a) Let $B$ be a 2x2 symmetric matrix and let $u$, $v$ be two eigenvectors of $B$ associated with the eigenvalues $w$ and $l$ respectively. Suppose $u$ = $$ \left[ \begin{array}{c} 1/\sqrt2\\ -1/\sqrt2 \end{array} \right] $$ and $v$ = $$ \left[ \begin{array}{c} 1/\sqrt2\\ 1/\sqrt2 \end{array} \right] $$ with $w$= 1 and $l$=3 respectively, find matrix $B$.

b) Let $C$ be another symmetric matrix of order n with a characteristic polynomial $(w-w_1)(w-w_2)...(w-w_n)$ where $w_1≤w_2≤...≤w_n$. Prove that for any nonzero vector $x$ that belongs in $R^n$, $w_1≤x^TCx/x^Tx≤w_n$.

What I have done: For part (a), converting $u$ and $v$ to orthogonal basis, I get $u_o$=$$ \left[ \begin{array}{c} 1\\ -1 \end{array} \right] $$ and $v_o$=$$ \left[ \begin{array}{c} 1\\ 1 \end{array} \right] $$. Letting $M$ = $$ \left[ \begin{array}{cc} 1&0\\ 0&3 \end{array} \right] $$ and $S$ = $$ \left[ \begin{array}{cc} 1&1\\ -1&1 \end{array} \right] $$, the matrix B is obtained by putting together $SMS^{-1}$ = $$ \left[ \begin{array}{cc} 2&1\\ 1&2 \end{array} \right] $$.

However, for part (b), while I can derive that $w_1,w_2,...,w_n$ are eigenvalues of C, I am unsure of how to proceed, hence may I get some help for this?

  • $\begingroup$ Use $Av=\lambda v$ to get system of linear equations and solve. $\endgroup$ Nov 26 '18 at 17:13

Suppose that $\lVert x\rVert=1$. Then $x^Tx=1$. On the other hand, let $(v_1,\ldots,v_n)$ an orthonormal basis of eigenvecors, such that the eigenvalu corresponding to $v_k$ is $w_k$. Then $x$ can be written as $\alpha_1v_1+\cdots+\alpha_kv_k$. So, $Cx=\alpha_1w_1v_1+\cdots+\alpha_nw_nv_n$ and so$$x^TCX=\lvert\alpha_1\rvert^2w_1+\cdots+\lvert\alpha_n\rvert^2w_n.$$Since $\lvert\alpha_1\rvert^2+\cdots+\lvert\alpha_n\rvert^2=1$, this number is between $w_1$ and $w_n$.

If $x$ is an arbitrary non-zero vector, you can apply the previus result to $\frac x{\lVert x\rVert}$.

  • $\begingroup$ Sir Why in particular is $\lvert\alpha_1\rvert^2+\cdots+\lvert\alpha_n\rvert^2=1$? $\endgroup$ Nov 26 '18 at 17:29
  • $\begingroup$ Because the basis is orthonormal and $\lVert x\rVert=1$. $\endgroup$ Nov 26 '18 at 17:30
  • $\begingroup$ Oh ok sir. got it ! $\endgroup$ Nov 26 '18 at 17:31

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