= Solution
{c}
The definition of an <eigenvalue> and the <kernel> criterion for a singular <square matrix> give
$$
\begin{aligned}
\lambda\text{ is an eigenvalue}
&\iff\ker(T-\lambda I)\ne\{0\}\\
&\iff\lambda I-A\text{ is not invertible}\\
&\iff\det(\lambda I-A)=0\\
&\iff\chi_T(\lambda)=0.
\end{aligned}
$$
For the second equivalence, a <linear map> from a finite-dimensional <vector space> to itself has zero <kernel> exactly when its <rank of a linear map> is the full <dimension>, by the <rank-nullity theorem>, and therefore exactly when it is invertible. This proves both directions.
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