Which of the following are complete eigenvalues (by complete,
corresponding eigenspace has the same dimension as its multiplicity)
for the indicated matrix? What is the dimension of the associated
eigenspace?a.) $\pmatrix{2&1\\1&2}$
b.) $\pmatrix{-1&-4&-4\\0&-1&0\\0&4&3}$
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For a, I got the eigenvalues of $\lambda = 3,1$ and the corresponding eigenvectors are $v = \pmatrix{1\\1}$, $v_1 = \pmatrix{1\\-1}$ but the answer says it is complete and has $dim = 1$ with basis of $\pmatrix{1\\1}$.
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For b, I got the eigenvalues of $\lambda = -1,-1,3$, its corresponding eigenvectors is $\pmatrix{0\\-1\\1}$ for $\lambda = -1$ and $\pmatrix{1\\0\\1}$ for $\lambda = 3$ but then the $dim = 2$ with basis $\pmatrix{1\\0\\0}$, $\pmatrix{0\\-1\\1}$ and it is complete. How did they get that?
Also, why does $\pmatrix{1&3\\3&1}$ (which has eigenvalues of -2,4 and
eigenvectors $\pmatrix{-1\\1}$, $\pmatrix{1\\1}$) form a basis in
$\mathbb{R^2}$ and in $\mathbb{C^2}?$
Best Answer
(A) Let me check what you're doing for eigenvals/eigenvecs...
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
Which implies completeness.
(B) Probably similar confusion. Again correct, $\lambda_1=\lambda_2=-1$ and $\lambda_3=3$.
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
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 $\Bbb C$.