAy\rangle$ and we find that $A^tA=I$.

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  • $\begingroup$ Doesn't this hold for non-square matrices $A\colon \mathbb R^n \to \mathbb R^m$ as well? $\endgroup$ – Rahul Jan 17 '12 at 0:49
  • $\begingroup$ @Rahul: Yes, if you interpret the inner products in the appropriate spaces. $\endgroup$ – Arturo Magidin Jan 17 '12 at 3:20

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