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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 $y^*xne 0$.

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 $exists mune 0, mu ne lambda$. I don’t know where to go from here.



3. Going forwards: $A$ is diagonalizable. Then $exists textnonsingular S: S^-1AS=D$, 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 $y^*xne 0$, then A has an eigenvalue not equal to zero. Then can I make a statement about the similarity of $A$ to some diagonal matrix?

Suppose $xneq0$. 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
$$ A=xy^*=(P^-1a)(P^-1b)^*=P^*(ab^*)P $$
which implies that $A$ and $ab^*$ have the same eigenvalues. Note that
$$ e_1b=left[beginmatrixa_1b_1&a_1b_2&cdots&a_1b_n\
0&0&cdots&0\
vdots&vdots&cdots&vdots\
0&0&cdots&0\
endmatrixright] $$
which has eigenvalues $a_1b_1$ and $0$ (the multiplicity $n-1$) and also note that
$$ a_1b_1=e_1^*b=(a,b)=(P^-1a,P^-1y)=(x,y)=x^*y=lambda. $$
Thus $A$ has eigenvalues $lambda$ and $0$ (the multiplicity $n-1$).

Let $x^*ynot=0$, choose an orthonormal matrix $P$ such that $Px=|x|e_1$ where $e_1=(1,0,cdots,0)^*$. Let $$barb=e_1+kP^*y $$
where $k$ is such that
$$ (e_1,barb)=1+k(e_1,Py)=1+frack(Px,Py)=1+frackx^*y=0$$
namely, $k=-fracx^*y$. Let $e_2=fracbarbbarb$ and choose $e_3,cdots,e_n$ such that $e_1,e_2,cdots,e_n$ are orthonormal. Then
$$ PAP^*=Pxy^*P^*=|x|e_1(Py)^*=fracke_1(barb-e_1)^*=-fracke_1e_1^*,$$
namely $xy^*$ is diagonalizable.

Hints:

1. Consider $xy^*x$.


2. Assume $xy^*v=lambda vne 0$, 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 $y^*xne 0$, we have $dimker(x^*y) =n-1$, choose a basis there, extend by $x$, and in that basis the matrix of $x^*y$ is diagonal with a single nonzero entry $y^*x$.

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

$textchar(Bbb K) = 0 tag 0$

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

$z = dfracy^ast zmux = alpha x, ; alpha = dfracy^ast zmu ne 0; tag 7$

$z$ 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

$lambda = y^ast x ne 0, tag10$

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

$phi_y: V to Bbb K, ; phi_y(z) = y^ast z in Bbb K, ; z in V, tag11$

satisfies

$dim ker phi_y = n - 1, tag12$

where

$dim_Bbb K V = n; tag13$

therefore there exist $n - 1$ linearly independent vectors

$w_1, w_2, ldots, w_n - 1 in ker phi_y, tag14$

each of which satisfies

$y^ast w_i = phi_y(w_i) = 0, ; 1 le i le n - 1; tag15$

then

$Aw_i = (xy^ast)w_i = x(y^ast w_i) = 0, 1 le i le n - 1; tag16$

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

$S = [x ;w_1 ; w_2 ; ldots ; w_n - 1]; tag17$

that is, the columns of $S$ are the vectors $x$, $w_1$, $w_2$, and so forth; then

$AS = [Ax ; Aw_1 ; Aw_2 ; ldots ; Aw_n - 1] = [(y^ast x)x ; 0 ; 0 ; ldots ; 0]; tag18$

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

$S^-1S = S^-1[x ;w_1 ; w_2 ; ldots ; w_n - 1] = [S^-1x ; S^-1w_1 ; S^-1w_2 ; ldots ; S^-1w_n - 1] = I; tag19$

from (18) and (19) we infer that

$S^-1AS = [S^-1(y^ast x)x ; S^-10 ; S^-10 ; ldots ; S^-10] = [y^ast x S^-1x ; 0 ; 0 ; ldots ; 0]; tag20$

now inspection of (19) reveals that

$S^-1x = e_1 = beginpmatrix 1 \ 0 \ 0 \ vdots \ 0 endpmatrix; tag21$

therefore (20) becomes

$S^-1AS = [y^ast x e_1 ; 0 ; 0 ; ldots ; 0], tag22$

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 $A$.