Help with finding eigenvectors and polynomial factorization I have a real and invertible matrix
$$ A = \left( \begin{matrix} 5 & 4 \\
  4&5 \end{matrix} \right)$$
and I need to find the eigenvectors.
The characteristic polynomial is $(5-\lambda)^2 - 16$. According to Wolfram Alpha this factors to $(1-\lambda)(9-\lambda)$ (How does one get there, are there any good tutorials on polynomial factorization?), so the eigenvalues are $1$ and $9$.
Calculating the eigenvector to the eigenvalue $1$ given by $(A-I)\vec x = \vec 0$.
I get that $\vec x$ is the null vector $=\left(
\begin{array}{c}
0\\
0\\
\end{array}
\right)$(can that be right?).
For the second eigenvector $(A-9I)\vec x = 0$ I get 
$$x_{1} = 4x_{2}$$
$$-x_{2} = 5x_{1}$$
Is this right, and how can I write the second eigenvector with this information??
 A: Everyone else has focussed on the polynomial side of things, so I'll focus on the linear algebra part.
Firstly, the null vector, $\vec x=\vec 0$, is not an eigenvector of any matrix.
Why? By definition, $\vec x$ is an eigenvector of $A \in \mathbb{M}_{m,n}$ iff $\vec x$ is a nonzero vector such that $A \vec x = \lambda \vec x$ (for some scalar $\lambda$).
Also, I note that you've asked "is this the eigenvector for this eigenvalue?"
For any eigenvalue, there is no unique eigenvector. Indeed, if $\left(
\begin{array}{c}
1\\
1\\
\end{array}
\right)$ is an eigenvector of $A$, then so is $\left(
\begin{array}{c}
2\\
2\\
\end{array}
\right)$, $\left(
\begin{array}{c}
3\\
3\\
\end{array}
\right)$ or any scalar multiple of $\left(
\begin{array}{c}
1\\
1\\
\end{array}
\right)$.
A: Well $(5-\lambda)^2-4^2$ is a difference of squares, but you really need to learn factoring.
$$\begin{pmatrix}
4&4\\4&4\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix}=0$$
gives $x=-y$ so $\begin{pmatrix}1\\-1\\\end{pmatrix}$ is one eigenvector,
and $$\begin{pmatrix}
-4&4\\4&-4\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix}=0$$
gives $x=y$ so $\begin{pmatrix}1\\1\\\end{pmatrix}$ is the other eigenvector,
A: The characteristic polynomial is, as you correctly state, $(5 - \lambda)^2 - 16$. Expanding this we get
$
25 - 10\lambda + \lambda^2 - 16
$
i.e.
\begin{eqnarray}
p(\lambda) = \lambda^2 - 10 \lambda + 9
\end{eqnarray}
Now, this is a quadratic polynomial in $\lambda$, so you can use the quadratic formula to find the values for $\lambda$ such that $p(\lambda) = 0$. But, in this case it is "easy" (when you get the hang of it) to see directly from the coefficients what $\lambda$ should be, i.e. $\lambda$ should be either $9$ or $1$. To see this, consider the product
\begin{eqnarray}
(\lambda - a)(\lambda - b) = \lambda^2 - (a + b)\lambda + ab
\end{eqnarray}
now which $a$ and $b$ should we choose to get the coefficients in $p(\lambda)$? Well $ab$ should be $9$ and $a + b$ should be $10$. So $a = 9$ and $b = 1$ works.
To find the eigenvector of $\lambda = 1$ we must find $x$ such that
\begin{eqnarray}
(A - 1I)x = 0
\end{eqnarray}
So lets write out what $A - 1I$ is, namely the matrix
\begin{eqnarray}
A - 1I = 
\begin{pmatrix}
4 & 4 \\
4 & 4
\end{pmatrix}
\end{eqnarray}
Thus let $x= (a,b)$ be such that $(A - 1I)x = 0$ then
\begin{eqnarray}
x = 4
\begin{pmatrix}
a + b \\
a + b
\end{pmatrix}
\end{eqnarray}
so for this vector to be $0$ we must have that $b = - a$, so $x$ is a multiple of the vector $(1,-1)$. 
Now, try to do the same for $\lambda = 9$.
A: For the polynomial factorization, expand it all then factor.
$$\begin{align}&(5-\lambda)^2-16\\=&25-10\lambda+\lambda^2-16\\=&9-10\lambda+\lambda^2\\=&(1-\lambda)(9-\lambda)\end{align}$$
A: The two vectors are not correct.
The first: Since $A-I=4J$ where $J=\begin{pmatrix}1&1\\1&1\end{pmatrix}$ you have:
$$(A-I)x=0 \iff x_1+x_2=0$$
$\vec0$ is never an egen vector. An egen vector is by definition one vector $x$ such that $x \neq 0$ and $(A-I)x=0$
The second: Since $A-9I=4K$ where $K=\begin{pmatrix}-1&1\\1&-1\end{pmatrix}$:$$(A-9I)x=0 \iff x_1-x_2=0$$
A: Observe that the sum of each row is same so $A\begin{pmatrix}1\\1\end{pmatrix}=9\begin{pmatrix}1\\1\end{pmatrix}$. Hence $\lambda=9$ is an eigen value and $\begin{pmatrix}1\\1\end{pmatrix}$ is an eigen vector. To get the other eigen value, you may use determinant is product of eigen values (or trace is sum of eigen values). Hence the other eigen value is $\lambda=1$.
