# Complete matrix

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.) $\pmatrix{-1&-4&-4\\0&-1&0\\0&4&3}$

• 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 $\pmatrix{1\\1}$.

• 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 $\mathbb{C^2}?$

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(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 $\Bbb C$.

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Excellent answer, thanks a lot!!! Last question, for part c. for $\mathbb{C^n}$ how is it a scalar multiple of it? What will make a basis both in $\mathbb{R^n}$ and $\mathbb{C^n}$? Also, what will make the basis in $\mathbb{R^n}$ but not in $\mathbb{C^n}$ and vice-versa? – Q.matin Nov 21 '12 at 20:52
For $\Bbb C$ it depends how you are looking at your basis. If you are looking over the complex numbers or over the real numbers. The standard basis for $\Bbb R^n$ over $\Bbb R$ is the same as the standard basis for $\Bbb C^n$ over $\Bbb C$, i.e. since you can manufacture anything in $\Bbb C^2$ with $$x_1\pmatrix{0\\1}+x_2\pmatrix{0\\1}$$ where $x_1, x_2 \in \Bbb C$. Since $x_1$ and $x_2$ can be imaginary, we can produce any element in $\Bbb C^2$ with just those two basis vectors. – KUSH Nov 21 '12 at 22:18
If you need a basis of $\Bbb C^2$ over $\Bbb R$ you will also need vectors that provide $i$ to your mix. In this case you would need $$x_1\pmatrix{1\\0}+x_2\pmatrix{0\\1}+x_3\pmatrix{i\\0}+x_4\pmatrix{0\\i}$$ where each $x_k \in \Bbb R$. Now the first two are the "real basis" and the second two are the "imaginary basis". – KUSH Nov 21 '12 at 22:20
The "standard basis" for any field $F^n$ is usually considered to be over $F$, thus making it simply $$e_1=\pmatrix{1\\0\\0\\...}, e_2=\pmatrix{0\\1\\0\\...}, ... e_n=\pmatrix{...\\0\\0\\1}$$ – KUSH Nov 21 '12 at 22:32
You're very welcome! – KUSH Nov 21 '12 at 23:20