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

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

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 $A$.

 
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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.


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