# 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 $$\lambda I$$.

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 $$M = \lambda I$$.