(A) Let me check what you're doing for eigenvals/eigenvecs...
- $det\pmatrix{{2-\lambda}&1\\1&{2-\lambda}}=(2-\lambda)^2-1=\lambda^2-4\lambda+3=(\lambda-3)(\lambda-1)=0$. So you are correct, $\lambda_1=1$ and $\lambda_2=3$.
- $M-\lambda_1I_2=\pmatrix{{1}&1\\1&{1}}$ which leads to this equation $x_1+x_2=0$ for both which means this eigenvector $v_1=\pmatrix{1\\-1}$.
- $M-\lambda_2 I_2=\pmatrix{{-1}&1\\1&{-1}}$ which leads to the equation $x_1-x_2=0$ for both which means this eigenvector $v_2=\pmatrix{1\\1}$. Everything you did is correct.
Perhaps what you are confused on is associating the eigenvector with an eigenspace. The dimension of $v_1$ and $v_2$ are both $1$ which corresponds to the multiplicity of $\lambda_1$ and $\lambda_2$ which are both $1$ $\Rightarrow$ complete.
Summary: Correspondence
- $dim(v_1)=1$ and $mult(\lambda_1)=1$
- $dim(v_2)=1$ and $mult(\lambda_2)=1$
Which implies completeness.
(B) Probably similar confusion. Again correct, $\lambda_1=\lambda_2=-1$ and $\lambda_3=3$.
- $M-\lambda_1 I_3=\pmatrix{0&-4&-4\\0&0&0\\0&4&4}$ which leads to the system of equations $x_2+x_3=0$. But you are forgetting that $x_1$ can be anything. So your eigenvectors for $\lambda_1$ are $\pmatrix{0\\-1\\1}$ AND $\pmatrix{1\\0\\0}$. So the dimension of the eigenspace is $2$ (the span of $\pmatrix{0\\-1\\1}, \pmatrix{1\\0\\0}$) which corresponds to the multiplicity of $\lambda_1$ which is $2$ since it appears twice.
Also you are correct for $dim (v_3)=1$ corresponding to $mult (\lambda_3)=1$ $\Rightarrow$ complete.
If you want to do a little more study on evals, evecs and espaces, check this link out.
(C) Your eigenvecs are $\pmatrix{-1\\1}$ and $\pmatrix{1\\1}$. If you can show that the standard basis for $\Bbb R^2$ which is $\pmatrix{1\\0}$ and $\pmatrix{0\\1}$ is within your eigenvecs then you show that it is a basis. Notice
- $\frac 1 2 \pmatrix{-1\\1}+\frac 1 2 \pmatrix{1\\1} = \pmatrix{0\\1}$
- $\frac {-1} 2 \pmatrix{-1\\1}+\frac 1 2 \pmatrix{1\\1} = \pmatrix{1\\0}$
So since you can form the standard basis using just a linear combination of your eigenvectors, these eigenvectors also serve as a basis for $\Bbb R^2$.
$\Bbb C^2$ is basically the same, as you can have your scalars be in $\Bbb C$.