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I thought that this question is simple, and asked it at Stackexchange. To my surprise, no one was able to answer it there. Now have to elevate it to Overflow.

I thought that this question is simple, and asked it at Stackexchange. To my surprise, no one was able to answer it there. Now have to elevate it to Overflow.

In finite dimensions, the noninvertibility of $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is equivalent to $\,c\,$ being an eigenvalue of the matrix $\,{\mathbb{M}}\,$. In infinite dimensions, this is not necessarily so. When $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is noninvertible (while the linear operator $\,{\mathbb{M}}\,$ is bounded), $\,c\,$ is said to belong to the spectrum of $\,{\mathbb{M}}\,$ -- which does not necessitate it being an eigenvalue. An operator on an infinite-dimensional space may have a nonempty spectrum and, at the same time, lack eigenvalues.

In finite dimensions, the noninvertibility of $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is equivalent to $\,c\,$ being an eigenvalue of the matrix $\,{\mathbb{M}}\,$. However, in infinite dimensions this is not necessarily so. When $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is noninvertible (while the linear operator $\,{\mathbb{M}}\,$ is bounded), $\,c\,$ is said to belong to the spectrum of $\,{\mathbb{M}}\,$ -- which does not necessitate it being an eigenvalue. An operator on an infinite-dimensional space may have a nonempty spectrum and, at the same time, lack eigenvalues.

I present its formulation for group representations, because this is the language understandable to a physicist. (Notions like moduli are difficult to us.)

• Suppose that $\,V\,$ is a countable-dimension vector space over $\,{\mathbb{C}}\,$ and that $\,{\mathbb{A}}\,$ is a group representation acting irreducibly on $\,V\,$. If the intertwiner $\,{\mathbb{M}}\in\,$Hom$\,_C(V, V )\,$ commutes with the action of $\,{\mathbb{A}}\,$, then $\,{\mathbb{M}}\,$ is a scalar multiple of the identity operator.

We begin with the following observation:

if $\,V\,$ is a countable-dimensional vector space over $\,{\mathbb{C}}\,$ and $\,{\mathbb{M}}\in\,$Hom$\,_C(V, V )\,$, then there is a number $\,c\in{\mathbb{C}}\,$ for which $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is not invertible on the space $\,V\,$.

To employ reductio ad absurdum, start with an assumption that the map $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is invertible for $\,\forall c\in {\mathbb{C}}\,$. Then, for any non-zero polynomial $$\,P(x)\,=\,(x-p_1)\,(x-p_2)\,.\,.\,.\,(x-p_N)\,\;,$$ invertible is the map $$\,P({\mathbb{M}})\,=\,({\mathbb{M}}\,-\,p_1\,{\mathbb{I}})\,({\mathbb{M}}\,-\,p_2\,{\mathbb{I}})\,.\,.\,.\,({\mathbb{M}}\,-\,p_N\,{\mathbb{I}})\,\;,$$ because the composition of invertible maps is invertible.

What now remains is to consider a number $\,c\,$ for which the map $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is not invertible on the space $\,V\,$, and to prove that this map is zero. The latter statement follows from Schur's second lemma. (Recall that $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is an intertwiner of an irreducible representation $\,{\mathbb{A}}\,$ to itself.)

${\mathbb{QED}}$

Now, my question.

Does this circumstance act as an impediment to the ending of the above proof?

Was it really legitimate of me to say that, if

I present its formulation for group representations, because this is the language understandable to a physicist.

• Suppose that $\,V\,$ is a countable-dimension vector space over $\,{\mathbb{C}}\,$ and that $\,{\mathbb{A}}\,$ is a group representation acting irreducibly on $\,V\,$. If the intertwiner $\,{\mathbb{M}}\in\,$Hom$\,_C(V, V )\,$ commutes with the action of $\,{\mathbb{A}}\,$, then there exists a number $\,c\in{\mathbb{C}}\,$ for which $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is not invertible on the space $\,V\,$.

To employ reductio ad absurdum, start with an assumption that the map $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is invertible for $\,\forall c\in {\mathbb{C}}\,$. Then, for any non-zero polynomial $$\,P(x)\,=\,(x-p_1)\,(x-p_2)\,.\,.\,.\,(x-p_N)\,\;,$$ invertible is the map $$\,P({\mathbb{M}})\,=\,({\mathbb{M}}\,-\,p_1\,{\mathbb{I}})\,({\mathbb{M}}\,-\,p_2\,{\mathbb{I}})\,.\,.\,.\,({\mathbb{M}}\,-\,p_N\,{\mathbb{I}})\,\;,$$ because the composition of invertible maps is invertible.

${\mathbb{QED}}$

Now, my question.

We have proven that, for some $\,c\in{\mathbb{C}}\,$, the operator $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is not invertible.

Can we now use Schur's second lemma, to state that $\,{\mathbb{M}}\,$ is a scalar multiple of the identity operator?

Despite this circumstance, will it be legitimate to say that, if