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I'm aware that perhaps this proof has been analized and discussed a lot here but there's something that isn't clear to me from what I've been reading from Axler's text. How can we assert the equality of factorizations when we evaluate $z= T$ in $c(z - \lambda_1)\cdots(z - \lambda_m)$ to get $c(T - \lambda_1 I)\cdots(T - \lambda_m I)$? Because to me it seems that the fact that we have to add the identity to each term as to make it consistent changes the factorization altogether.

I know that $c(T - \lambda_1 I)\cdots(T - \lambda_m I)$ is an operator but it seems a bit forced that he is able to find the exact form we need to conclude the proof of the theorem just like that.

Axler's proof of the existence of eigenvalues for operators defined in complex vector spaces

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    $\begingroup$ If I recall correctly, Axler shows earlier that the operation of substitution into a polynomial is a homomorphism. $\endgroup$ Commented Oct 14 at 16:36
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    $\begingroup$ Multiplication by $\lambda$ in a vector space is achieved by applying the linear map $\lambda I$. $\endgroup$ Commented Oct 14 at 16:50

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To elaborate on blargoner's comment, notice that when $T$ is an operator and $I$ is the identity operator, the ring $R$ of operators generated by the complex multiples of $T, I$ is not only a ring but a commutative ring (with identity). (If we take $T^0$ to be $I$, then its elements are polynomials in $T$, and powers of $T$ commute with each other.)

By the universal property of polynomial rings, then, since this ring is commutative, there exists a homomorphism $f: \mathbb C[z]\to R$ sending $z\mapsto T$. This homomorphism is what allows the polynomial factorization identity to be carried over.

You're right to be cautious, though! Polynomial factorization identities involving two or more variables do not necessarily carry over because rings generated by multiple non-commuting operators do not generate commutative rings and hence do not give rise to any such homomorphism from $\mathbb C[z,w]$. Hence, even very basic multi-variable factorization identities such as difference of squares $x^2-y^2 = (x+y)(x-y)$ can fail to be true with operators. Indeed, if $TS \ne ST$, then $(S+T)(S-T) = TS-ST \ne 0$.

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  • $\begingroup$ So this means, as Ted Shifrin said, that this homomorphism maps $\lambda \mapsto \lambda I$? $\endgroup$ Commented Oct 14 at 17:27
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    $\begingroup$ I did not write this, but the homomorphism is also guaranteed to fix the base ring, in this case $\mathbb C$. This guarantees that the complex numbers themselves are preserved (and of course, identity elements are unique and must also be preserved) and hence $\lambda$ is mapped to $\lambda I$. See here: en.wikipedia.org/wiki/Polynomial_ring#Polynomial_evaluation $\endgroup$ Commented Oct 14 at 17:30

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