Orthonormal Basis, Eigenvectors and Eigenvalues in $\mathbb{R}^n$

Question

Let $n$ be a positive integer and let $c_1,..., c_n$ be a list of real numbers. Let $\{v_1, . . . , v_n\}$ be an orthonormal basis for $\mathbb{R}^n$, let $d = \text{min}\{c_1,...,c_n\}$. For each $1 \leq i \leq n$, set $d_i= \sqrt{c_i-d}$ and let $w_i = d_iv_i$. Let $B \in \mathcal{M}_{n\times n} (\mathbb{R})$ be the matrix the columns of which are $w_1,...,w_n$ and let $A = BB^T +dI$. For $1 \leq i \leq n$, show that $v_i$ is an eigenvector of $A$ associated with the eigenvalue $c_i$.

This one has me stumped. Any help, suggestions or direction will be greatly appreciated. Thanks

Answer 1

By definition $$ B_{ij}=(w_j)_i,\qquad B^T_{ij}=(w_i)_j $$ Then \begin{align} (Av_k)_i&=A_{ij}(v_k)_j=B_{ih}B^T_{hj}(v_k)_j+d\delta_{ij}(v_k)_j\\ &=(w_h)_i(w_h)_j(v_k)_j+d(v_k)_i\\ &=(w_h)_id_h\delta_{hk}+d(v_k)_i\\ &=d_k^2(v_k)_i +d(v_k)_i\\ &=(d_k^2+d)(v_k)_i=c_k(v_k)_i \end{align}

Answer 2

A long-winded version of enzotib's answer:

The $j$th row of $B^T$ is $w_j=d_jv_j$, so the $j$th element of $B^Tv_i$ is $d_j(v_i\cdot v_j)$, but the $v_i$ are orthonormal, so $B^Tv_i=d_i\mathbf e_i$, where $\mathbf e_i$ is the standard basis vector. The columns of $B$ are the images of the standard basis vectors, so $B(d_i\mathbf e_i)=d_iw_i=d_i^2v_i=(c_i-d)v_i$, therefore $Av_i=(c_i-d)v_i-dv_i=c_iv_i$.

Answer 3

Define a matrix ${\bf{D}}=diag(d_1, \dots, d_n)$. Further define a matrix ${\bf{B}} = ({\bf{w}}_1,{\bf{w}}_2, \dots, {\bf{w}}_n)$. Note that ${\bf{B}}=\bf{VD}$. Hence,

\begin{equation} {\bf{B}} =\left ( {\bf{v}}_1, {\bf{v}}_2, \dots, {\bf{v}}_n \right ) \begin{pmatrix} d_1 &0 &\dots &0 \\ 0&d_2 &0 &\vdots \\ \vdots &\vdots &\ddots &\vdots \\ 0& 0& \dots&d_n \end{pmatrix}. \end{equation}

Taking into account that ${\bf{V}}{\bf{V}}^T = {\bf{I}}$ we can write

\begin{equation} {\bf{A}} = {\bf{B}} {\bf{B}}^T+ d{\bf{I}} = {\bf{V}}{\bf{D}}{\bf{D}}^T{\bf{V}}^T + d{\bf{V}} {\bf{V}}^T. \end{equation}

Since, ${\bf{D}}$ is diagonal ${\bf{D}}{\bf{D}}^T = {\bf{D}}^2 = diag(d_1^2, \dots, d_n^2)$ so we write

\begin{equation} {\bf{A}} ={\bf{V}}{\bf{D}}^2{\bf{V}}^T + d{\bf{V}} {\bf{V}}^T = {\bf{V}} \left [ {\bf{D}}^2 + d {\bf{I}} \right ]{\bf{V}}^T, \end{equation}

but $ {\bf{D}}^2 + d {\bf{I}}=diag(c_1,\dots,c_n)$ so we get the eigen-decomposition of ${\bf{A}}$

\begin{equation} {\bf{A}} = {\bf{V}} \begin{pmatrix} c_1 &0 &\dots &0 \\ 0&c_2 &0 &\vdots \\ \vdots &\vdots &\ddots &\vdots \\ 0& 0& \dots&c_n \end{pmatrix}{\bf{V}}^T, \end{equation}

hence, ${\bf{v}}_i$ is an eigenvector of ${\bf{A}}$ associated with the eigenvalue $c_i$.