Proof that the same eigenvalues are associated to orthogonalized eigenvectors The normal matrix ${\bf m}$ has eigenvectors $v_i$ with corresponding eigenvalues $\lambda_i$
$$\lambda_iv_i={\bf m}v_i.\tag{1}$$
$u_i$ is the Gram-Schmidt orthogonalized eigenvector that corresponds to $v_i$.
How to show that this also implies $$\lambda_iu_i={\bf m}u_i\tag{2} \,\,\,?$$
Incomplete proof :
Gram-Schmidt orthogonalization can be written as
$$u_i=v_i+\sum_{k=1}^{i-1}c_kv_k\quad\text{with}\,\, c_k\in \mathbb{R}.\tag{3}$$
After setting eq.(3) into eq.(1) we get
$$\lambda_i\left(u_i-\sum_{k=1}^{i-1}c_kv_k\right)={\bf m}\left(u_i-\sum_{k=1}^{i-1}c_kv_k\right)\tag{4}$$
and it follows
$$\lambda_i u_i-{\bf m}u_i=\lambda_i \sum_{k=1}^{i-1}c_kv_k-{\bf m}\sum_{k=1}^{i-1}c_kv_k\tag{5}.$$
It remains to show that $\lambda_i$ is an eigenvalue to the vector $\displaystyle{\sum_{k=1}^{i-1}c_kv_k}$. How can this be done?
 A: Since ${\bf m}$ is normal, $v_i$ and $v_j$ are normal if $\lambda_i\ne \lambda_j$.
Consider the start of the Gram-Schmidt process. $$u_1=v_1, u_2=v_2+c_1v_1\quad\text{with}\,\, c_1\in \mathbb{R}.$$
Then, by Gram-Schmidt, $u_1.u_2=0$.
By our opening observation, either $u_1.v_2=0$ and then $c_1=0$ or $\lambda_1= \lambda_2$. In both cases, $\lambda_2u_2={\bf m}u_2 $.
Proceeding in this way through the Gram-Schmidt process we see that each $u_i$ is a linear combination of $v_j$ with the same eigenvector as $v_i$ and so each each $u_i$ has eigenvalue $\lambda_i$.
The general step of Gram-Schmidt
$$u_i=v_i-\sum_1^{i-1}\frac{u_k.v_i}{u_k.u_k}u_k$$
where $u_1, ..., u_{i-1}$ are the previously generated orthogonal eigenvectors. The $u_k$ which correspond to eigenvalues other than $\lambda_i$ satisfy $u_k.v_i=0$ and so the only $u_k$ with non-zero coefficients are those with eigenvalue $\lambda_i$. Therefore
$${\bf m}.u_i={\bf m}.v_i-\sum_1^{i-1}\frac{u_k.v_i}{u_k.u_k}{\bf m}.u_k=\lambda_iv_i-\sum_1^{i-1}\frac{u_k.v_i}{u_k.u_k}\lambda_iu_k =\lambda_iu_i $$
