Show some eigenvalue properties for $A=xy^*$

eigenvalues-eigenvectorslinear algebramatricesmatrix analysis

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?

Best Answer

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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