Find the inverse and determinant of A=(aI +T), where is $a\ne 0$, $T$ has rank-one and zero trace.
I just verified that a rank-one matrix has at most one non-zero eigenvalue.
Now since T is of rank-one and has zero trace, that means all of its eigenvalues are $0$, and so T is nilpotent. (And $T^2 = 0$.)
But I'm not sure how this helps with computing the inverse and determinant of $A = aI + T$.
Thanks in advance,
 A: Recall that for a $n\times n$ matrix B with eigenvalues $\lambda_i$,
$$\det B = \prod_i \lambda i $$
and
$$\operatorname{Tr} B = \sum_i\lambda_i. $$
Since $\det T=\operatorname{Tr} T = 0$, it follows that $0$ is the only eigenvalue of $T$, and so
$$\det(T-\lambda I)=0\iff \lambda = 0. $$
Since
$$A - \lambda I = aI+T - \lambda I = T - (\lambda-a)I,$$
we see that $A$ has eigenvalues $a$. Therefore the determinant of $A$ is simply $a^n$, and since $T^2=0$, we see that
$$\frac1{a^2}(aI-T)(aI+T) = I, $$
so that $$\left(aI+T\right)^{-1} = \frac1{a^2}(aI-T). $$
A: Here is another approach:
If $T$ has rank one, then there are $u,v$ such that $T = u v^*$. If $\operatorname{tr} T = 0$, then $\sum_k u_k \overline{v}_k = 0$ and so
we see that $v^* u = 0$ or, in other words, $v \bot u$.
Let $A=aI+T = aI + u v^*$.
Suppose $Ax = y$, then $a x + u v^* x = y$. Premultiplying by $v^*$ gives
$a v^* x = v^* y$ and so we have
$a x + {1 \over a} u v^* y = y$ or $x = {1 \over a}(I - {1 \over a}u v^*)y$,
from which we see that $A$ is invertible and
$A^{-1} = {1 \over a}(I - {1 \over a}u v^*) = {1 \over a}(I - {1 \over a}T ) $.
There are various ways of computing the determinant of $A$. Here is one
way: Choose $w_2,...,w_n$ such that ${1 \over \|v\|} v, w_2,...,w_n$ forms
an orthonormal basis. Then we see that
$A w_k = a w_k $, and
$A ({1 \over \|v\|} v) = a {1 \over \|v\|} v + \|v\| u$.
Since $u = ({1 \over \|v\|} v)^* u ({1 \over \|v\|} v) + \sum_k (w_k^* u) w_k$,
we have
$A ({1 \over \|v\|} v) = ( a + v^* u )({1 \over \|v\|} v) + \|v\| \sum_k (w_k^* u) w_k $. Hence, in the basis ${1 \over \|v\|} v, w_2,...,w_n$, $A$
has the form
$\begin{bmatrix} a + v^* u & 0 & \cdots & 0 \\
\|v\| w_1^* u & a & \cdots & 0 \\
\|v\|w_2^* u & 0 & \cdots & 0 \\
\vdots & \vdots & & \vdots\\
\|v\|w_n^* u & 0 & \cdots & a\end{bmatrix}$,
and so we see that 
$\det A = a^{n-1} (a+ v^* u)$. Since $v^*u = 0$ we have
$\det A = a^n$.
This is a special case of the fact that, for appropriate dimensioned matrices,
we have $\det (I + AB) = \det (I + BA)$.
