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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 $A\vec{v_j} = λ\vec{v_j} , j = 1, 2, . . . , k)$. Show that in this basis the matrix of the operator $A$ has block triangular form $$ \begin{bmatrix} &λI_k &∗ \\ &0 &B \end{bmatrix} $$ where $I_k$ is $k × k$ identity matrix and $B$ is some $(n − k) × (n − k)$ matrix.

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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 $B$.

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  • $\begingroup$ Thank you! I think I got confused because the author refers to both the original matrix A, and the new matrix B (in the given basis) as A in a subsequent chapter (I think it was the one about diagonalization), which made me go back and try this problem. $\endgroup$ – sarpu Feb 25 '18 at 7:00

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