I have an $n \times n$ matrix, $A$, which is to be concatenated with a row and column of ones as such: $$B = \begin{bmatrix} & & & 1\\ & A & & ...\\ & & & 1\\1 & ... & 1 & 0\end{bmatrix}$$ In the most general case, if we know $rank(A) = k \le n$, can we say that $rank(B) = k + 1$? This is my intuition because the concatenation essentially "adds" a rank to the full matrix $B$, which should not be dependent on the structure of $A$ due to the zero in the lower right, but I can't think of how to prove it.
If this is not the case, then consider three more (increasingly specific) scenarios:
1) $rank(A) = n$
2) $A = A^T$ and (1) holds
3) $A$ diagonal and (1) holds
What can we say about each of these cases? Any hints or nudges in the right direction are appreciated.
$rank(B)$ is not necessarily $k+1$.
For example, \begin{bmatrix} 1 & 2\\ 2 & 3\\ \end{bmatrix} and \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 0 \end{bmatrix} both have rank 2
This just proves that condition 1 does not satisfy the result.
Condition 2): $A = A^T$ and $rank(A)=n$ together cannot guarantee the result either.
Let $$A = \begin{bmatrix} 1 & 0\\ 0 & -1\\ \end{bmatrix}$$
then we have $$rank(A) = rank(B) = 2$$
We can use the same example to disprove condition 3.
Actually this example can disprove all three conditions.