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?

Solution 1:

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

Solution 2:

Absolutely! Although mvw's answer is correct, it might be a little more than you are looking for. Such a technique is actually a really big theorem in Linear Algebra. All we do is apply elementary row operations to you matrix: $$\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} \rightarrow_{(1/3)R_2} \begin{bmatrix} 1 & 2 \\ 1 & 4/3 \\ \end{bmatrix} \rightarrow_{R_2 - R_1} \begin{bmatrix} 1 & 2 \\ 0 & (4/3)-(6/3) \\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 2 \\ 0 & (-2/3)\\ \end{bmatrix} \rightarrow_{(-6)R_2} \begin{bmatrix} 1 & 2 \\ 0 & 4 \\ \end{bmatrix}$$ From here, it should seem like a sinch to comlplete the steps. We note that each of the row operations is expressible as an elementary matrix of type, 1, 2, or 3.