Find the eigenvalues of the matrix and give the bases for each of the corresponding eigenspaces

Question

I'm having issues with this problem. I have solved for the eigenvalues but am having trouble finding the bases for both eigenvalues. The pictures below contain my work for solving for the eigenvalues and I solved for one of the basis but I think it is incorrect so I stopped. Can someone please help me. Thanks

Answer 1

I think I know where you made a mistake; when you calculated $A-(2+\sqrt{3}*i)I$ , you wrote that the term on the up-left was $(\sqrt{3}*i-1)$ whereas it was $(-\sqrt{3}*i-1) $, you could have tried anything from there it would have been confusing at best :).

I'll show you a practical way of dealing with this:

Calculate $B=A-(2+\sqrt{3}*i)I $

B = $\left( \begin{array}{ccc} {-1-i*\sqrt{3}}&2\\ -2&{1-i*\sqrt{3}}\\ \end{array}\right) $ = $ \left( \begin{array}{ccc} {-2*e^{\frac{i\pi}{3}}}&2\\ -2&{2*e^{\frac{-i\pi}{3}}}\\ \end{array}\right) $

$ -2*e^{\frac{i\pi}{3}}*e^{\frac{-i\pi}{3}} = -2*e^{0} = -2 $

$ -2*e^{\frac{-i\pi}{3}} = -(2*e^{\frac{-i\pi}{3}}) $

Let $C_1$ be the left column of B, $C_2$ the other. An eigenvector of A relatively to this eigenvalue is a vector of ker(B) by definition. So if this vector is written:

$ x_o = \begin{bmatrix} a \\ b \\ \end{bmatrix} $

Then you have : $aC_1 + bC_2 = B*x = 0$

$ e^{\frac{-i\pi}{3}}*C_1 = -C_2 $ from the two calculus above so an eigenvector for the eigenvalue 2+$i*\sqrt{3}$ is : $x_1 = \begin{bmatrix} e^{\frac{-i\pi}{3}} \\ 1 \\ \end{bmatrix}$

The same calculus goes for the other eigenvalue, with :

$ B = \left( \begin{array}{ccc} {-1+i*\sqrt{3}}&2\\ -2&{1+i*\sqrt{3}}\\ \end{array}\right) $ = $ \left( \begin{array}{ccc} {2*e^{\frac{2i\pi}{3}}}&2\\ -2&{2*e^{\frac{i\pi}{3}}}\\ \end{array}\right) $

you get : $x_2 = \begin{bmatrix} e^{\frac{i\pi}{3}} \\ 1 \\ \end{bmatrix} $

I think this technique is not very complicated, and quite useful because often you can see visually what will be the good combination that gives a$C_1$ +b$C_2$ =0 Trouble comes when the eigenvalue has a subspace of dimension 1 when the eigenvalue has an order of 2 in the characteristic polynomial for instance. In such cases, you can look for the same technic but using the transposed matrix and a null combination of the ligns of the matrix because of orthogonal spaces, especially when the size of the matrix is 3.

Answer 2

Your eigenvalues, $2 + i\sqrt{3}, 2 - i\sqrt{3}$ are correct. Now to solve the system in your picture above, multiply the 1st row by $\frac{1}{2}(-1-i\sqrt{3})$ and add it to the second row. This will turn the $-2$ in the second row to $0$ and you can then find your eigenvector(s).

Answer 3

The eigenvalues $\lambda_1=2+i\sqrt{3}$ and $\lambda_2=2-i\sqrt{3}$ are correct.

Now you have to solve $$ (A-\lambda_1I_2)x=0 $$ which is the system having matrix $$ \begin{bmatrix} 1-\lambda_1 & 2 \\ -2 & 3-\lambda_1 \end{bmatrix} = \begin{bmatrix} -1-i\sqrt{3} & 2 \\ -2 & 1-i\sqrt{3} \end{bmatrix} $$ Since you know this matrix has rank $1$, the only needed equation is $$ (-1-i\sqrt{3})x_1+2x_2=0 $$ so an eigenvector is $$ \begin{bmatrix} 2 \\ 1+i\sqrt{3} \end{bmatrix} $$ (set $x_1=2$ and compute $x_2$).

Do similarly for the other eigenvalue.