Help understanding proof in simultaneous diagonalization

When $lambda_1$ is an eigenvalue of a matrix and $v_1$ and $v_2$ are the components of the corresponding eigenvector, then the following equation holds:

$begin{pmatrix} a-lambda_1 & b c &d-lambda_1 end{pmatrix}begin{pmatrix} v_1 v_2 end{pmatrix}=begin{pmatrix} 0 0 end{pmatrix}$

Now when you scale up the eigenvector (say by three) it looks like this:

$begin{pmatrix} a-lambda_1 & b c &d-lambda_1 end{pmatrix}begin{pmatrix} 3v_1 3v_2 end{pmatrix}=begin{pmatrix} 0 0 end{pmatrix}$

This you can write as

$3 begin{pmatrix} a-lambda_1 & b c &d-lambda_1 end{pmatrix}begin{pmatrix} v_1 v_2 end{pmatrix}=begin{pmatrix} 0 0 end{pmatrix}$

But the matrix multiplied with the eigenvector still yields the zero vector!

This is what he meant when he said "Since this vector is unique up to a scale.": any scaled up eigenvector of a matrix is still an eigenvector. And how does the last equation follow from it?

When you write $$OmegaLambdaleft|omega_irightrangle=omega_iLambdaleft|omega_irightrangle$$ then you know that $Lambdaleft|omega_irightrangle$ gives you a vector which is an eigenvector of $Omega$. But you said that $Omega$ is nondegenerate, so for any $omega_i$ there is a unique $left|omega_irightrangle$. What this means is that this eigenvector you get by applying $Lambdaleft|omega_irightrangle$ must be $left|omega_irightrangle$. Luckily, any eigenvector which is scaled up (here by $lambda_i$) is still an eigenvector, so that you get

$$Omegalambda_ileft|omega_irightrangle=omega_i lambda_i left|omega_irightrangle$$

or

$$Lambdaleft|omega_irightrangle=lambda_ileft|omega_irightrangle $$