# Problem on rank of a matrix-Mathematics Competition

Hi, this problem is from a previous year question paper of a math competition which is held in our college annually. I hope it is allowed to discuss these problems here, if not please let me know.

Let $$m,n\in \mathbb N$$. Find the rank of the block matrix $$C$$, where $$C=\begin{pmatrix} (J-I)_{m\times m} & J_{m\times n}\\ J^T_{n\times m} & A_{n\times n}\end{pmatrix}$$. Here $$J$$ is the matrix with all $$1$$, and $$A$$ has rank $$n$$ with $$A\neq J, I, J-I$$.

My try: The eigenvalues of $$(J-I)_{m\times m}$$ are $$m-1$$ with multiplicity $$1$$ and $$-1$$ with multiplicity $$m-1$$.

I am guessing that Rank($$C)=m+n$$, but I dont know any means to prove it. Does there exist any particular result by which rank of block matrices can be determined.

I have just started reading matrix theory, so I am not quite sure on how to proceed with these type of problems. Can someone please point me towards any text or provide me some hints on how to complete this problem?

Apologies for giving the problem wrong. It has been edited now.

Let $$\mathbf e_m$$ the vector with all one of size $$m$$, then $$\mathbf J_{n\times m} = \mathbf e_n\mathbf e_m^T$$.

$$\begin{bmatrix} \mathbf I_{m\times m} & -\mathbf J_{m\times n}A^{-1}\\ \mathbf O & \mathbf I_{m\times m} \end{bmatrix} C = \begin{bmatrix} \mathbf J_{m\times m} - \mathbf I_{m\times m} -\mathbf J_{m\times n}A^{-1}\mathbf J_{n\times m}& \mathbf O\\ \mathbf J_{n\times m} & A \end{bmatrix}.$$ So: $$\text{rank}(C) = \text{rank}(A) + \text{rank}\left(\mathbf J_{m\times m} - \mathbf I_{m\times m} -\mathbf J_{m\times n}A^{-1}\mathbf J_{n\times m}\right)$$

\begin{align}\mathbf J_{m\times m} - \mathbf I_{m\times m} -\mathbf J_{m\times n}A^{-1}\mathbf J_{n\times m} &= \mathbf e_m \mathbf e_m^T - \mathbf I_{m\times m} - \mathbf e_m\mathbf e_n^TA^{-1}\mathbf e_n \mathbf e_m^T\\ &= \left(1 - \mathbf e_n^TA^{-1}\mathbf e_n\right) \mathbf J_{m\times m} - \mathbf I_{m\times m}\end{align}

So the $$\text{rank}\left(\mathbf J_{m\times m} - \mathbf I_{m\times m} -\mathbf J_{m\times n}A^{-1}\mathbf J_{n\times m}\right) = \left\{\begin{array}{cl} m-1 &\text{if 1 - \mathbf e_n^TA^{-1}\mathbf e_n = \frac{1}{m}}\\ m & \text{otherwise} \end{array}\right.$$

And you can conclude after that.

• I checked your edits, I am stuck on the calculation. How did you write $\mathbf{e_me_n^T}A^{-1}\mathbf{e_ne_m^T}=\mathbf{e_n^T}A^{-1}\mathbf{e_n}J_{m\times m}$ Sep 25, 2021 at 5:01
• $$\mathbf e_m \mathbf e_n^TA^{-1} \mathbf e_n \mathbf e_m^T = \mathbf e_m \underset{\in \mathbb R}{\left(\mathbf e_n^TA^{-1}\mathbf e_n\right)}\mathbf e_m^T = \left(\mathbf e_n^TA^{-1}\mathbf e_n\right)\underset{=\mathbf J_{m\times m}}{\left(\mathbf e_m\mathbf e_m^T\right)}$$ Sep 25, 2021 at 15:46
$$\left( \begin{array}{cc|c} 0&1&1 \\ 1&0&1 \\ \hline 1&1&2 \end{array} \right)$$
In this one, the lower right part, which is $$A,$$ does not have full rank. The whole business supports the idea that the rank of the big matrix is either $$m+n$$ or $$m+n-1$$
$$\left( \begin{array}{ccc|cc} 0&1&1&1&1 \\ 1&0&1&1&1 \\ 1&1&0&1&1 \\ \hline 1&1&1&\frac{3}{2}&\frac{3}{2} \\ 1&1&1&\frac{3}{2}&\frac{3}{2} \\ \end{array} \right)$$