For which values of $\ a, T $ is an isomorphism

Question

$\ T: \mathbf R^n \rightarrow \mathbf R^n $, $\ T(v) = Av $

$\ A = \begin{bmatrix} 1 & a & a & \cdots& a \\ a & 1 &a & \cdots & a \\ \vdots & & \ddots & & \vdots \\ \\ a & \cdots & \cdots & \cdots & 1\end{bmatrix} $

for what values of $\ a$ , the transformation will be isomorphism. So looking for cases where $\ \ker T = \{ 0 \} $ , one of them is $\ a = 0 $ but maybe better to try find all cases where $\ T$ isn't isomorphism ?

So Im trying to find $\ Ax = 0 $ where $\ x \not = 0 $ and I can see that $\ a =1 $ satisfies that but I'm not sure what is the way to find all the solutions to this equation? I was thinking of writing the transformation as follows:

$\ T(x_1,x_2,..,x_n) = (x_1 + ax_2+\dots+ax_n, ax_1+x_2+ax_3+\dots+ax_n,\dots,ax_1+\dots+ax_{n-1}+x_n) = \\ (x_1 + a(x_2+\dots+x_n). x_2+ a(x_1+x_3+\dots+x_n), \dots , x_n+a(x_1+\dots+x_{n-1})) $

Answer 1

Hint:

Try to evaluate $detA$ and find $a:detA\not=0$

Hint for evaluating $detA$:

Add all rows to the first one

Answer 2

Hint: What happens if you sum up all the equation given by $Ax=0$ ? And then try to look at a single equation!

Answer 3

Let $J$ be the matrix of all $1$s. Then, supposing $a \neq 0$, $$A = aJ + (1 - a)I = a \left(J + \frac{1 - a}{a} I\right).$$ Hence, $\lambda$ is an eigenvalue of $J$ if and only if $a(\lambda + \frac{1 - a}{a}) = a\lambda + 1 - a$ is an eigenvalue of $A$.

So, what are the eigenvalues of $J$? There's an eigenvector $(1, 1, \ldots, 1)$, corresponding to eigenvalue $\lambda = n$. The rest of the eigenvalues are $0$, as the rows/columns of $J$ are all multiples of each other, and hence the matrix is of rank $1$. So, the eigenvalues of $A$ are $a \cdot 0 + 1 - a = 1 - a$ and $an + 1 - a$.

In order for $T$ to not be an isomorphism, i.e. for $A$ not to be invertible, we require that one of its eigenvalues be $0$. This will only be the case when $a = 1$ or when $a = \frac{-1}{n - 1}$. Any other values will make $T$ an isomorphism.