Proving that left and right eigenvector are equal using singular values Let $A$ be a square (complex) non-Hermitian matrix. Let us note by $s_i$ its singular values (in decreasing order) and by $a_i$ its eigenvalues (also in decreasing order, wrt. their absolute value).
If $|a_1| = s_1$ and $s_2 < s_1$, then we can see that $|a_2| < |a_1|$, therefore the eigenspace associated to $a_1$ is of dimension $1$. Let us denote by $v$ an element of this eigenspace.
I am trying to prove that if the conditions above are fulfilled then we also have 
$$v^H A = a_1v^H .$$
Where $v^H$ denotes the Hermitian conjugate.
It is a similar question to this one in subject, but not in conditions. Thanks a lot for the help!
 A: making using of the hint, where $B = U^H A U =  \pmatrix{a_1 & \mathbf x^H\\\mathbf 0&B_0}$ 
First consider that the diagonal elements of a Hermitian Positive (Semi)Definite matrix $C$ are a lower bound on the maximal eigenvalue.  This result comes from majorization, but also more simply  
with $\big \Vert \mathbf v \big \Vert_2 := 1$
$\lambda_\text{max} C  = \text{max: } \mathbf v^H C \mathbf v \geq \mathbf e_k^H C \mathbf e_k = c_{k,k}$
where $\mathbf e_k$ is a standard basis vector  
now select 
$C:= BB^H$
observe $c_{1,1} = \vert a_1 \vert^2 + \big \Vert \mathbf x \big \Vert_2^2$
but the maximal eigenvalue of $C$ is the squared maximal singular value of $B$ so 
$\vert a_1 \vert^2 = s_1^2 \geq \mathbf e_1^H C \mathbf e_1 = \vert a_1 \vert^2 + \big \Vert \mathbf x \big \Vert_2^2\longrightarrow 0 \geq \big \Vert \mathbf x \big \Vert_2^2\longrightarrow 0 = \big \Vert \mathbf x \big \Vert_2^2\longrightarrow \mathbf 0 = \mathbf x $ 
by the positive definiteness of the (squared) 2 norm.  
so it is in fact
$B = U^H A U =  \pmatrix{a_1 & \mathbf 0^H\\\mathbf0&B_0}$
as desired  
to confirm, this implies
$\pmatrix{a_1\mathbf u_1 \vert *}= U\pmatrix{a_1 & \mathbf 0^H\\\mathbf0&B_0} =UB =  A U$
so $A\mathbf u_1  =a_1\mathbf u_1$
i.e.it is a (right) eigenvector
and
$\pmatrix{a_1 \mathbf u_1^H  \\*} = \pmatrix{a_1 & \mathbf 0^H\\\mathbf 0&B_0}U^H = BU^H = U^H A $
so $\mathbf u_1^H A = a_1 \mathbf u_1^H$, i.e. it is a left eigenvector  
