How to show the matrix has Rank $\le 5$

Question

I want to show that the following matrix has Rank $\le 5$.

The matrix is \begin{bmatrix} 2&1&1&1&0&1&1&1\\ 1&2&1&1&1&0&1&1\\ 1&1&2&1&1&1&0&1\\ 1&1&1&2&1&1&1&0\\ 0&1&1&1&2&1&1&1\\ 1&0&1&1&1&2&1&1\\ 1&1&0&1&1&1&2&1\\ 1&1&1&0&1&1&1&2 \end{bmatrix}

I found that there is a submatrix in the matrix which has rank $ =4$ given by

$[2,1,1,1],[1,2,1,1],[1,1,2,1],[1,1,1,2]$.

I need to show the given matrix has at least 3 zero rows in order to show that Rank $\le 5$..

But I dont know how to show it. Can someone help.

Answer 1

The given matrix is equal to $A+B$ where $$ A=\begin{bmatrix} 1&1&1&1&1&1&1&1\\ 1&1&1&1&1&1&1&1\\ 1&1&1&1&1&1&1&1\\ 1&1&1&1&1&1&1&1\\ 1&1&1&1&1&1&1&1\\ 1&1&1&1&1&1&1&1\\ 1&1&1&1&1&1&1&1\\ 1&1&1&1&1&1&1&1 \end{bmatrix},\quad B=\begin{bmatrix} 1&0&0&0&-1&0&0&0&\\ 0&1&0&0&0&-1&0&0&\\ 0&0&1&0&0&0&-1&0&\\ 0&0&0&1&0&0&0&-1&\\ -1&0&0&0&1&0&0&0&\\ 0&-1&0&0&0&1&0&0&\\ 0&0&-1&0&0&0&1&0&\\ 0&0&0&-1&0&0&0&1& \end{bmatrix}. $$ These matrices have rank $1$ and $4$ respectively, so their sum has rank at most $1+4$.

Answer 2

Let $$ A = \begin{bmatrix} 2 & 1 & 1 & 1 \\ 1 & 2 & 1 &1 \\ 1 &1&2&1\\1&1&1&2\end{bmatrix}, \quad B = \begin{bmatrix} 1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1 \end{bmatrix} = vv^{T} $$ where $v = [1, 1, 1 ,1]^{T}$. It is easy to see that $rank(B) = 1$. Now our original matrix is $$ X = \begin{bmatrix} A & 2B - A\\ 2B -A & A\end{bmatrix} $$ and we have $$ \begin{bmatrix} I & I \\0&I \end{bmatrix}X\begin{bmatrix} I&0\\I&I \end{bmatrix} = \begin{bmatrix} 4C&2C \\ 2C & A\end{bmatrix} $$ so the rank of $X$ is same as the rank of the above matrix. (Rank of matrix is preserved by multiplying invertible matricies) Since all 4 colums and 4 rows are the same, Gaussian elimination (elementary row and column operation) gives a matrix with same rank $$ \begin{bmatrix} 0_{3\times 3} & 0_{3 \times 5} \\ 0_{5\times 3} & C \end{bmatrix} $$ where $$ C = \begin{bmatrix} 4&2&2&2&2 \\ 2&2&1&1&1\\2&1&2&1&1\\2&1&1&2&1\\2&1&1&1&2\end{bmatrix} $$ which is invertible by Gaussian elemination again.

Answer 3

We will show that the rank of the given matrix is exactly $5$.

The matrix can be written as a block matrix such that (see LU-decomposition), $$M:=\begin{bmatrix} A & B \\B &A \end{bmatrix}=\begin{bmatrix} I & 0 \\BA^{-1}&I \end{bmatrix}\cdot \begin{bmatrix} A&0\\0&A-BA^{-1}B \end{bmatrix}\cdot\begin{bmatrix} I&A^{-1}B \\ 0 & I\end{bmatrix}.$$ with $$A = \begin{bmatrix} 2 & 1 & 1 & 1 \\ 1 & 2 & 1 &1 \\ 1 &1&2&1\\1&1&1&2\end{bmatrix}, \quad B = \begin{bmatrix} 0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0 \end{bmatrix}.$$ Now $$A^{-1}=\frac{1}{5}\begin{bmatrix} 4 & -1 & -1 & -1 \\ -1 & 4 & -1 &-1 \\ -1 &-1&4&-1\\-1&-1&-1&4\end{bmatrix}\quad \text{and}\quad A-BA^{-1}B=\frac{4}{5}\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 &1 \\ 1 &1&1&1\\1&1&1&1\end{bmatrix}.$$ Therefore we may conclude that $$\text{rank}(M)=\text{rank}\left(\begin{bmatrix} A&0\\0&A-BA^{-1}B \end{bmatrix}\right)=\text{rank}(A)+\text{rank}(A-BA^{-1}B) =4+1=5.$$

Answer 4

Call column $j$ of this matrix $C_j$, and denote by $\def\2{\mathbf 2}\2$ the column vector with all $8$ entries equal to $2$. Note that for $j=1,2,3,4$ one has $C_j+C_{j+4}=\2$. Thus $C_6=\2-C_2=C_1+C_5-C_2$, and similarly $C_7=C_1+C_5-C_3$ and $C_8=C_1+C_5-C_4$. The last three columns being in the span of the first five, the rank of the matrix is at most$~5$.

Answer 5

We get rank exactly five from Sylvester's Law of Inertia:

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrrrrrr} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{ 1 }{ 2 } & \frac{ 1 }{ 3 } & 1 & 0 & 0 & 0 & 0 & 0 \\ \frac{ 1 }{ 2 } & \frac{ 1 }{ 3 } & \frac{ 1 }{ 4 } & 1 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 2 }{ 3 } & \frac{ 1 }{ 2 } & \frac{ 2 }{ 5 } & 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 2 } & - \frac{ 1 }{ 3 } & \frac{ 1 }{ 2 } & \frac{ 2 }{ 5 } & 1 & 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & \frac{ 1 }{ 3 } & - \frac{ 1 }{ 2 } & \frac{ 2 }{ 5 } & 1 & 0 & 1 & 0 \\ \frac{ 1 }{ 2 } & \frac{ 1 }{ 3 } & \frac{ 1 }{ 4 } & - \frac{ 3 }{ 5 } & 1 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrrrrrr} 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 4 }{ 3 } & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{ 5 }{ 4 } & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{ 4 }{ 5 } & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrrrrrr} 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } & \frac{ 2 }{ 3 } & - \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } \\ 0 & 0 & 1 & \frac{ 1 }{ 4 } & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ 0 & 0 & 0 & 1 & \frac{ 2 }{ 5 } & \frac{ 2 }{ 5 } & \frac{ 2 }{ 5 } & - \frac{ 3 }{ 5 } \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrrrrr} 2 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 2 & 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 2 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 2 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 & 2 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 0 & 1 & 1 & 1 & 2 \\ \end{array} \right) $$

Answer 6

By inspection, the vectors $(1,-1,1,-1,1,-1,1,-1)$, $(1,0,-1,0,1,0,-1,0)$, and $(0,1,0,-1,0,1,0,-1)$ are all in the nullspace, and are linearly independent, so the rank is at most $5$.