Adding the constant column increases the rank of symmetric non-full ranked matrix

Question

Assume we have some symmetric Laplacian matrix $\mathbf{A} \in \mathbb{R}^{n\times n}$ with $rank(A) = n-1$.

Then, let us consider matrix $\mathbf{B} \in \mathbb{R}^{n+1\times n}$, which is matrix $\mathbf{A}$ with one more constant non zero row added.

Will we have then $rank(B) = n$? Below, I have a simple R code for the A matrix

$$ A = \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix} $$ then, adding the row with all elements equal to one increases the rank. I do not understand.

data = c(2,-1, -1, -1, 2, -1, -1, -1, 2)
A = matrix(data,nrow=3,ncol=3,byrow=TRUE) 
print(A)
rankMatrix(A)
B = rbind(A, rep(1,3))
print(B)
rankMatrix(B)

Answer 1

Adding a row increases the dimension of the output vector by $1$, however the matrix would only "increase" in rank if this row was linearly independent to the $n-1$ independent rows. The rank can't be continually increased however. The largest possible rank for an $m\times n$ matrix is $\min(m,n)$ since the rank cannot be larger than the dimension of output vector nor can the matrix "increase" the "size" of the domain (the "input" vector space) by making the rank larger than the domain dimension. This is because linear maps are well-defined and have only one output vector per input vector

Answer 2

If the sum of the entries of your new row is $0$ then $\text{rank}(B) = \text{rank}(A) = n-1$.

To prove that statement:

• Prove that the nulspace of $A$, $N(A)$ is generated by the vector $\left(1,\ldots,1\right)$.
• Since $A$ is symmetric the row space of $A$ is the orthogonal complementary of $N(A)$.

Answer 3

When we take Matrix A & "ADD" Column 2 & Column 3 to Column 1 , we see that Column 1 is totally ZERO.
Hence Column 3 is not "INDEPENDENT" of the other Columns.
Hence Matrix A is not full rank.

When we "APPEND" a new row of $(0,0,0)$ to get Matrix B , adding Column 2 & Column 3 to Column 1 will still give Column 1 Zero hence it will still not be full rank.

When we "APPEND" a new row of $(1,1,1)$ to get Matrix B , adding Column 2 & Column 3 to Column 1 will not make Column 1 Zero , because the last element in that Column will be $1+1+1=3$.
In other words , the new row makes the Column 1 "INDEPENDENT" of the other Columns.
Hence the rank of matrix B will INCREASE by 1.

ADDENDUM :
When we alternatively append a row like $(0,+a,-a)$ , we will end up with Zero Column 1 when we add the 2 Column & Column 3 to Column 1 , because $0+a-a=0$
More generally , we might append a row like $a+b,-a,-b$ , where we still have $(a+b)-a-b=0$
It will not Change the rank.

All other types of rows will Increase the rank by 1.