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?
  • $\begingroup$ 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

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.



  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

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

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

$\mu z = Az = A(\alpha x) = \alpha Ax = \alpha \lambda x = \lambda(\alpha x) = \lambda z; \tag 8$


$(\mu - \lambda)z = 0 \Longrightarrow \mu = \lambda, \tag 9$

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

$\phi_y: V \to \Bbb K, \; \phi_y(z) = y^\ast z \in \Bbb K, \; z \in V, \tag{11}$


$\dim \ker \phi_y = n - 1, \tag{12}$


$\dim_{\Bbb K} V = n; \tag{13}$

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

$y^\ast w_i = \phi_y(w_i) = 0, \; 1 \le i \le n - 1; \tag{15}$


$Aw_i = (xy^\ast)w_i = x(y^\ast w_i) = 0, 1 \le i \le n - 1; \tag{16}$

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 $A$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.