Dtyjlui

I'm doing an assignment in linear algebra that involves a notation that I don't understand it is simply two ones ($mathbf{11}$). I understand that a $mathbf{1}$ in linear algebra means a sum vector but does that mean that $mathbf{11}$ is simply the matrix counterpart "sum matrix"?.

For example,
$$mathbf{1}_{2} = begin{bmatrix} 1 \ 1end{bmatrix}.$$
Would it be correct to assume that
$$begin{bmatrix}mathbf{1}_{2}mathbf{1}_{2}end{bmatrix} = begin{bmatrix} 1 & 1 \ 1 & 1end{bmatrix}?$$

I'm asking because there is an equation that involves the transpose of $mathbf{11}$ but since it is all ones and it is square that doesn't do anything.

The question is a bit vague; best guess follows.
The vector $mathbf{1}_{n}$ is a column vector of length $n > 0$ with identical unit entries:
$$
mathbf{1}_{3} = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}
$$
This is useful for creating constant vectors with the scalar product $cmathbf{1}$. And, as alluded, when used in a dot product, it produces a summation. For example, given a (real) vector $x$ of dimension 3,
$$
xcdotmathbf{1}_{3} = mathbf{1}_{3} cdot x = sum_{k=1}^{3}x_{k}.
$$
To address your specific question,
$$
mathbf{1}_{n} cdot mathbf{1}_{n} = sum_{k=1}^{n}1 = n.
$$

Many instructors encourage students to think of matrices as a combination of column vectors. For example, using two copies of $mathbf{1}_{3}$ yields
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 \ 1 & 1 \ 1 & 1 \end{bmatrix}
$$
In this instance, the transpose matrix is
$$
mathbf{A}^{T} = begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 end{bmatrix}
$$

As you suspected, the transpose of $n$ copies of the constant vector of length $n$ is equivalent to the original matrix:
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
$$
The transpose matrix is a composition of row vectors
$$
mathbf{A}^{T}
= begin{bmatrix} mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
=mathbf{A}
$$

An important class, symmetric matrices, are defined by the property
$$mathbf{A}=mathbf{A}^{T}$$