Finding matrix given eigenvalues and eigenvectors.

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?

• Use $Av=\lambda v$ to get system of linear equations and solve. – Yadati Kiran 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 $$\frac x{\lVert x\rVert}$$.
• Sir Why in particular is $\lvert\alpha_1\rvert^2+\cdots+\lvert\alpha_n\rvert^2=1$? – Yadati Kiran Nov 26 '18 at 17:29
• Because the basis is orthonormal and $\lVert x\rVert=1$. – José Carlos Santos Nov 26 '18 at 17:30