Determine the rank of the linear map given by a $n \times n$ matrix , dependend on n. Proof by induction

Question

The task:

Let $$ A:= \begin{pmatrix} 1 & a & a & ... & a\\ a & 1 & a & ... &a \\ a & a & ... & a & a\\ ... & ... &... & 1 & a \\ a & a & a & a & 1 \end{pmatrix} $$

a $ n \times n$ matrix / linear mapping. Determine the rank of A over $\mathbb{R}$ for $a\not= 1$.

What i have:

The rank is defined as $dim(Im(f))$. In our case our linear mapping maps $\mathbb{R}^n \rightarrow \mathbb{R}^n$

as $ dim V = dim (ker(f)) + dim (Im(f))$

we know that

$ n - dim(ker(f)) = dim (Im(f)) = r_k (f)$

For n = 1 we have a matrix just containing 1, so the kernel will be trivial. So the rank for $n=1$ is $1-0 = 1$.

For n = 2 i wrote down $$ \begin{pmatrix} 1 & a \\ a & 1 \end{pmatrix} $$ If you scale the first line by a and then substract the first from the second we have:

$$ \begin{pmatrix} 1 & a \\ 0 & (1-a^2) \end{pmatrix} $$ So i came to the conclusion that for $ a = \pm 1 $ the kernel will be one - dimensional, and the rank will be $2-1 = 1$

If $a\not= \pm 1 $ the kernel is trivial, so the rank is $2-0 = 2$.

I think that

\begin{equation} r_k(A) = \begin{cases} n & \text{for } a\not= \pm 1 \\ n -1 & \text{for } a= \pm 1 \end{cases} \end{equation}

I want to prove it by induction now and here is where i am stuck: How can i show that for every n i will get the above result?

Answer 1

you can write this matrix as rank one perturbation of identity. that is $$ A = (1-a)I + auu^\top, u = (1,1,\cdots, 1)^\top.$$ it is not hard to verify that that $A$ has eigenvalues $$(1-a) +na, 1-a, 1-a, \cdots, 1-a$$ and the corresponding eigenvectors are $\{u, u^\perp\}.$ therefore the determinant of $A$ is $$(1-a)^{n-1}(1-a+na). $$

Answer 2

As $\det A=(1-a)^{n-1}\bigl(1+(n-1)a\bigr)$, which one can check by a direct row reduction, one sees that $\,\operatorname{rank} A=n\,$ if $a\neq 1, -\dfrac1{n-1}$.

It is also clearly equal to $1$ if $\,a=1$.

Claim: if $a= -\dfrac1{n-1}$, $\,\operatorname{rank} A=n-1$.

Indeed, adding all above rows to the last one, then multiplying each row with $1-n$, you get the equivalent matrix:

$$\begin{bmatrix} 1-n&1&1&\dots&1&1\\ 1&1-n&1&\dots&1&1\\ 1&1&1-n&\dots&1&1\\ \vdots\\ 1&1&1&\dots&1-n&1\\ 0&0&0&\dots&0&0 \end{bmatrix}$$ All we have to prove is, for instance, the $(n-1)\times(n-1)$ minor: $$\begin{vmatrix} 1&1&\dots&1&1\\ 1-n&1&\dots&1&1\\ 1&1-n&\dots&1&1\\ \vdots\\ 1&1&\dots&1-n&1 \end{vmatrix}$$ is non-zero. But that is easy: substracting the last column from the $n-2$ first, one obtains: $$\begin{vmatrix} 0&0&\dots&0&1\\ -n&0&\dots&0&1\\ 0&-n&\dots&0&1\\ \vdots\\ 0&0&\dots&-n&1 \end{vmatrix}=(-1)^{n-2}\begin{vmatrix} -n&0&\dots&0\\ 0&-n&\dots&0\\ \vdots\\ 0&0&\dots&-n \end{vmatrix}=n^{n-2}.$$

Summary: $$A\enspace\text{has rank equal to}\enspace\begin{cases} 1&\text{if}\enspace n=1\\ n-1&\text{if}\enspace n=-\dfrac1{n-1}\\ n&\text{otherwise} \end{cases} $$