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.
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$$Tv=\lambda v=\lambda v+0\times v_2+\cdots+0\times v_k.$$