Find the inverse and determinant of A=(aI +T),

Question

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,

Answer 1

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 $$\left(aI+T\right)^{-1} = \frac1{a^2}(aI-T). $$

Answer 2

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 $\det (I + AB) = \det (I + BA)$.