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

  1. $\lambda$ is an eigenvalue of $A$.
  2. If $\lambda \ne 0$, it will be the only nonzero eigenvalue of $A$.
  3. Explain why $A$ is diagonalizable iff $y^*x\ne 0$.

My approach thusfar:

  1. 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.

  2. 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.

  3. 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?
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  • $\begingroup$ Hint for 1+2: $\lambda$ is an eigenvalue of $A$ with corresponding eigenvector $q$ if and only if $A q = \lambda q$. What happens for $q=x$? $\endgroup$ – Florian Mar 6 '19 at 18:17
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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.

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Hints:

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

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