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

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?

• $z$ 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
• math.stackexchange.com/questions/3737949/… – 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 math.stackexchange.com/questions/2440218/… -- but they don't offer an analytic solution. Maybe there is some trick here. – Peter Franek Aug 19 '20 at 20:12
• 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

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 $$4.92$$.

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 $$4.04074$$.

• "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