# Better Understanding of the Proof : $P(A)$ has no other eigenvalues than $P(\lambda_i)$ where $\lambda_i$ is an eigenvalue of $A$

Assume that $P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$ is a polynomial such that the coefficients are complex numbers and $A \in M_n(\mathbb C)$. We can define $P$ on $A$ like this :
For example if $P(x)=x^2+x+2$ then $P(A)=A^2+A+2I$ .

Assume that $\gamma$ is an eigenvalue of $P(A)$.
We want to show that there exists $\lambda_i$ such that $\lambda_i$ is an eigenvalue of $A$ and $P(\lambda_i)=\gamma$.
This is what our teacher wrote ( And he mentioned that he was writing a brief version of it so he may have emitted some parts ) :

$P(x)-\gamma=a_m(x-\lambda_1)(x-\lambda_2)\dots(x-\lambda_m)$
$\implies P(A)-\gamma I_n=a_m(A-\lambda_1 I_n)(A-\lambda_2 I_n)\dots(A-\lambda_m I_n)$
We know that $P(A)-\gamma I_n$ is not invertible. So, there exists $\lambda_i$ such that $A-\lambda_i I_n$ is not invertible. So, $\lambda_i$ is an eigenvalue of $A$.
Thus we have :
$P(\lambda_i)-\gamma=0 \implies P(\lambda_i)=\gamma$

My problem :

1. How on earth do we know that $P(x)-\gamma=a_m(x-\lambda_1)(x-\lambda_2)\dots(x-\lambda_m)$ ?
I know that $P(x)$ has eigenvalues of $A$ as roots ( which are in the form of $\lambda_i$'s ). But why $P(x)-\gamma$ has them as roots too? ( Please don't just mention the fundamental theorem of linear algebra. Explain it more because i'm new to linear-algebra. )

2. What is $a_m$? Why is the last factor $A-\lambda_m I_n$ not $A-\lambda_n I_n$?

3. How does this proof guarantee that $P(A)$ has no other eigenvalues? ( Someone may say that there is another eigenvalue which doesn't come from a $\lambda_i$ )

Note : I don't want the proof cause i have it. I want to understand it.

1. You make a confusion : $P$ is any polynomial, not a cancelling polynomial of $A$. What your teacher says is that $P(X)-\lambda$ is a polynomial of degree $m$, so it has to have $m$ roots in $\mathbb C$, which he calls $\lambda_1,\dots,\lambda_m$.
2. $a_m$ is the leading coefficient of $P(X)-\lambda$, which is, if $P$ is not a constant, the leading coefficient of $P$.
3. The only thing this demonstration shows is that if $\lambda$ is an eigenvalue of $P(A)$, then there exists an eigenvalue $\mu$ of $A$ such that $\lambda=P(\mu)$. You may be confused by the similarity of the notations $\lambda$ and $\lambda_i$.
• So, the way of proof is this : Theorem said eigenvalues of $P(A)$ =$\{P(\lambda):\lambda$ is an eigenvalue of $A\}$ ... we proved this part ...we showed that $P(\lambda) I_n - P(A)$ is not invertible. What i wrote as his proof in this question, shows that if we have an eigenvalue of $P(A)$ like $\gamma$, there exists an eigenvalue of $A$ like $\lambda_i$ such that $\gamma=P(\lambda_i)$ ... and thus, the theorem is proved ... am i right ? – Arman Malekzadeh Oct 27 '16 at 11:21