# 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:

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?
• 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 $q=x$? – Florian Mar 6 '19 at 18:17

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.

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 $$y^*x$$.

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

$$(y^\ast z) \ne 0, \; x \ne 0, \tag 6$$

$$z = \dfrac{y^\ast z}{\mu}x = \alpha x, \; \alpha = \dfrac{y^\ast z}{\mu} \ne 0; \tag 7$$

$$z$$ 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 $$A$$.