Question on Schur's first lemma From my written notes, (purposely not typeset via Mathjax this time), I have a proof for Schur's first lemma:

My only question is regarding the red circled part, the author claims that for any $\vec y$ such that $$\underline{\underline{M}}\vec y=\lambda \vec y$$ which immediately implies that the matrix, $\underline{\underline{M}}$ is proportional to the unit matrix $\underline{\underline{I}}$.
But from linear algebra, I was taught that in general: $$\underline{\underline{M}}\vec y= \lambda \vec y\implies \big(\underline{\underline{M}}-\lambda \underline{\underline{I}}\big)\vec y=\vec 0$$
The proof (in the image) above states that for this construction, $\underline{\underline{M}}\propto \underline{\underline{I}}$.
But, I can find a matrix $\underline{\underline{M}}$, that is not proportional to the unit matrix for arbitrary $\vec y$, here is one such example: consider $\underline{\underline{M}}=\begin{pmatrix}5&-1\\-1&5\\ \end{pmatrix}$. Which, can be solved by method of $\mathrm{det}\big(\underline{\underline{M}}-\lambda \underline{\underline{I}}\big)=0$, and if this method is carried out it can be shown that there is an eigenvalue $\lambda=4$ with  eigenvector $\vec y_1=c_1\begin{pmatrix}1\\1\\ \end{pmatrix}$ where $c_1 \ne 0$, and an eigenvalue $\lambda=6$ with eigenvector $\vec y_2=c_2\begin{pmatrix}-1\\1\\ \end{pmatrix}$ where $c_2 \ne 0$
I know that I am working in $\mathbb{R}^2$ and the author was working in $\mathbb{C}^n$, but the point is that I have found a matrix, $\underline{\underline{M}}$ that is not proportional to the unit matrix (for arbitrary $\vec y$). But according to the proof by the author, I shouldn't be able to find such a matrix. So what is the problem here?

You are right that there are matrices $M$ which are not proportional to the unit matrix.  The proof you included shows that if $D(g)M = MD(g)$ for every $g$, then $M$ cannot be a matrix like
$$\begin{pmatrix} 5 & -1 \\ -1 & 5 \end{pmatrix}.$$
Instead, $M$ MUST look like $\lambda I$ for some scalar $\lambda$.  Why?  Because in the proof, they show that if $Mx = \lambda x$ for SOME nonzero vector $x$, then in fact $Mx = \lambda x$ for EVERY vector $x$.  This forces $M$ to take the form $\lambda I$.
