# is matrix transpose a linear transformation?

this was the question posed to me. does there exist a matrix $A$ for which $AM$ = $M^T$ for every $M$.the answer to this is obviously no as i can vary the dimension of $M$. but now this lead me to think , if i take , lets say only $2\times2$ matrix into consideration. now for a matrix $M$,$A=M^TM^{-1}$ so $A$ is not fixed and depends on $M$, but the operation follows all conditions of a linear transformation and i had read that any linear transformation can be represented as a matrix. so is the last statement wrong or my argument flawed?

• The question "is $M\mapsto M^T$ linear" is not the same as (nor even related to) "does there exist $A$ such that $M^T=AM$. – Marc van Leeuwen Feb 11 '15 at 15:03
• @MarcvanLeeuwen all linear transformation can be represented by matrices right? – avz2611 Feb 11 '15 at 15:04
• @avz2611 Yes, but the dimension of the transformation $M \mapsto M^T$ is different from the dimension of $M$, so $A$ cannot both represent this transformation and be multiplied by $M$. – 211792 Feb 11 '15 at 15:06
• Yes all all linear transformations (between finite dimensional spaces $V,W$) can be represented by matrices, with respect to chosen bases of $V,W$. Matrix multiplication operates on a column vector of coordinates of the argument with respect to the first basis. But $M$ cannot be taken to be the column vector of its own coordinates for any basis, simply because $M$ is not a column to begin with. Note that the representing matrix always operates by matrix$\times$column product, whi is not what you are doing. – Marc van Leeuwen Feb 11 '15 at 15:08
• @NowIGetToLearnWhatAHeadIs: Yes I know that, even if the hypothesis turns out to be false for all cases except $n\times n$ matrices for $n\in\{0,1\}$, so it is saying nothing more than $P\to Q$ where $P$ is almost always false and $Q$ is always true. But the confusion of OP is clearly that the reverse implication would hold. – Marc van Leeuwen Feb 12 '15 at 5:50

The operation that transposes "all" matrices is, itself, not a linear transformation, because linear transformations are only defined on vector spaces.

Also, I do not understand what the matrix $A=M^TM^{-1}$ is supposed to be, especially since $M$ need not be invertible. Your understanding here seems to be lacking...

However:

The operation $\mathcal T_n: \mathbb R^{n\times n}\to\mathbb R^{n\times n}$, defined by $$\mathcal T_n: A\mapsto A^T$$

is a linear transformation. However, it is an operation that maps a $n^2$ dimiensional space into itself, meaning that the matrix representing it will have $n^2$ columns and $n^2$ rows!

• So one could, for example, represent all $2 \times 2$ matrices as 4-dimensional column vectors. Assuming your vector is $\langle a, b, c, d \rangle$, the $4 \times 4$ "transpose matrix" would then be the matrix which swaps $b$ and $c$, which is pretty easy to construct. – 211792 Feb 11 '15 at 15:00
• @5xum i was saying if an invertible matrix is given that value of $A$ will give me $M^T$ – avz2611 Feb 11 '15 at 15:03
• @avz2611 Yes, but the matrix $A$ is still a matrix, meaning it is a transformation of vectors, not matrices. If $M$ is a $n$ by $n$ matrix, then $A=M^TM^{-1}$ is a mapping from $\mathbb R^n$ onto itself. The "transpose" is a mapping from the space of matrices onto itself. – 5xum Feb 11 '15 at 15:05

The vector space of $n\times n$ matrices is $n^2$-dimensional, hence the matrix representation of the (indeed) linear map $X\mapsto X^T$ would have to be $n^2\times n^2$ (and you better rearrange/interprete the given $n\times n$ matrices into $n^2\times 1$ column vectors).

In addition to the other answers, you certainly have a matrix representation of transpose if you treat $n\times n$ matrices as large vectors ($n^2\times 1$ vector). So that you will have to translate for example a matrix $M=[m_{ij}]_{ij}$ into $(m_{11},\dots,m_{n1},\cdots,m_{1n},\dots,m_{nn})^T$. The transpose sending $m_{ij}$ to $m_{ji}$ is simply the matrix: $$\begin{pmatrix} \overbrace{1~0~\cdots~0}^{n} \\ & 1~0~\cdots~0 \\ && \ddots\\ &&&1~0~\cdots~0 \\ 0~1~\cdots~0 \\ & 0~1~\cdots~0 \\ && \ddots\\ &&&0~1~\cdots~0 \\ \vdots&\vdots&&\vdots\\ 0~\cdots~0~1 \\ & 0~\cdots~0~1 \\ && \ddots\\ &&&0~\cdots~0~1 \\ \end{pmatrix}$$

Even if we restrict $A$ and $M$ to be $n \times n$ matrices, there is no such matrix $A$.

Indeed, taking $M=I$, we get $A=I$. But then, taking a non-symmetric matrix $M$, we can't have $AM=M^T$.

While current answers clearly indicate why the reasoning is not correct, I would like to add one point that has not been mentioned so far. In trying to define $A=M^TM^{-1}$, you are making a matrix $A$ that depends on the argument $M$ that the map is being applied to. That can never be right. While describing a linear transformation as a map involves, like for any map, an expression that gives the result in terms of the argument (as here $M\mapsto M^T$), the matrix representing the linear map must by definition contain constant entries, values that do not depend on the argument (here $M$) the linear map is potentially going to be applied to. This is similar to the coefficients of a polynomial function of$~x$: these coefficient by definition cannot involve$~x$. See also this answer to a related confused question.

Of course this takes nothing away from the other arguments, mostly that $A$ is not the right type of matrix to be considered the matrix representing the linear map $M\mapsto M^T$ (and not to mention the problem you would have for the case of non-square matrices, for which $M\mapsto M^T$ is still a linear transformation).

I thought I would clarify that the transpose is a linear operation by explicitly giving the set of linear operations that need to be performed on the original matrix to get its transpose. I will give the expression for the case of a square matrix $M_{n \times n}$ but this can be extended to arbitrary matrices too.

Let $I_n$ denote the $n \times n$ identity matrix whose $i^{\rm th}$ column is the standard basis $e_i$. Let $S_{n,n}$ denote the $n^2 \times n^2$ perfect shuffle matrix corresponding to writing an $n^2 \times 1$ vector into an $n \times n$ matrix column-wise and then reading it row-wise. Then, it can be shown that

$$M^T = \sum_{i=1}^{n} \left( e_{i}^{T} \otimes I_n \right) S_{n,n} \left( I_n \otimes M \right) \left( \sum_{j=1}^{n} e_j \otimes e_j \right) e_{i}^T,$$

where $\otimes$ denotes the Kronecker product of two matrices. It should be noted that $\left( \sum_{j=1}^{n} e_j \otimes e_j \right) = vec(I_n)$ is just the vectorized form of $I_n$, $\left( I_n \otimes M \right) vec(I_n) = vec(M)$ is the vectorized form of $M$ and $S_{n,n} vec(M) = vec(M^T)$ is the vectorized form of $M^T$. Hence, this also shows that vectorization is a linear transformation.

Answering the title question: "the transpose is a linear map from the space of m × n matrices to the space of all n × m matrices".