How to find eigenvalues of matrix $\begin{bmatrix} 3& a+1\\a+1&3 \end{bmatrix}$ I want to find the eigenvalues of the following matrix
$\begin{bmatrix} 3& a+1\\a+1&3 \end{bmatrix}$ expressed in $a$ using $\begin{bmatrix} \lambda - 3& a+1\\a+1&\lambda-3 \end{bmatrix}$.
But the $a$-term makes it difficult for me to find it. I hope someone can show me how to do this. Thanks in advance !
EDIT:
Determinant = $4a^2+8a+4$
Using the abc rule I get
$(6\overset{+}{-} \sqrt{4a^2+8a+4})/2$
 A: Given

$$
\begin{bmatrix} 3& a+1\\a+1&3 \end{bmatrix}. \tag 1
$$

So you need to solve

$$
\det \begin{bmatrix} 3 - \lambda& a+1\\a+1&3 - \lambda \end{bmatrix} = 0.
\tag 2
$$

Then you get

$$
\Big( 3 - \lambda \Big)^2 - \Big( a + 1 \Big)^2 = 0. \tag 3
$$

Thus

$$
3 - \lambda = \pm ( a + 1 ). \tag 4
$$

So

$$
\lambda = 3 \mp ( 1 + a ). \tag 5
$$

$$
\Downarrow
$$

$$
\bbox[16px,border:2px solid #800000] { \lambda = 4 + a \vee \lambda = 2 - a. }
$$

A: Hint. The the characteristic polynomial is thus
$$
P(\lambda)=(\lambda-3)^2-(a+1)^2=(\lambda-a-4)(\lambda+a-2)
$$ then you found
$$
\lambda=a+4 \qquad \lambda=-a+2
$$ as candidates.
A: You could do this through the characteristic polynomial, as someone will surely show, but to expand a little, you could use some tricks:


*

*Trace of the matrix equals the sum of the eigenvalues

*Determinant of the matrix equals to the product of eigenvalues


This means that the sum is $\lambda_1 + \lambda_2 = 6$ and the product is $\lambda_1 \lambda_2 = -a^2-2a+8$. Using this information, you can get the result of $\lambda_1 = 2-a$ and $\lambda_2 = a+4$
Furthemore, since the sums of the columns are equal to each other, the value of the sum also has to be an eigenvalue (prove this!). This gives you one eigenvalue with no work: $3 + a + 1= a+4$
In this case, using the columns trick and the trace trick, you can get the eigenvalue virtually without having to do anything. 
A: \begin{align}
\left| \begin{matrix} \lambda -3 & a+1 \\
a+1 &\lambda-3 \end{matrix} \right| & = (\lambda-3)(\lambda-3) - (a+1)(a+1) \\
&= \lambda^2-6\lambda+9-(a^2+2a+1) \\
&= \lambda^2-6\lambda+9-a^2-2a-1 \\
&= \lambda^2-6\lambda-a^2-2a+8 \\
&= \lambda^2-6\lambda-(a^2+2a-8)\\
&= \lambda^2-6\lambda-(a+4)(a-2) \\
&= (\lambda -a-4)(\lambda +a-2) =0\\
\end{align}
Thus $\lambda = a+4$ and $\lambda = -a+2$.
A: Since both rows have the same sum, $\begin{bmatrix} 1 \\1 \end{bmatrix}$ is an eigenvector.
As 
$$\begin{bmatrix} 3& a+1\\a+1&3 \end{bmatrix}\begin{bmatrix} 1 \\1 \end{bmatrix}=\begin{bmatrix} a+4 \\a+4 \end{bmatrix}=(a+4) \begin{bmatrix} 1 \\1 \end{bmatrix}$$
it follows that $a+4$ is one of the eigenvalues. 
The second one can be found from $6=tr(A)$ is the sum of eigenvalues, or by observing that $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ is also an eigenvector, by observing that the differences on the rows are negative eachother.
A: Right so you want the determinant of your matrix to be zero, so you have
$(\lambda-3)^2-(a+1)^2=0$
$\lambda^2-6\lambda+9-a^2-2a-1=0$
Now you can use the quadratic formula to find the "roots" of this equation. Your value for lambda will be in terms of $a$, of course, which may be what is tripping you up.
