matrices - Is there a way to extract the diagonal from a matrix with simple matrix operations

I have a square matrix A. Is there a way I can apply operations like addition, subtraction, matrix multiplication, matrix inverse and transpose to get the diagonal of the matrix. For example having:

$$\begin{pmatrix}1&2\\3&4\end{pmatrix}$$
I would like to get $(1,4)$.

P.S. based on the conversation with mvw, here is a better description:

I am on board of an alien space ship and the board computer allows only matrix operations but access to the individual matrix elements is blocked. I can only use addition, subtraction, matrix multiplication, matrix inverse and transpose. No access to individual row/column/element. I can only create matrices of any dimension $(1 x n)$, $(n x 1)$, $(n x 2n)$ that have all zeros or all ones. Is there a way for me to get a diagonal vector?

Answer

Note: This solution is not working for the updated question.

$$D = \text{diag}(a_{11}, \ldots, a_{nn}) = \sum_{i=1}^n P_{(i)} A P_{(i)}$$
where $P_{(i)}$ is the projection on the $i$-th coordinate:

$$(P_{(i)})_{jk} = \delta_{ij} \delta_{jk} \quad (i,j,k \in \{1,\ldots,n\})$$
and $\delta$ is the Kronecker delta ($1$ for same index values, otherwise $0$).

Transforming the diagonal matrix $D$ into a row vector can be done by

$$d = u^T D$$
where each of the $n$ components of $u$ is $1$.

$$u = (1,1,\ldots,1)^T$$
Combining both gives
$$d = \sum_i u^T P_{(i)} A P_{(i)} = \sum_i e_i^T A P_{(i)}$$
where $e_i$ is the $i$-th canonical base vector.

Example:

octave> A, P1, P2, u
A =
   1   2
   3   4

P1 =
   1   0
   0   0


P2 =
   0   0
   0   1

u =
   1
   1

octave> u'*(P1*A*P1+P2*A*P2)
ans =

   1   4