# Eigenvalues of $2\times 2$ Symmetric Matrices are Real by Considering Characteristic Polynomials ## Problem 609

Let $A$ be a $2\times 2$ real symmetric matrix.
Prove that all the eigenvalues of $A$ are real numbers by considering the characteristic polynomial of $A$. Add to solve later

## Proof.

Let $A=\begin{bmatrix} a& b \\ c& d \end{bmatrix}$.
Then as $A$ is a symmetric matrix, we have $A^{\trans}=A$.
This implies that
$\begin{bmatrix} a& c \\ b& d \end{bmatrix}=\begin{bmatrix} a& b \\ c& d \end{bmatrix}.$ Hence we have $b=c$ by comparing entries.

Now, we find the characteristic polynomial $p(t)$ of $A$.
We have
\begin{align*}
p(t)&=\det(A-t I)=\begin{vmatrix}
a-t & b\\
b& d-t
\end{vmatrix}\6pt] &=(a-t)(d-t)-b^2\\ &=t^2-(a+d)t+ad-b^2. \end{align*} Note that the eigenvalues of A are roots of the characteristic polynomial p(t). Hence, it suffices to show that the roots of p(t) are real numbers. The quadratic polynomial has only real roots if and only if its discriminant is non-negative. The discriminant of p(t) is given by \begin{align*} (a+d)^2-4(ad-b^2)&=a^2+2ad+d^2-4ad+4b^2\\ &=a^2-2ad+d^2+4b^2\\ &=(a-d)^2+4b^2. \end{align*} Observe that the last expression is the sum of two squares of real numbers. Hence the discriminant of p(t) is nonnegative. We conclude that every 2\times 2 symmetric matrix has only real eigenvalues. ### Remark We also could find the eigenvalues directly. By the quadratic formula, the eigenvalues of A are \[\frac{a+d\pm\sqrt{(a+d)^2-4(ad-b^2)}}{2}=\frac{a+d\pm \sqrt{(a-d)^2+4b^2}}{2} and as the number inside the square root (discriminant) is positive, we conclude that the eigenvalues are real. Add to solve later

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