Row reducing matrices to obtain the eigenvector

Find all distinct (real or complex) eigenvalues of $$A$$. Then find the basic eigenvectors of $$A$$ corresponding to each eigenvalue. For each eigenvalue, specify the number of basic eigenvectors corresponding to that eigenvalue.

$$A = \begin{bmatrix}-16 & -10\\17& 10\end{bmatrix}$$

So $$A -\lambda I = \begin{bmatrix}-16-\lambda & -10\\17& 10-\lambda\end{bmatrix}$$

Which means $$det(A-\lambda I) = \lambda^2 + 6\lambda - 10$$

Therefore, using the quadratic formula, the eigenvalues should be:

$$\lambda = -3\pm \sqrt(19)i$$

So now I get the two eigenvectors which are:

$$B = \begin{bmatrix}-13 - \sqrt19i& -10\\17& 13 - \sqrt19i\end{bmatrix}$$

$$C = \begin{bmatrix}-13 + \sqrt19i& -10\\17& 13 + \sqrt19i\end{bmatrix}$$

Now the part I am struggling is finding out how to row reduce these matrices in order to obtain the answer:

For example, with $$B$$:

$$Bx = 0$$

$$Bx = \begin{bmatrix}-13 - \sqrt19i& -10\\17& 13 - \sqrt19i\end{bmatrix} \begin{bmatrix}x1\\x2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$$

I know I have to do row operations to first get zeroes in the bottom row, and then to get a 1 in the first row/column, but I don't know how. This requires dealing with complex numbers, and I am not sure how to do it in this scenario. Could someone give me a detailed walkthrough on how to do this?

• Eigenvectors are vectors, not matrices. Your $B$ and $C$ are the matrices $A-\lambda I$, not the eigenvectors of $A$. – amd Nov 30 '18 at 2:44

$$(\lambda - \lambda_1)(\lambda - \lambda_2) = \lambda^2-(\lambda_1+\lambda_2)\lambda+\lambda_1\lambda_2=\lambda^2-trace(A)\lambda +\det(A)$$

$$trace(A)=-16+10=-6$$

$$\det(A)=-160+170=10$$

The characteristic polynomial is $$\lambda^2+6\lambda \color{red}{+} 10$$.

$$\lambda = \frac{-6 \pm\sqrt{6^2-40}}{2}$$

Upon trying to solve for the equation $$(A-\lambda I)v=0$$

If $$\lambda$$ is indeed an eigenvalue, then the matrix is rank $$1$$ for this $$2 \times 2$$ matrix. You can just focus on a non-zero row and ignroe the other constraint as it doesn't give you extra information.