Show some eigenvalue properties for $A=xy^*$ Let $x,y$ be given vectors of dimension $n \times 1$, $A=xy^*$, and $\lambda=y^*x$. I’m trying to demonstrate the following:


*

*$\lambda$ is an eigenvalue of $A$. 

*If $\lambda \ne 0$, it will be the only nonzero eigenvalue of $A$. 

*Explain why $A$ is diagonalizable iff $y^*x\ne 0$.


My approach thusfar:


*

*NTS $\det(A-\lambda I)=0$.
I’d really rather not expand $\det(xy^*-y^*xI)$ , but even doing so, I don’t see how doing so would help. 

*I can verbally logic this: Since A is the product of a pair of vectors, it’s obvious that each column/row will be a scalar multiple of eachother. By proof: Suppose $\exists \mu\ne 0, \mu \ne \lambda$. I don’t know where to go from here.

*Going forwards: $A$ is diagonalizable. Then $\exists \text{nonsingular} S: S^{-1}AS=D$, where $D$ is a diagonal matrix similar to $A$. but if $A$ has only the eigenvalue zero, then $S$ is singular, a contradiction.
Going the other direction, if $y^*x\ne 0$, then A has an eigenvalue not equal to zero. Then can I make a statement about the similarity of $A$ to some diagonal matrix?

 A: Hints:


*

*Consider $xy^*x$.

*Assume $xy^*v=\lambda v\ne 0$, then it also equals to $x(y^*v)$, so $v$ is a scalar multiple of $x$.

*By the above, if $y^*x=0$, the only eigenvalue of $xy^*$ is $0$, so if it's diagonalizable, it must be similar to the diagonal matrix with the eigenvalues, though $xy^*\ne 0$ (unless $x=y=0$).
On the other hand, if $y^*x\ne 0$, we have $\dim\ker(x^*y) =n-1$, choose a basis there, extend by $x$, and in that basis the matrix of $x^*y$ is diagonal with a single nonzero entry $y^*x$. 

A: Suppose $x\neq0$. Let $P$ be an orthonormal matrix (i.e., $ P^*P=I$)
such that $ Px=a$, where $a=(a_1,0,\cdots,0)^*$. Let $Py=b=(b_1,b_2,\cdots,b_n)^*$. Then 
$$ A=xy^*=(P^{-1}a)(P^{-1}b)^*=P^*(ab^*)P $$
which implies that $A$ and $ab^*$ have the same eigenvalues. Note that
$$ e_1b=\left[\begin{matrix}a_1b_1&a_1b_2&\cdots&a_1b_n\\
0&0&\cdots&0\\
\vdots&\vdots&\cdots&\vdots\\
0&0&\cdots&0\\
\end{matrix}\right] $$
which has eigenvalues $a_1b_1$ and $0$ (the multiplicity $n-1$) and also note that
$$ a_1b_1=e_1^*b=(a,b)=(P^{-1}a,P^{-1}y)=(x,y)=x^*y=\lambda. $$
Thus $A$ has eigenvalues $\lambda$ and $0$ (the multiplicity $n-1$). \
Let $x^*y\not=0$, choose an orthonormal matrix $P$ such that $Px=\|x\|e_1$ where $e_1=(1,0,\cdots,0)^*$. Let $$\bar{b}=e_1+kP^*y $$
where $k$ is such that
$$ (e_1,\bar{b})=1+k(e_1,Py)=1+\frac{k}{\|x\|}(Px,Py)=1+\frac{k}{\|x\|}x^*y=0$$
namely, $k=-\frac{\|x\|}{x^*y}$. Let $e_2=\frac{\bar{b}}{\|\bar{b}\|}$ and choose $e_3,\cdots,e_n$ such that $e_1,e_2,\cdots,e_n$ are orthonormal. Then
$$ PAP^*=Pxy^*P^*=\|x\|e_1(Py)^*=\frac{\|x\|}{k}e_1(\bar{b}-e_1)^*=-\frac{\|x\|}{k}e_1e_1^*,$$
namely $xy^*$ is diagonalizable.
A: Since the underlying field $\Bbb K$ over and $\Bbb K$-vector space $V$ on which $A$ operates, $A \in \mathcal L(V)$, are unspecified, I am going to assume that
$\text{char}(\Bbb K) = 0 \tag 0$
for the remainder of this answer. 
To show that
$\lambda = y^\ast x \tag 1$
is an eigenvalue of
$A = xy^\ast, \tag 2$
we need merely consider
$Ax = (xy^\ast)x = x(y^\ast x) = x(\lambda) = \lambda x, \tag 3$
which, assuming $x \ne 0$, shows that $\lambda$ is an eigenvalue of $A$ with corresponding eigenvector $x$.
Now if $\mu \ne 0$ is any other eigenvalue of $A$, then
$\exists z \ne 0, \; Az = \mu z; \tag 4$
since $\mu \ne 0$ and $z \ne 0$, we have
$0 \ne \mu z = Az = (xy^\ast)z = x(y^\ast z) = (y^\ast z)x; \tag 5$
from this we infer that
$(y^\ast z) \ne 0, \; x \ne 0, \tag 6$
which leads us to
$z = \dfrac{y^\ast z}{\mu}x = \alpha x, \; \alpha = \dfrac{y^\ast z}{\mu} \ne 0; \tag 7$
$z$ is thus a scalar multiple of $x$, whence
$\mu z = Az = A(\alpha x) = \alpha Ax = \alpha \lambda x = \lambda(\alpha x) = \lambda z; \tag 8$
thus,
$(\mu - \lambda)z = 0 \Longrightarrow \mu = \lambda, \tag 9$
and we see that $\lambda \ne 0$ is the only non-zero eigenvalue of $A$.
Last but by no means least, if
$\lambda = y^\ast x \ne 0, \tag{10}$
then as we have seen above, $\lambda$ is the sole non-vanishing eigenvalue of $A = xy^\ast$, and furthermore, $\lambda$ is of geometric and algebraic multiplicity $1$; we can see that this is true via the observation that the kernel of the  linear map
$\phi_y: V \to \Bbb K, \; \phi_y(z) = y^\ast z \in \Bbb K, \; z \in V, \tag{11}$
satisfies
$\dim \ker \phi_y = n - 1, \tag{12}$
where
$\dim_{\Bbb K} V = n; \tag{13}$
therefore there exist $n - 1$ linearly independent vectors
$w_1, w_2, \ldots, w_{n - 1} \in \ker \phi_y, \tag{14}$
each of which satisfies
$y^\ast w_i = \phi_y(w_i) = 0, \; 1 \le i \le n - 1; \tag{15}$
then
$Aw_i = (xy^\ast)w_i = x(y^\ast w_i) = 0, 1 \le i \le n - 1; \tag{16}$
from (11)-(16) we see that the dimension of the kernel of $A$, that is, the dimension of the $0$-eigenspace, is $n - 1$; from this fact we conclude that the dimension of the $\lambda$-eigenspace is precisely $1$, i.e., $\lambda$ is of geometric and algebraic multiplicity $1$ as asserted above.
We may now build the matrix $S$ as
$S = [x \;w_1 \; w_2 \; \ldots \;  w_{n - 1}]; \tag{17}$
that is, the columns of $S$ are the vectors $x$, $w_1$, $w_2$, and so forth; then
$AS = [Ax \; Aw_1 \; Aw_2 \; \ldots \; Aw_{n - 1}] = [(y^\ast x)x \; 0 \; 0 \; \ldots \; 0]; \tag{18}$
now the $w_i$ are linearly independent from one another, and $x$ is linearly independent from the $w_i$ since they are eigenvectors associated to different eigenvalues; thus the matrix $S$ is non-singular and we may form $S^{-1}$ such that
$S^{-1}S = S^{-1}[x \;w_1 \; w_2 \; \ldots \;  w_{n - 1}] = [S^{-1}x \; S^{-1}w_1 \; S^{-1}w_2 \; \ldots \; S^{-1}w_{n - 1}] = I; \tag{19}$
from (18) and (19) we infer that
$S^{-1}AS = [S^{-1}(y^\ast x)x \; S^{-1}0 \; S^{-1}0 \; \ldots \; S^{-1}0] = [y^\ast x S^{-1}x \; 0 \; 0 \; \ldots \; 0]; \tag{20}$
now inspection of (19) reveals that
$S^{-1}x = e_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}; \tag{21}$
therefore (20) becomes
$S^{-1}AS = [y^\ast x e_1 \; 0 \; 0 \; \ldots \; 0], \tag{22}$
which has only one non-zero entry, $y^\ast x$, in the top left-hand corner.  It is manifestly diagonal and every diagonal entry besides $y^\ast x$ is zero, as we expect based on what we have uncovered regarding the eigenvalues of $A$.
