Eigenvectors of the matrix I'm getting frustrated with a question. I'm trying to find the eigenvector of $\left( \begin{array}{cc} 1/2 & 0 \\ 0& 2\end{array} \right)$ and it's been a few good years since I've had to touch eigenvectors.
I get the characteristic polynomial is $\left( \begin{array}{cc} 1/2-\lambda & 0 \\ 0& 2-\lambda \end{array} \right)$, and solving this gives me $2$ eigenvalues of $\lambda = 1/2$ and $\lambda = 2$.
Solving for $\lambda = 1/2$, I get the matrix $\left( \begin{array}{cc} 0 & 0 \\ 0& 3/2\end{array} \right)$
When solving the above matrix, I get a system of equations:
$$\left\lbrace\begin{array}{l}  0x + 0y = 0
\\  0x + 3/2y = 0 \end{array} \right.$$
And thus solving this yields me $x = 0$, and $y = 0$ which is incorrect. The correct answer is $(1,0)$ for $\lambda = 1/2$.
Where am I going wrong?
Thanks.
 A: $$\begin{cases} 0x+0y=0 \\ 0x+\frac 32y=0\end{cases} \iff \begin{cases} 0=0 \\ \frac 32y=0\end{cases}$$
That first equation doesn't give us any information.  It's simply a tautology (something of the form $a=a$).  Solving the second equation however, we see that $y=0$. But you can see that there aren't any conditions on the value of $x$.  So the solutions to this matrix equation are $$\pmatrix{x \\ y} = \pmatrix{t \\ 0}$$ for any arbitrary real number $t$.  But while there are an infinite number of solutions, they're all just scalar multiples of each other so we generally just choose one of them and call it "the" eigenvector.  A good choice might be $t=1$, but you can choose any other value you like.  Then you'll get your eigenvector $\pmatrix {1 \\ 0}$.
A: The condition
$$
\frac{3}{2}y=0
$$
gives $y=0$. However, there are no more conditions and so $x$ may take any real value.
A: are you trying to find the eigenvalues and eigenvectors of the diagonal matrix $$A = \pmatrix{1/2&0\\0&2}? $$ we have $$Ae_1 = \frac 1 2 e_1, Ae_2 = 2e_2 \text{ where } e_1 \ \pmatrix{1\\0}, e_2=\pmatrix{0\\1}.\tag 2$$ the equation already tells you that the eigenvalues of $A$ are $1/2, 2.$
