I understand why this idiom (which is common in math writing) might seem confusing, but what the author is saying is correct. When he says that
$\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ equals one of the $\lambda_j$'s
he means that
$\lambda$ is an eigenvalue of $T$ if and only if $\lambda\in\{\lambda_1,\ldots,\lambda_n\}$
or, to phrase it another way,
$\lambda$ is an eigenvalue of $T$ if and only if $\lambda=\lambda_1$, or $\lambda=\lambda_2$, ..., or $\lambda=\lambda_n$
Thus, if I set $\lambda$ equal to $\lambda_1$, the right side of the biconditional is true, so that $\lambda$ is an eigenvalue of $T$ when $\lambda=\lambda_1$; and similarly with all of the diagonal entries $\lambda_1,\ldots,\lambda_n$.
If $\Bbb F$ is any field, and $A \in \Bbb F^{m, m}$, and $\lambda \in \Bbb F$ is an eigenvalue of $A$ so that
$Av = \lambda v, \tag{1}$
for $0 \ne v \in \Bbb F^m$, then
$A^2 v = A(Av) = A(\lambda v) = \lambda Av = \lambda^2 v, \tag{2}$
and if
$A^kv = \lambda^k v, \tag{3}$
then
$A(A^k v) = A(\lambda^k v) = \lambda^k (Av) = \lambda^{k + 1} v, \tag{4}$
which, given (1) and (2) basically completes a very simple inductive proof that
$A^n v = \lambda^n v \tag{5}$
for all positive $n \in \Bbb Z$. Taking things one step further, we have
$a_n A^n v = a_n \lambda^n v \tag{6}$
for any $a_n \in \Bbb F$. From (6) we deduce that if $p(x) \in \Bbb F[x]$ with, say,
$p(x) = \sum_0^N a_i x^i, \tag{7}$
then
$p(A) v = \sum_0^N a_i A^i v = \sum_0^N a_i \lambda^i v = p(\lambda) v, \tag{8}$
so that if $p(A) = 0$, we conclude that $p(\lambda) v = 0$, whence, since $v \ne 0$, we have $p(\lambda) = 0$.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
Best Answer
4 is $\color{red}{\text{false}}$, since if $M(M-2)=0$, then $x(x-2)$ is the minimal polynomial of $M$ and since the characteristic polynomial must be a multiple of $x(x-2)$ and therefore at least one of the eigenvalues is zero which means $M$ is not invertible ($M$ is singuar), $\color{red}{\text{unless }M=2I}$ and that's why this statement is false, the only exception is $M=2I$.
2 is $\color{red}{\text{false}}$, since $p(x)=x^5+bx^4+cx^3+dx^2+ex+f$ is a monic polynomial of degree 5. with the given conditions we get two equations
$$\begin{cases}17a+9b+5c+3d+2e+33=0 \\ 97a+35b+13c+5d+2e+275=0\end{cases}$$
by solving for $(d,e)$ and substituting in the characteristic polynomial $p(x)$ we get $$p(x)=x^5+ax^4+bx^3+cx^2+(-4c-13b-40a-121)x+(\frac{7}{2}c+15b+\frac{103}{2}a+165)$$ This equation is very general and by a simple substitution of $a=b=c=0$ we get the polynomial $p(x)=x^5-121x+165$ which has two complex roots, so in general the second statement is false.