If $\lambda$ is an eigenvalue of $T$, then $ |\lambda| \leq n \ \text{max} \{|\mathcal{M}(T)_{j,k}| : 1 \leq j,k \leq n\} $

The following exercise comes from "Linear Algebra Done Right", Sheldon Axler, 4th Edition, Section 5A, Exercise 16.

Suppose $v_1, \ldots, v_n$ is a basis of $V$ and $T \in \mathcal{L}(V)$. Prove that if $\lambda$ is an eigenvalue of $T$, then$$|\lambda| \leq n \ \text{max} \{|\mathcal{M}(T)_{j,k}| : 1 \leq j,k \leq n\}$$

where $\mathcal{M}(T)_{j,k}$ denotes the entry in row $j$, column $k$ of the matrix of $T$ with respect to the basis $v_1, \ldots, v_n$.

My initial attempt at this exercise followed a similar proof to the Gershgorin circle theorem, but it doesn't lead me anywhere near to the above result.

At this point in the text, we did not cover norms yet, so I cannot use that here. Could someone give a hint in the right direction?