The probability value of the three matrices

Question

Determine the probability value that the following three matrices have real eigenvalues. For example, if the random variables $A$ and $B$ are given by a uniform distribution $(0,1)$, with $A$ and $B$ independent $$P=\left[\begin{array}{cc} A & A+B\\ A-B & -B \end{array}\right]; Q=\left[\begin{array}{cc} A & - B\\ A-B & B \end{array}\right]; R=\left[\begin{array}{cc} A & A\\ A-B & B \end{array}\right]$$ Will the probabilities of the three matrices also change if the interval changes to $(-1,1)?$

My work: (Please correct my work is it correct? Are there any other ideas? Please help me, thank you)

Matrix P:

The characteristic polynomial of matrix P is given by:

$$det(\lambda I - P) = \lambda^2 - 2\lambda B + (A^2 - B^2)$$

For real eigenvalues, the discriminant of this quadratic equation must be non-negative:

$$(-2B)^2 - 4(A^2 - B^2) \geq 0$$

This simplifies to:

$$A^2 \geq 4B^2$$

This condition implies that the eigenvalues of matrix P are real if and only if $A \geq 2B$ or $A \leq -2B$.

The probability that $P$ has real eigenvalues is equal to the probability that either $A \geq 2B$ or $A \leq -2B$. Since $A$ and $B$ are independent and uniformly distributed on $(0,1)$, this probability can be calculated as follows:

$$P(\text{real eigenvalues}) = P(A \geq 2B) + P(A \leq -2B)$$

$$= \int_{0}^{1/2} \int_{0}^{2b} dx dy + \int_{0}^{1/2} \int_{2b}^{1} dx dy$$

$$= \frac{1}{2} + \frac{1}{2} = 1$$

Therefore, the probability that matrix P has real eigenvalues is 1.

Matrix Q:

The characteristic polynomial of matrix Q is given by:

$$det(\lambda I - Q) = \lambda^2 - 2\lambda B - (A^2 + B^2)$$

For real eigenvalues, the discriminant of this quadratic equation must be non-negative:

$$(-2B)^2 + 4(A^2 + B^2) \geq 0$$

This simplifies to:

$$A^2 \geq -4B^2$$

This condition has no real solutions, indicating that matrix Q always has complex eigenvalues.

Matrix R:

The characteristic polynomial of matrix R is given by:

$$det(\lambda I - R) = \lambda^2 - 2\lambda B + (A^2 - 2B + B^2)$$

For real eigenvalues, the discriminant of this quadratic equation must be non-negative:

$$(-2B)^2 - 4(A^2 - 2B + B^2) \geq 0$$

This simplifies to:

$$A^2 \geq B^2 - 2B$$

This condition implies that the eigenvalues of matrix R are real if and only if $A \geq B$.

The probability that R has real eigenvalues is equal to the probability that $A \geq B$. Since $A$ and $B$ are independent and uniformly distributed on $(0,1)$, this probability can be calculated as follows:

$$P(\text{real eigenvalues}) = P(A \geq B)$$

$$= \int_{0}^{1} \int_{0}^{x} dx dy$$

$$= \frac{1}{2}$$

Therefore, the probability that matrix R has real eigenvalues is 1/2.

Answer 1

Your general approach is correct, but the details of the computation are strangely incorrect. Here is a correct approach for the matrix $P$.

We compute the characteristic polynomial of $P$ to be $$ p(\lambda) = \lambda^2 + (B - A)\lambda + (B^2 - AB - A^2). $$ The discriminant of this polynomial is $$ (B - A)^2 - 4(B^2 - AB - A^2) = 5A^2 + 2AB - 3B^2 = (5A - 3B)(A + B). $$ Thus, the condition that the eigenvalues of $P$ are real reduces to $5A - 3B \geq 0$. That is, we are looking for the probability of a point in $[0,1]\times [0,1]$ being selected that lies to the right of the line $x = \frac 35 y$ (where $x$ plays the role of $A$ and $y$ the role of $B$). This leads to the integral $$ \int_0^{1}\int_{(3/5) y}^1 1\,dx\,dy = 1 - \frac 12 \cdot \frac 35 = \frac 7{10}, $$ where the above computation is simplified by noting that we are simply looking for the complement of the area of a right triangle with legs of length $1$ and $3/5$. Therefore, the probability that $P$ has real eigenvalues is $7/10$.