I was studying linear algebra from Treil's LADW, and this is a problem from the text that I couldn't come up with an answer for. Why does a matrix like this have this form?
Let $\vec{v}_1 , \vec{v}_2 , . . . , \vec{v}_n$ be a basis in a vector space $V$. Assume also that the first $k$ vectors $\vec{v}_1, \vec{v}_2 , . . . , \vec{v}_k$ of the basis are eigenvectors of an operator A, corresponding to an eigenvalue $λ$ (i.e. that $A\vec{v}_j = \lambda\vec{v}_j , j = 1, 2, . . . , k)$. Show that in this basis the matrix of the operator $A$ has block triangular form $$ \begin{bmatrix} \lambda I_k & ∗ \\ 0 & B \end{bmatrix} $$ where $I_k$ is $k \times k$ identity matrix and $B$ is some $(n − k) \times (n − k)$ matrix.
You need to remember how we write the matrix $A$ in terms of the given basis.
In general, given a basis $v_1,\ldots,v_n$ (I am suppressing the $\vec{}$ in the vectors), then we apply $A$ in each $v_j$ and write the result as a linear combination of this basis. So, if $$ Av_j=a_{1j}v_1+\ldots+a_{nj}v_j, $$ then the $j$-th column of the matrix $A$ will have the entries $a_{1j},\ldots,a_{nj}$.
For your particular case, we have \begin{align*} Av_1&=\lambda v_1+0v_2+\ldots+0v_n,\\ Av_2&=0v_1 + \lambda v_2+\ldots+0v_n,\\ &\phantom{123456}\vdots\\ Av_k&=0v_1+\ldots+\lambda v_k+0v_{k+1}+\ldots+0v_n. \end{align*} Then the first $k$ columns have the stated form. For the remaining vectors, it does not really matter what will be: \begin{align*} Av_{k+1}&=a_{1,k+1}v_1+\ldots+a_{n,k+1}v_n,\\ &\phantom{1234}\vdots\\ Av_n&=a_{1,n}v_1+\ldots+a_{n,n}v_n, \end{align*} Hence the author just represented by a star $*$ and a $(n-k)\times(n-k)$ matrix $B$.