Looking for what it means when the transpose of a vector matrix multiplied(i think) with itself is 1

Question

Suppose u and v are vectors such that $u^T v = 1$. Define $A =uv^T$.

Find an eigenvalue of A, also is A invertible if u and v are orthogonal.

I think the starting point is looking at

$1 = 1^T =(u^Tv)^T$

not sure where to go from there.

Answer 1

$u$ is an eigenvector with eigenvalue 1: $$A u=u v^T u=u (u^T v)^T=u (1)^T=u$$

Answer 2

I can answer the first question:

If $\mathbf{u} \mathbf{v}^T = 1$ and $A = \mathbf{v} \mathbf{u}^T$, by the properties of eigenvalues and the characteristic equation, the eigenvalues of $A$ will be the same as those of $A^T$. It suffices to prove that both $A$ and $A^T$ have the same characteristic equation:

We know that the following properties are true:

$(A+B)^T = A^T + B^T$

$I^T = I$

$\operatorname{det}(A) = \operatorname{det}(A^T)$

Then

$$\operatorname{det}(A-\lambda I) = \operatorname{det}((A-\lambda I)^T) = \operatorname{det}(A^T - \lambda I^T) = \operatorname{det}(A^T - \lambda I)$$

Since the characteristic equation of $A$ is the same as that of $A^T$, they have the same eigenvalues. Then if we find an eigenvalue for $A^T$, it will also be an eigenvalue of $A$.

We have that

$A^T = (\mathbf{u} \mathbf{v}^T)^T = (\mathbf{v}^T)^T = \mathbf{u}^T = \mathbf{v} \mathbf{u}^T$

Then $A^T \mathbf{v} = \mathbf{v} \mathbf{u}^T \mathbf{v} = \mathbf{v} · 1 = 1 · \mathbf{v} = \lambda \mathbf{v}$, where $\lambda =1$

Hence, $\lambda = 1$ is an eigenvalue of $A^T$, so it is also an eigenvalue of $A$.

If $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, then $\mathbf{u} · \mathbf{v} = \mathbf{u}^T \mathbf{v} = 0 \not = 1$, so I am not sure about what to do in the second part.

Answer 3

Assume that $\mathbf u\ne0$, otherwise $A=0$ as well. The image of $A$ is spanned by $\mathbf u$: for any vector $\mathbf w$, $A\mathbf w = \mathbf u\mathbf v^T\mathbf w = (\mathbf v^T\mathbf w)\mathbf u$. This also shows that $\mathbf u$ is mapped to a scalar multiple of itself, so it’s an eigenvector of $A$, and I’ll leave to you to compute the corresponding eigenvalue. You also know from the preceding that $A$’s kernel is nontrivial and that it consists of the orthogonal complement of $\mathbf v$. Therefore, $0$ is also an eigenvalue and the rank-nullity theorem tells us that the two eigenspaces found so far account for the entire parent space, so there are no other eigenvalues. I hope that you can tell from all of this whether or not $A$ is invertible.