Krdytkyu

I'm currently taking a university class on linear algebra. In some proofs, a given matrix $A in mathbb{R}^{mtimes n}$ is said to be able to be represented as a $1times n$ row vector of $m times 1$ column vectors, i.e.:
$$
A = [vec{a_1}quadvec{a_2}quaddotsquad vec{a_n}]
$$

with $vec{a_i}$ the ith column of $A$. Naturally, the transpose of $A$, as used in most such proofs, would then be given by an $ntimes 1$ column vector of $1times m$ row vectors:
$$
A^T = [vec{a_1^T}quadvec{a_2^T}quaddotsquad vec{a_n^T}]^T
$$
When performing matrix-vector multiplication of the form $Avec x$, I know the vector's amount of rows has to match the matrix's amount of columns. (In this case, $vec x$ would be an $n times 1$ column vector, and the result would be an $[m times n][n times 1] = [m times 1]$ column vector.)

Now, when considering $A$ as a row vector as in the first equation, we can see that this holds up, as the result would be $[1 times n][n times 1] = [1 times 1]$, though containing a sum of scaled $m times 1$ column vectors, so, after scaling and adding, an $m times 1$ column vector.
$$
Avec x = [vec{a_1}quadvec{a_2}quaddotsquad vec{a_n}][x_1quad x_2 quaddotsquad x_n]^T = vec{a_1}cdot x_1 + vec{a_2}cdot x_2 + dots + vec{a_n}cdot x_n in mathbb{R}^{m times 1}
$$
My problem, then, arises when considering the same product, but swapping $A$ with $A^T$ (Edit: For clarity, this is not transposing $Ax$, but rather interchanging $A$ to see what happens.), thus $A^Tvec x$ with $vec x$ the same $n times 1$ column vector. Conventionally, this would be the product of an $n times m$ matrix with an $n times 1$ vector: impossible. However, when considering $A^T$ as an $n times 1$ column vector like in the second equation, this does seem to become possible, namely as the dot of two vectors of equal dimensions:
$$
A^Tboldsymbol{cdot}vec x = [vec{a_1^T}quadvec{a_2^T}quaddotsquad vec{a_n^T}]^Tboldsymbol{cdot}[x_1quad x_2 quaddotsquad x_n]^T = vec{a_1^T}cdot x_1 + vec{a^T_2}cdot x_2 + dots + vec{a^T_n}cdot x_n in mathbb{R}^{1times n}
$$
This, to me, seems to be a paradox, as there obviously is a mismatch in the dimensions of $vec x$ (i.e. $n$) and the dimensions of the input space of $A^T$ (i.e. $m$), yet, rewriting $A^T$ as a vector, as is done with $A$, eliminates said mismatch, as it seems. Is it wrong to assume one can eliminate this mismatch? Is it wrong to assume the matrix-vector product of a matrix in $mathbb{R}^{ntimes 1}$ with a vector in $mathbb{R}^n$ to be equal to the dot product of two vectors in $mathbb{R}^n$? Is it wrong to write a matrix as a vector of vectors, or perhaps to transpose such a construction?

There are 2 wrong things:

1) Although you can write an $n times m$ matrix as a table with $m $ columns each of which is a column vector of $n $ entries, this does not mean you are allowed to consider that its dimension is $1 times m $. It is still $n times m$.

2) the dot product of vectors $a $ and $b $ is carried out as $a^T , b $ so dimensions must not be equal, the must be $1 times n $ and $n times 1$.