Exercise 4.1.8

Question

Let $\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n$ be a basis in a vector space $V$. Assume also that the first $k$ vectors $\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_k$ of the basis are eigenvectors of an operator $A$, corresponding to an eigenvalue $\lambda$ (i.e. that $A \mathbf{v}_j = \lambda \mathbf{v}_j, j = 1, 2, ..., k$). Show that in this basis the matrix of the operator $A$ has block triangular form
%%%%%%%%%%%%%%%%%
\begin{equation*}
    \begin{bmatrix}
    \lambda I_k  &*  \
    \mathbf{0}  &B
    \end{bmatrix}
\end{equation*}
%%%%%%%%%%%%%%%%%

Jing Huang · Accepted Answer

\begin{proof} $
ewcommand{\T}{^\mathsf{T}}$
$A_{VV} = [I]_{VS} A_{SS} [I]_{SV}$, where $S$ represents the standard basis. $[I]$ is the coordinate change matrix and $[I]_{SV} = [\mathbf{v}_1\T, \mathbf{v}_2\T, ..., \mathbf{v}_n\T]$. $[I]_{SV} A_{VV} = A_{SS} [I]_{SV} = A_{SS} [\mathbf{v}_1\T, \mathbf{v}_2\T, ..., \mathbf{v}_n\T] = [\lambda \mathbf{v}_1\T, \lambda \mathbf{v}_2\T, ..., \lambda \mathbf{v}_k\T, ..., A_{SS}\mathbf{v}_n\T]$. Denote the $i$-th column of $A_{VV}$ with $\mathbf{a}_i$. Consider $\mathbf{a}_1$, then $[I]_{SV} \mathbf{a}_1 = \lambda \mathbf{v}_1\T$. Since $\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n$ is a basis, then $\mathbf{a}_1$ can only be the form $\mathbf{a}_1 = [\lambda, 0, 0, ..., 0]\T$. Similarly, check the first $k$ columns of $A_{VV}$, they are $\lambda$ times the first $k$ standard base vector. So $A_{SS}$ has the block triangular form above. 
\end{proof}

Exercise 4.1.8

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Exercise 4.1.8

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