From my written notes, (purposely not typeset via Mathjax this time), I have a proof for Schur's first lemma: Proof of 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 $\lambda I$.


D_S's answer is completely correct. I just wanted to add some additional points that were too long to fit in a comment.

The layout and wording of the proof in your notes is very confusing, I would say. Here's a brief outline of how the proof goes:

(i) $M$ has at least one eigenvector, with eigenvalue $\lambda$. This is a standard result from linear algebra.

(ii) the $\lambda$-eigenspace $E_\lambda\subset\mathbb{C}^n$ (i.e. the set $E_\lambda = \{x\in\mathbb{C}^n \mid Mx = \lambda x\}$) is an actual vector subspace. This is also a standard result, usually not proved as part of Schur's Lemma, although you prove it above. It's slightly incorrect to say that eigenvectors corresponding to eigenvalue $\lambda$ form a subspace, since eigenvectors by definition must be non-zero, while $E_\lambda$ also contains the zero vector.

(iii) $E_\lambda$ is $G$-invariant. This is true since if $x\in E_\lambda$, i.e. $Mx=\lambda x$, then for any $g\in G$, $MD(g)x = D(g)Mx = D(g)(\lambda x) = \lambda (D(g)x)$, which implies that $D(g)x\in E_\lambda$ also. Note you don't ever actually say this subspace is invariant/$G$-invariant in your proof.

(iv) since $D$ is irreducible, the only invariant subspaces of $\mathbb{C}^n$ are $\{0\}$ and $\mathbb{C}^n$. We know that $E_\lambda\ne\{0\}$ (since there is at least one eigenvector corresponding to $\lambda$, and this vector is nonzero by definition). Therefore we must have that $E_\lambda = \mathbb{C}^n$. This implies that in fact $M = \lambda I$.


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