Minimising the largest eigenvalue of the matrix

Question

I have the following $5 \times 5$ matrix $A(x)$

$$A(x) = x \begin{bmatrix} 17 & 24 & 1 & 8 & 15 \\ 23 & 5 & 7 & 14 & 16 \\ 4 & 6 & 13 & 20 & 22 \\ 10 & 12 & 19 & 21 & 3 \\ 11 & 18 & 25 & 2 & 9 \end{bmatrix} + (1-x) \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 3 & 6 &10 & 15 \\ 1 & 4 & 10 & 20 & 35 \\ 1 & 5 & 15 & 35 & 70 \end{bmatrix}$$

When I calculated the matrix $A$,

$$A(x) = \begin{bmatrix} 16x + 1 & 23x + 1& 1& 7x + 1& 14x + 1 \\ 22x + 1& 3x + 2& 4x + 3& 10x + 4& 11x + 5 \\ 3x + 1& 3x + 3& 7x + 6& 10x + 10& 7x + 15 \\ 9x + 1& 8x + 4& 9x + 10& x + 20& 35 - 32x \\ 10x + 1& 13x + 5& 10x + 15& 35 - 33x& 70 - 61x \end{bmatrix} $$

I would like to find the value of $x \in [0,1]$ that minimises the largest eigenvalue of the matrix $A(x)$.

I don't understand what means the "minimising the largest eigenvalue of a matrix".

Please help me to solve it. Thanks a lot.

Answer 1

To the OP. Do you realize that you have an unpleasant attitude towards people who want to help you? For some time, on this website, young (or not) people of a modest level appear very arrogant. The coronavirus may not work for them, but in this case, they may think about getting treatment.

Let $U(x)=xA+(1-x)B$. Since $U(x)$ is a positive matrix (the $u_{i,j}$ are $>0$), $\rho(U)=\max_{\lambda\in spectrum(U)}|\lambda|$ is an eigenvalue of $U$; moreover it's a single eigenvalue and it's the sole eigenvalue, the modulus of which, is $\rho(U)$.

We consider the characteristic polynomial of $U(x)$: $p(x,y)=$

Note that $p(x,\rho(U(x)))=0$. Now we seek $x_0$ s.t. $y_0=\rho(U(x_0))$ reaches $\min_{x\in[0,1]}\rho(U(x))$. Since $\rho(U)$ is always a single eigenvalue, we deduce that $\dfrac{\partial p}{\partial x}(x_0,y_0)=0$ where

Finally $(x_0,y_0)$ is in the intersection of the implicit curves $p(x,y)=0,\dfrac{\partial p}{\partial x}(x,y)=0$.

By drawing the graphs of the functions, we see that the intersection point with maximum $y$ is obtained for $x\approx 0.8$.

Using a zoom, we obtain this approximation: $x_0\approx 0.796035,y_0\approx 63.378642$.

With a software, we can do better

EDIT. Anwer to the OP. Method 1. You calculate the minimum for $x\in [0,1]$ of the function $\rho(U(x))$; unfortunately, there is no explicit formula for $\rho(U(x))$ because it's a root of a polynomial of degree $5$.

Method 2. You solve the system $p(x,y)=0,\dfrac{\partial p}{\partial x}(x,y)=0$ by choosing the initial point well. Example with Maple

fsolve({$p(x,y),\dfrac{\partial p}{\partial x}(x,y)$},{$x=0.8,y=63$});

There must be a similar procedure in Matlab.