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 $A$.

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.

There is at Least One Real Eigenvalue of an Odd Real Matrix
Let $n$ be an odd integer and let $A$ be an $n\times n$ real matrix.
Prove that the matrix $A$ has at least one real eigenvalue.
We give two proofs.
Proof 1.
Let $p(t)=\det(A-tI)$ be the characteristic polynomial of the matrix $A$.
It is a degree $n$ […]

Eigenvalues and their Algebraic Multiplicities of a Matrix with a Variable
Determine all eigenvalues and their algebraic multiplicities of the matrix
\[A=\begin{bmatrix}
1 & a & 1 \\
a &1 &a \\
1 & a & 1
\end{bmatrix},\]
where $a$ is a real number.
Proof.
To find eigenvalues we first compute the characteristic polynomial of the […]

Eigenvalues of a Hermitian Matrix are Real Numbers
Show that eigenvalues of a Hermitian matrix $A$ are real numbers.
(The Ohio State University Linear Algebra Exam Problem)
We give two proofs. These two proofs are essentially the same.
The second proof is a bit simpler and concise compared to the first one.
[…]

Positive definite Real Symmetric Matrix and its Eigenvalues
A real symmetric $n \times n$ matrix $A$ is called positive definite if
\[\mathbf{x}^{\trans}A\mathbf{x}>0\]
for all nonzero vectors $\mathbf{x}$ in $\R^n$.
(a) Prove that the eigenvalues of a real symmetric positive-definite matrix $A$ are all positive.
(b) Prove that if […]

A Matrix Having One Positive Eigenvalue and One Negative Eigenvalue
Prove that the matrix
\[A=\begin{bmatrix}
1 & 1.00001 & 1 \\
1.00001 &1 &1.00001 \\
1 & 1.00001 & 1
\end{bmatrix}\]
has one positive eigenvalue and one negative eigenvalue.
(University of California, Berkeley Qualifying Exam Problem)
Solution.
Let us put […]

Transpose of a Matrix and Eigenvalues and Related Questions
Let $A$ be an $n \times n$ real matrix. Prove the followings.
(a) The matrix $AA^{\trans}$ is a symmetric matrix.
(b) The set of eigenvalues of $A$ and the set of eigenvalues of $A^{\trans}$ are equal.
(c) The matrix $AA^{\trans}$ is non-negative definite.
(An $n\times n$ […]

Maximize the Dimension of the Null Space of $A-aI$
Let
\[ A=\begin{bmatrix}
5 & 2 & -1 \\
2 &2 &2 \\
-1 & 2 & 5
\end{bmatrix}.\]
Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix.
Your score of this problem is equal to that […]

A Square Root Matrix of a Symmetric Matrix
Answer the following two questions with justification.
(a) Does there exist a $2 \times 2$ matrix $A$ with $A^3=O$ but $A^2 \neq O$? Here $O$ denotes the $2 \times 2$ zero matrix.
(b) Does there exist a $3 \times 3$ real matrix $B$ such that $B^2=A$ […]