Proving that $\text{tr}^2(A) - 4\det(A) \geq 0$

Question

I'm trying to figure out how to prove that when given a diagonal matrix

\begin{align} A= \begin{bmatrix} \lambda & 0\\ 0 & \mu \end{bmatrix}, \end{align} with a positive determinant $\det(A) > 0$, the following statement is true $$\text{tr}^2(A) - 4\det(A) \geq 0$$ Thank you in advance.

Answer 1

$\begin{align} tr(A)^2-4 \det(A)\geq 0 \end{align} \iff (\lambda + \mu)^2 \ge 4 \lambda \mu \iff \lambda^2 -2 \lambda \mu + \mu^2 \ge 0 \iff (\lambda- \mu)^2 \ge 0$.

Answer 2

Hint :

Simply using the definitions of $\text{tr}$ and $\det$ :

The trace of the matrix $A$, is :

$$\text{tr}(A) = \lambda + \mu $$

The determinant of $A$, simply is :

$$\det(A) = \lambda \mu$$

Can you now follow a simple algebraic path to $\text{tr}^2(A) - 4\det(A)$ and conclude about its sign ? Use also the condition that as you stated, the determinant is positive.

Answer 3

More generally, the characteristic polynomial of a $2 \times 2$ matrix $A$ is $x^2-\operatorname{tr}(A)x + \det(A)$. The roots of this polynomial are the eigenvalues of $A$ and so are real iff the discriminant $\operatorname{tr}(A)^2 -4\det(A)$ is nonnegative.