How to calculate eigenvectors from an identity matrix with eigenvalue $= 1$? When determining the eigenvectors from matrix $A$:
$$
      \begin{pmatrix}
      1 & 0 \\
      0 & 1 \\
      \end{pmatrix}
$$
I found the eigenvalue to be $ \lambda = 1 $ 
calculating  $ (A - 1*\lambda) $  gives me the matrix :
$$ \left[
    \begin{array}{cc|c}
      0&0&0\\
      0&0&0
    \end{array}
\right] $$
Which eigenvector would this produce?
 A: For $\lambda=1$, the Eigenvector equation is
$$0v=0.$$
This clearly means that any vector is a solution.

By the way, this can be found directly by identifying
$$Av=\lambda v$$ and $$Iv=v.$$
A: You made a small error when calculating your eigenvalues. If you solve for the the determinant equal to zero you end up with the following equation:
$(1-\lambda)(1-\lambda) = 0$.
Solving this you get both eigenvalues of $\lambda_1 = \lambda_2 = 1$. You can see from this how a diagonal matrix greatly simplifies your calculations. In a diagonal matrix the diagonal terms are your eigenvalues. Try recalculcating your eigenvectors with these eigenvalues. You will get $[0,\ 1]^T$ and $[1,\ 0]^T$.
Hope this helps.
A: This is fairly obvious, and can be solved with a bit of intuition without even touching an equation
The basis vector $\vec{i}$ equals $[1,0]^T$
and the basis vector $\vec{j}$ equals $[0,1]^T$.
This means this matrix
$(\vec i,\vec j)$ is the identity matrix. Nothing is changed
because no vector is changed; every (nonzero) vector is an eigenvector with an eigenvalue of $1$.
(Visual Intuition for what's going on)
