So I'm trying to solve this -
Let $A\in M_{5\times 6}(\mathbb{R})$, $A$ has rank $4$.
Let $D=\begin{pmatrix} I_5 & A\\ A^T & 0 \end{pmatrix}$.
I need to find the rank of $D$.
My solution so far:
First, $D$ has $11$ rows and $11$ columns.
It is given that $\rho(A)=\rho(A^T)=4$.
I claim, that since $\rho(A^T)=4$, $D$ has 9 rows which are not-zeroes and LI, hence the rank of $D$ is $9$.
This seems too simple. Am I missing something? Are there any other options?
On
As 5xum has pointed out in a comment, you need to show that the four linearly independent rows in $A^T$ are linearly independent from the first five rows of $D$. In general, this is just not true. Here is a counterexample: $$ \underbrace{\left[\begin{array}{ccccc|cccccc} 1&0&0&0&0 &1&0&0&0&0&0\\ 0&1&0&0&0 &0&1&0&0&0&0\\ 0&0&1&0&0 &0&0&1&0&0&0\\ 0&0&0&1&0 &0&0&0&1&0&0\\ 0&0&0&0&1 &0&0&0&i&0&0\\ \hline 1&0&0&0&0 &0&0&0&0&0&0\\ 0&1&0&0&0 &0&0&0&0&0&0\\ 0&0&1&0&0 &0&0&0&0&0&0\\ 0&0&0&1&i &0&0&0&0&0&0\\ 0&0&0&0&0 &0&0&0&0&0&0\\ 0&0&0&0&0 &0&0&0&0&0&0\\ \end{array}\right]}_D \left[\begin{array}{ccccc|cccccc} 0&0&0\\ 0&0&0\\ 0&0&0\\ 1&0&0\\ i&0&0\\ \hline 0&0&0\\ 0&0&0\\ 0&0&0\\ -1&0&0\\ 0&1&0\\ 0&0&1\\ \end{array}\right]=0 $$ The nullity of $D$ is $3$ here. Hence the rank of $D$ is only $8$.
The field in the counterexample above is complex. This suggests that you need to use the condition that the ground field in your question is real.
On
We are to find the null space of $D$. Now let a vector $(x_1,x_2,...,x_{11})$ be in null space of $D$ & $c_1,c_2,..,c_6$ & $r_1,r_2,..,r_5$ are the columns & rows of $A$. Now note that, taking into consideration the lower half of the $D$, $x_1r_1+x_2r_2+..+x_5r_5=0$ and then such $(x_1,x_2,..,x_5)$ is either 0 and a non null vector because $N(A^T)=(5-4)=1$.
Now for $x_1=x_2=..=x_5=0$ note that $x_6c_1+x_7c_2+..x_{11}c_6=0$ choice of this non null vector will be 2 as $N(A)=6-4=2$.
Now for non null, let, $x_1r_1+x_2r_2+..+x_5r_5=0$ then $x'=(x_1,x_2,..,x_5) \in N(A^T)$ then $x'=x_6c_1+x_7c_2+..x_{11}c_6\in C(A)$ now $x'\in N(A^T)\cap C(A)$ implies $x'=0$
Hence null space of D contains 2 non null vector so rank of D is 9
[Edited and repaired after seeing @user1551 's example.]
First, use elementary column operations to see that $$\rho(\begin{pmatrix} I & A \\A^T & O\end{pmatrix})= \rho(\begin{pmatrix} I & A - IA \\A^T & -A^{T}A\end{pmatrix})= \rho(\begin{pmatrix} I & O \\A^T & -A^{T}A\end{pmatrix}) $$ and then elementary column operations to see that $$ \rho(\begin{pmatrix} I & O \\A^T & -A^{T}A\end{pmatrix})= \rho(\begin{pmatrix} I & O \\A^T - A^{T}I & -A^{T}A\end{pmatrix})= \rho(\begin{pmatrix} I & O \\O & -A^{T}A\end{pmatrix}). $$
Hence $\rho(D)=\rho(I)+\rho(A^{T}A)$.
Now treat the calculation of $\rho(A^{T}A)$ as a separate problem.
Note first that $Ax=0$ certainly implies $A^{T}Ax=0$. But the converse is true since the field is real: $A^{T}Ax=0$ implies $x^{T}A^{T}Ax=0$ implies $Ax=0$. Hence the null spaces of $A$ and $A^{T}A$ coincide. By the rank-nullity theorem the nullity of $A$ is $6-4=2$, so the nullity of $A^{T}A$ is 2, and so the rank of $A^{T}A$ is $6-2=4$.
Had the base field been $\mathbb{C}$ then the result would have been true on replacing $A^{T}$ by $\bar{A}^{T}$.