Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am given the $2\times2$ matrix $$A = \begin {bmatrix} -2&-1 \\\\ 15&6 \ \end{bmatrix}$$

I calculated the Eigenvalues to be 3 and 1. How do I find the vectors? If I plug the value back into the character matrix, I get $$B = \begin {bmatrix} -5&1 \\\\ 15&3 \ \end{bmatrix}$$

Am I doing this right? What would the eigenvector be?

share|improve this question

3 Answers 3

Almost right, only the $1$ in the upper right hand corner of $B$ should be a $-1$. Can you find the eigenvectors now?

share|improve this answer

Solve the equation $Bx = 0$. If $Bx=0$ then $(A-\lambda I)x = 0$, so $Ax = \lambda I x = \lambda x$, so $x$ is an eigenvector for $\lambda$.

share|improve this answer

Remember what the word "eigenvector" means. If $3$ is an eigenvalue, then you're looking for a vector satisfying this: $$A = \begin {bmatrix} -2&-1 \\\\ 15&6 \ \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} = 3\begin{bmatrix} x \\ y\end{bmatrix}$$

Solve that. You'll get infinitely many solutions since every scalar multiple of a solution is also a solution.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.