Diagonalization and change of basis

Question

My question arises from a quantum mechanical problem about perturbation theory, but it is of algebraic nature. I'm a bit forgotten about some basic notions of algebra and I need help in the following:

I have the following matrix, written on a certain basis $\left|1\right\rangle$ and $\left|2\right\rangle$:

$$\begin{bmatrix} E_0 &-A \\ -A & E_0 \end{bmatrix}$$

Eventually I found the matrix eigenvalues $E_I=E_0-A$ and $E_{II}=E_0+A$ and eigenvectors $\left|I\right\rangle = \begin{bmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{bmatrix}$ and $\left|II\right\rangle=\begin{bmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} \end{bmatrix}$. I found out in the solutions of further problems that I can write these vectors as $\left|I\right\rangle=\frac{1}{\sqrt{2}}\left|1\right\rangle+\frac{1}{\sqrt{2}}\left|2\right\rangle$ and $\left|II\right\rangle=\frac{1}{\sqrt{2}}\left|1\right\rangle-\frac{1}{\sqrt{2}}\left|2\right\rangle$. Why is that? Thank you in advance

Answer 1

Because$$|I\rangle=\begin{bmatrix}\frac1{\sqrt{2}}\\\frac1{\sqrt{2}}\end{bmatrix}=\frac1{\sqrt2}\begin{bmatrix}1\\0\end{bmatrix}+\frac1{\sqrt2}\begin{bmatrix}0\\1\end{bmatrix}=\frac1{\sqrt2}|1\rangle+\frac1{\sqrt2}|2\rangle$$and$$|II\rangle=\begin{bmatrix}\frac1{\sqrt{2}}\\-\frac1{\sqrt{2}}\end{bmatrix}=\frac1{\sqrt2}\begin{bmatrix}1\\0\end{bmatrix}-\frac1{\sqrt2}\begin{bmatrix}0\\1\end{bmatrix}=\frac1{\sqrt2}|1\rangle-\frac1{\sqrt2}|2\rangle.$$

Answer 2

The coordinates of a vector relative to some basis are just the coefficients of the corresponding linear combination of the basis vectors. Taking your first eigenvector, its coordinates relative to the basis are $\frac1{\sqrt2}$ and $\frac1{\sqrt2}$. This means by definition that $$|I\rangle = \frac1{\sqrt2}|1\rangle+\frac1{\sqrt2}|2\rangle.$$