Let \[ A=\frac16 \begin{pmatrix} 13&-5&-2\\ -5&13&-2\\ -2&-2&10 \end{pmatrix} \] and let \[ v=\begin{pmatrix}x\\y\\z\end{pmatrix}\in\mathbb R^3. \] As \(v\) ranges over the set satisfying \[ v^TAv=1, \] show that \(\|v\|\) is bounded and determine its least upper bound.
Solution
1. The vectors\[ u_1=(1,1,1)^T, \qquad u_2=(1,1,-2)^T, \qquad u_3=(1,-1,0)^T \]are pairwise orthogonal eigenvectors of \(A\) with eigenvalues \(1\), \(2\), and \(3\), respectively.
Direct multiplication gives \[ Au_1=u_1, \qquad Au_2=2u_2, \qquad Au_3=3u_3. \] Also \[ u_1^Tu_2=1+1-2=0, \qquad u_1^Tu_3=1-1=0, \qquad u_2^Tu_3=1-1=0. \] Since these are three nonzero orthogonal vectors in \(\RR^3\), they form an orthogonal basis.
2. For every \(v\in\RR^3\),\[ v^TAv\geq\norm{v}_2^2. \]
Let \[ e_j\coloneqq\frac{u_j}{\norm{u_j}_2}, \qquad j=1,2,3. \] By step 1, \(e_1,e_2,e_3\) is an orthonormal eigenbasis. Write \[ v=c_1e_1+c_2e_2+c_3e_3. \] Then \[ v^TAv = c_1^2+2c_2^2+3c_3^2 \geq c_1^2+c_2^2+c_3^2 = \norm{v}_2^2. \]
3. Every \(v\) satisfying \(v^TAv=1\) obeys\[ \norm{v}_2\leq1. \]
Step 2 gives \[ 1=v^TAv\geq\norm{v}_2^2. \] Taking square roots yields the bound.
4. The bound in step 3 is attained.
Take \[ v_0\coloneqq\frac1{\sqrt3}(1,1,1)^T. \] Then \(\norm{v_0}_2=1\), and step 1 gives \[ Av_0=v_0. \] Therefore \[ v_0^TAv_0=v_0^Tv_0=1. \] Thus \(v_0\) belongs to the prescribed level set and has norm \(1\).
5. The least upper bound of \(\norm{v}_2\) on the level set \(v^TAv=1\) is\[ \boxed{1}. \]
Step 3 shows that \(1\) is an upper bound, while step 4 shows that it is achieved. Hence it is the maximum and therefore the least upper bound.
6. Q.E.D.
Step 5 gives the required bound and its least upper bound.