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Let A be a 3 by 3 matrix $$\begin{pmatrix} 1 & -2 & -1\\ -2 & 1 & 1 \\ -1 & 1 & 4 \end{pmatrix}$$

Then we have a real-number vector $\vec{ z }= \left( \begin{array}{c} z_1 \\ z_2 \\ z_3 \end{array} \right)$ such that $$\vec{z}^{ \mathrm{ T } }\vec{z} = 1$$ $$z_1+z_2+z_3=1$$

What is the range of $\vec{z}^{ \mathrm{ T } }A\vec{z} $?

I have found that $A$'s eigenvalues are -1,2, and 5 and eigenvectors are $\left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right)$$\left( \begin{array}{c} 1 \\ -1 \\ 1 \end{array} \right)$$\left( \begin{array}{c} -1 \\ 1 \\ 2 \end{array} \right)$ for each.

Can anyone help me?

Rodrigo de Azevedo
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ohisamadaigaku
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  • $z$ is from $\mathbb{R}^3$? Are you asking for the set of all possible values for all $z$ such that $\bar{z}\cdot z=1$ and $z_1 + z_2 + z_3=1$? – Peter Franek Aug 19 '20 at 17:59
  • @PeterFranek Yes $Z$ is from $ \mathbb{ R }^3$. I'm asking for the possible range of $\vec{z}^{ \mathrm{ T } }A\vec{z}$, not the value of elements or vector. – ohisamadaigaku Aug 19 '20 at 18:11
  • https://math.stackexchange.com/questions/3737949/the-plane-xyz-1-intersects-the-curve-x2y2z2-1-at-a-circle-c-find-th – Kumar Aug 19 '20 at 20:04
  • So you are searching for a minimum and maximum of a 2-dimensional quadratic function in a circle (equivalently, in a disc). This might be related https://math.stackexchange.com/questions/2440218/maximization-of-quadratic-form-over-unit-euclidean-sphere-not-centered-at-the-or -- but they don't offer an analytic solution. Maybe there is some trick here. – Peter Franek Aug 19 '20 at 20:12
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    The method of Lagrange multipliers gives a polynomial system. The conditional extrema are the two real roots of $117 x^4 - 860 x^3 + 1480 x^2 - 256 x - 688$. – Maxim Aug 20 '20 at 12:14

2 Answers2

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The $\vec z$ that satisfy the two constraints are the points on the circle formed by the intersection of the unit sphere $\vec z^\mathrm T\vec z = 1$ and the plane $z_1 + z_2 + z_3 = 1$. Describe that circle using the parametric form explained at https://math.stackexchange.com/a/1184089/389981 : The circle passes through points $(1,0,0),(0,1,0),(0,0,1)$ which are evenly spaced on the circle, so its center is their average $(1,1,1)/3$ from which it follows that the radius is $\sqrt{2/3}$. By symmetry, the vector $(1,1,1)$ is normal to the plane in which the circle lies, so two orthogonal vectors in that plane are $(1,-1,0),(1,1,-2)$; normalized, they are $(1,-1,0)/\sqrt 2,(1,1,-2)/\sqrt 6$. Hence, a parametric description of points on the circle is\begin{align}(1,1,1)/3 & + \sqrt{2/3}\cos\theta\,(1,-1,0)/\sqrt 2\\ & + \sqrt{2/3}\sin\theta\,(1,1,-2)/\sqrt 6\end{align} which simplifies to $$(1,1,1)/3 +\cos\theta\,(1,-1,0)/\sqrt 3 +\sin\theta\,(1,1,-2)/3.$$ Now apply $\vec z^\mathrm T A\vec z$ and simplify to get $$2(9 +\cos2\theta - 10\sin\theta + 2\sqrt3(\sin2\theta - \cos\theta))/9.$$ I used my graphing calculator to minimize and maximize that expression and found that the range of $\vec z^\mathrm T A\vec z$ is approximately $[-0.529741,4.9184228]$ where the minimum and maximum values occur at approximate values of $\theta$ of $1.78286$ and $4.04074$.

user0
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    "Now apply $z^t A z$ to each of the three vector"... well, it is a quadratic function, not a linear one, or am I missing something?... – Peter Franek Aug 19 '20 at 19:53
  • @PeterFranek I thank you for catching my error. I edited my solution. – user0 Aug 20 '20 at 18:32
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If we eliminate $z_3$ by replacing it with $1-z_1-z_2$, you want to find the minimum and maximum of $$\{z^TAz + b^Tz + c : zQz+q^Tz = 0\}$$

with $$A=\begin{pmatrix}7 & 2 \\ 2 & 3\end{pmatrix}, \; b=\begin{pmatrix}-10\\-6\end{pmatrix}, \; c=4, \; Q=\begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix}, \; q=\begin{pmatrix}-2\\ -2\end{pmatrix}.$$

Via the Lagrangian we find that an extremum must satisfy $2Az+b+\lambda(2 Qz + q)=0$ and $z^TQz+q^Tz = 0$, but I do not see an easy solution. The problem is now in a format that allows for this numerical procedure.

Instead I will go on and eliminate $z_2$ to get an unconstrained problem in $z_1$. The constraint is $2z_2^2+(2z_1-2)z_2+(2z_1^2-2z_1)=0$, so $z_2=\frac{1}{2}(1-z_1) \pm \sqrt{\frac{1}{4}-\frac{3}{4}z_1^2+\frac{1}{2}z_1}$. Plugging this into the objective function no longer gives a nice expression. Numerical analysis shows that the positive branch has a maximum of $41/9$ at $z_1=-1/3$ and a minimum of $\approx-0.53$ at $z_1\approx 0.538$ while the negative branch has a maximum of $\approx 4.92$ at $z_1 \approx -0.29$ and a minimum of $1$ at $z_1=1$.

So the range is approximately $-0.53$ to $4.92$.

LinAlg
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