Prove there is a basis of $V$ with respect to which the matrix of $T$ has 0’s everywhere in the first column, except possibly the first entry

Question

I'm reading Sheldon Axler's Linear Algebra Done Right. There is a paragraph on Eigenvectors I can't quite comprehend. The original text is below.

For some linear map $T$, an eigenvalue $\lambda$ of $T$ and a corresponding eigenvector $v$, we have $Tv = \lambda v$.
So we have (assuming the vector space is 4 dimensional to simplify the latex):
$$Tv = \lambda v$$
$$T\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
\end{bmatrix}=\begin{bmatrix}
\lambda x_1 \\
\lambda x_2 \\
\lambda x_3 \\
\lambda x_4 \\
\end{bmatrix}$$

I don't see how it forces the matrix for $T$ to be of this form below.
$$\begin{bmatrix}
\lambda & * & * & *\\
0 & * & * & *\\
0 & * & * & *\\
0 & * & * & *\\
\end{bmatrix}\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
\end{bmatrix}=\begin{bmatrix}
\lambda x_1 \\
\lambda x_2 \\
\lambda x_3 \\
\lambda x_4 \\
\end{bmatrix}$$

In particular, how does this matrix for $T$ ensure that once matrix multiplication is applied on $v$, we have the desired $\lambda x_1$ in the first row? Wouldn't we also need to make sure all the other entries in the first row of $T$ is zero?

Furthermore, what is the assurance that by choosing $v$ as one of the basis vectors, the entries below the $\lambda$ in the matrix for $T$ is zero?

The original text is below:

If V is a finite-dimensional complex vector space, then we already
know enough to show that there is a basis of V with respect to which
the matrix of T has 0’s everywhere in the first column, except
possibly the first entry. In other words, there is a basis of V with
respect to which the matrix of T looks like

here the denotes the entries in all the columns other than the first
column. To prove this, let $\lambda$ be an eigenvalue of T (one exists by
5.21) and let v be a corresponding eigenvector. Extend v to a basis of V. Then the matrix of T with respect to this basis has the form above.
Soon we will see that we can choose a basis of V with respect to which
the matrix of T has even more 0’s.

Answer 1

If the first vector of a basis $B$ of $V$ is $v$, then, since $Tv=\lambda v$, the first column of $[T]_B^B$ is$$\begin{bmatrix}\lambda\\0\\0\\\vdots\\0\end{bmatrix}.$$That's so because, if $B=\{v,v_1,\ldots,v_k\}$, then$$Tv=\lambda v=\lambda v+0\times v_2+\cdots+0\times v_k.$$