If $(A-\lambda I)x_0=0,~y_0^{T}(A-\lambda I)=0$ and $y_0^{T}x_0=0$, prove that eigenvalue $\lambda$ is not simple. Let $A\in M_{n\times n}(\Bbb R), \lambda\in\sigma(A)\cap\Bbb R\setminus\{0\}$.
If $x_0,\,y_0$ are real eigenvectors of $A$ such that $(A-\lambda I)x_0=0$ and $y_0^{T}(A-\lambda I)=0$ and $y_0^{T}x_0=0$, prove that eigenvalue $\lambda$ is not simple, i.e. has algebraic multiplicity $>1$.
Attempt. We just need to work with the case that geometric multiplicity of lambda is equal to $=1$.
Thanks in advance.
 A: For the sake of contradiction let's say $\lambda$ has multiplicity $1$. We are gonna prove that $y^T x \neq 0$.
For any matrix there exist an orthogonal matrix $V := \begin{bmatrix}x & U\end{bmatrix}$ such that
$$ A = \begin{bmatrix}x & U\end{bmatrix} \begin{bmatrix}\lambda & a^T \\ 0 & B\end{bmatrix} \begin{bmatrix}x^T \\ U^T\end{bmatrix}$$
Clearly $Ax=\lambda x$. Also since $\lambda$ has multiplicity $1$, $B$ does not have an eigenvalue of $\lambda$. Also note that $VV^T=xx^T+UU^T=I$. Now, expanding $A$ we obtain
$$ A = \lambda xx^T + xa^TU^T + UBU^T = \lambda (I-UU^T) + xa^TU^T + UBU^T = \lambda I + xa^TU^T + U(\lambda I - B)U^T $$
Let $y^T A = \lambda y^T$ and $y^T x = 0$. This implies $y^TU(\lambda I-B)U^T = 0$. So, either $y^TU=0$, in which case $y=\alpha x$ for some nonzero $\alpha$, or $\lambda$ is an eigenvalue of $B$ which contradicts the assumption.
A: Remark: I found that @obareey's answer is similar, but the reasoning is somewhat different.
Assume, for the sake of contradiction, that $\lambda$ is simple.
The real Schur decomposition of $A$ is given by
$$A = U \left(
          \begin{array}{cc}
            \lambda & b^T \\
            0 & B \\
          \end{array}
        \right)
 U^\mathsf{T}$$
where $U$ is orthogonal. Since $\lambda$ is simple, we have $\det (B - \lambda I) \ne 0$ or $B - \lambda I$ is non-singular.
Let $U^\mathsf{T}x_0 = z$ and $U^\mathsf{T}y_0 = w$. We have
$z\ne 0, w\ne 0$, and $w^\mathsf{T}z = y_0^\mathsf{T}x_0 = 0$.
From $(A - \lambda I) x_0 = 0$, we have
$$\left(
          \begin{array}{cc}
            0 & b^T \\
            0 & B - \lambda I \\
          \end{array}
        \right)
 z = 0. \tag{1}$$
Since $B - \lambda I$ is non-singular, from (1), we know that $z$ has the form of $[z_1, 0_{1\times (n-1)}]^\mathsf{T}$ with $z_1 \ne 0$.
Since $w^\mathsf{T}z = 0$, we know that $w$ has the form of $[0, \tilde{w}^\mathsf{T}]^\mathsf{T}$
with $\tilde{w}\ne 0$. From $y_0^T(A - \lambda I) = 0$, we have
$$w^\mathsf{T}\left(
          \begin{array}{cc}
            0 & b^T \\
            0 & B - \lambda I \\
          \end{array}
        \right) = 0 \tag{2}$$
which results in
$$\tilde{w}^\mathsf{T}(B - \lambda I)= 0$$
which contradicts the non-singularity of $B - \lambda I$. (Q.E.D.)
A: This result holds over any field so long as the one eigenvalue of interest, $\lambda_0$ exists in said field.  So I give a longer but elementary proof that works over any field.
0.)
the structure will be
$\lambda_0 \text{ is simple} \longrightarrow \mathbf y_0^T\mathbf x_0 \neq 0$
so assume $\lambda_0$ is simple
1.)
Per Cayley Hamilton we have
$\mathbf 0 = p(A) =  \big(A-\lambda_0I\big)q\big(A\big)$
being simple $q\big(\lambda_0\big)=\alpha \neq 0$ and
$q\big(A\big)\mathbf x_0 =\alpha \mathbf x_0$
this implies $\mathbf x_0$ is linearly independent of anything in $\ker\Big(q\big(A\big)\Big)$
application of Sylvester's Rank Inequality tells us
$n $
$= \dim \ker\Big(\big(A-\lambda_0I\big)q(A)\Big)$
$\leq  \dim \ker\Big(\big(A-\lambda_0I\big)\Big) +\dim \ker\Big(q(A)\Big)$
$= 1 +\dim \ker\Big(q(A)\Big)$
$\leq n$
recalling that there are at most n linearly independent vectors in our vector space. Thus
$\dim \ker\Big(q(A)\Big) = n-1$
We collect the $n-1$ linearly independent vectors $\in \ker \big(q(A)\big)$ and label them $\mathbf x_k$, $k\in \{1,2,...,n-1\}$.  Re-running the argument on $A^T$ gives us the same result for the left nullspace of $A$ and we have $\mathbf y_k$, $k\in \{1,2,...,n-1\}$
now collect all of these in 2 invertible matrices
$X:= \bigg[\begin{array}{c|c|c|c|c} \mathbf x_0& \mathbf x_1 & \mathbf x_2 &\cdots & \mathbf x_{n-1} \end{array}\bigg]$ and  $Y^T:= \begin{bmatrix}
 \mathbf y_0^T\\ 
 \mathbf y_1^T \\
\vdots\\ 
 \mathbf y_{n-1}^T \\ 
\end{bmatrix}$
2.)
$1 = \text{rank}\Big(q\big(A\big)\Big) = \text{rank}\Big(Y^T q\big(A\big)X\Big) $
where the first equality comes by rank-nullity and the second because multiplication by invertible matrices doesn't change rank.
computing the same thing two different ways:
i.) $Y^T q\big(A\big)X = Y^T \Big(q\big(A\big)X\Big) = Y^T \bigg[\begin{array}{c|c|c|c|c} \alpha \mathbf x_0& \mathbf 0 & \mathbf 0 &\cdots & \mathbf 0\end{array}\bigg]$
ii.) $Y^T q\big(A\big)X = \Big(Y^T q\big(A\big)\Big)X =  \begin{bmatrix}
\alpha \mathbf y_0^T\\ 
\mathbf 0^T \\
\vdots\\ 
 \mathbf 0^T \\ 
\end{bmatrix}X$
putting these together
$\alpha^{-1}\cdot  Y^T q\big(A\big)X = \begin{bmatrix}   \mathbf y_0^T\mathbf x_0 & \mathbf 0 \\ \mathbf 0 & \mathbf 0\end{bmatrix}$
thus
$1=\text{rank}\Big(Y^T q\big(A\big)X\Big) = \text{rank}\Big(\alpha^{-1} \cdot Y^T q\big(A\big)X\Big)\longrightarrow \mathbf y_0^T\mathbf x_0 \neq 0$
A: This is true over any field $\mathbb F$. Let $\{e_1,e_2,\ldots,e_n\}$ be the standard basis of $\mathbb F^n$. Since $y_0^Tx_0=0$, there exists a square matrix $Y$ whose second row is $y_0^T$ and whose last $n-1$ rows form a basis of the $(n-1)$-dimensional vector subspace $V=\{y:y^Tx_0=0\}$. Set the first row of $Y$ to a vector $y_1$ such that $y_1^Tx_0=1$. Then $y_1\not\in V$. Hence $Y$ is invertible. Let $X=Y^{-1}$. Then $x_0$ is the first column of $X$ because $Yx_0=e_1=YXe_1$.
Let $B=Y(A-\lambda I)X$. Then $Be_1=Y(A-\lambda I)Xe_1=Y(A-\lambda I)x_0=0$. Similarly, $e_2^TB=0$. In other words, the first column and the second row of $B$ are zero and we may write
$$
B=\pmatrix{0&\ast\\ 0&C}
$$
where the first row of $C$ is zero. Thus $0$ is an eigenvalue of $B$ of algebraic multiplicity $\ge2$ because it is an eigenvalue of $C$. Now, as $B=Y(A-\lambda I)Y^{-1}$, $\lambda$ is an eigenvalue of $A$ of multiplicity $\ge2$.
