1. Eigenvectors!

(i) Determine all the eigenvalues of A
(ii) For each eigenvalue of A, find the set of eigenvectors corresponding to

A = ( 1 2 )
( 3 2 )

I found eigenvalues of A to be 4 and -1.

I also found the eigenvectors to be (2/3,1) for =4 and (-1,1) for =-1

BUT the solution in the back of the book says (2,3) for =4 and (1,-1) for =1

I'm soo confusedd! Can someone tell me what's wrongg? Am I calculating the eigenvectors wrong?

Here's how i calculate eigenvector for =4

A-4I = (-3 2)
(3 -2)

then (-3 2 |0)
(3 -2 |0)

and i end up with x1 -(2/3) x2 =0. so x2 = t, and x1 = (2/3)t. eigenvector = (2/3,1) but book says (2,3)

and for =-1

A-(-1)I = (2 3 )
(2 3 )

then (2 3 |0)
(2 3 |0)

and i end up with x1+x2=0. x2= t and x1 = -t. eigenvector = (-1,1) but book says (1,-1)

2. You don't have a problem. Note that your eigenvectors are just multiples of the book's answer. This is fine!

3. Hi
$\texttt{det}(A-I_{2}X)=\texttt{det}\begin{pmatrix}
1-X & 2\\
3 & 2-X
\end{pmatrix}=(1-X)(2-X)-6=X^2-3X-4=(X-4)(X+1)$

and thus your eigenvalues are 4 and -1.
For 4
$A\begin{pmatrix}
x\\
y
\end{pmatrix}=4\begin{pmatrix}
x\\
y
\end{pmatrix}\Leftrightarrow \begin{pmatrix}
x+2y\\
3x+2y
\end{pmatrix}=$
$\begin{pmatrix}
4x\\
4y
\end{pmatrix}\Leftrightarrow \left\{\begin{matrix}
-3x+2y=0\\
3x-2y=0
\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x=\frac{2}{3}t\\
y=t,t\in \mathbb{R}
\end{matrix}\right.$

and thus the eigenvectors are of the form,
$x=x_1(\frac{2}{3},1),x_1\in \mathbb{R}\setminus \left \{ 0 \right \}$

4. The set of all eigenvectors of a matrix A, corresponding to a given eigenvalue, $\lambda$, form a subspace. In particular, any multiple of an eigenvector is also an eigenvector, corresponding to the same eigenvalue: If $Av= \lambda v$ and "r" is any scalar, then $A(rv)= r(Av)= r(\lambda v)= (r\lambda)v= \lambda (rv)$.