The rank of a $n\times n$ matrix

Question

$n \times n$ matrix:

So I need to determine the rank of the $n\times n$ matrix in dependence of $a$ and $b$. I guess for $a=0$, $b=0$ the rank is 0. For $b=0$, $a\neq0$ or $a=0$, $b\neq0$ the rank should be $n$ and for $a=b$ the rank should be 1.

I hope this is correct but I am not sure at all. But I still have no clue how to find out the rank for the case of $a\neq b, a > 0, b>0$. I would be very happy if anyone could help me with this.

Answer 1

I assume $n \ge 2$.

Let $\mathbf{1}$ denote the all ones vector so that $\mathbf{11}^\top$ is the all ones matrix.

$$A = b\mathbf{11}^\top + (a-b) I.$$

The eigenvalues of $b\mathbf{11}^\top$ are

• $bn$ with multiplicity $1$ (corresponding to eigenvector $\mathbf{1}$) and
• $0$ with multiplicity $n-1$ (the nullspace of $b\mathbf{11}^\top$ is the subspace orthogonal to $\mathbf{1}$).

[To be clear, I am still allowing $bn=0$ to be possible, in which case $0$ is an eigenvalue with multiplicity $n$.]

Thus the eigenvalues of $A$ are

• $bn+a-b=a+(n-1)b$ with multiplicity $1$ and
• $a-b$ with multiplicity $n-1$.

From here you can get all the possible ranks:

• rank $0$ if $a=b=0$
• rank $1$ if $a=b \ne 0$
• rank $n-1$ if $a=-(n-1)b$ and $b \ne 0$
• rank $n$ if $a \ne b$ and $a \ne -(n-1)b$.

With the extra restrictions $a\ge 0$ and $b\ge 0$, the possibilities are more limited.

• rank $0$ if $a=b=0$
• rank $1$ if $a=b \ne 0$
• rank $n$ otherwise