7
$\begingroup$

If I have a $2 \times 2$ matrix $A$, and then I find two eigenvalues $\lambda_1$ and $\lambda_2$ by subtracting $λI$ from $A$ and then taking the determinant=0(singular); to find $\lambda_1$ and $\lambda_2$.

So for a one eigenvalue $\lambda_1$, how many possibilities are there for eigenvectors? in another words, how many solutions are there?

  • 0
    If you now understand the situation, Binarylife, you can post an answer yourself; if no one finds anything wrong with your answer, then you can accept it. This will give you valuable practice in writing things up.2012-02-28

1 Answers 1

3

I hope you are asking for maximum possiblity of independent eigen vector: Below answer is for $2\times 2$ matrix.

We know that if $\lambda_1\neq\lambda_2$ then corresponding eigen vector will be independent. Below answer is based on this fact.

If $\lambda _1$ and $\lambda_2$ are different.... then there are only one independent eigen vector for corresponding eigen values.

If $\lambda_1$ and $\lambda_2$ are same then there may be two linear independent eigen vector.

  • 2
    @Binarylife , as other mention, if $v$ is eigen vector then $av$ will be eigen vector.. So for an eigen value, eigen vector is a vector space... Not just one... So if $(1,1)$ is eigen vector then $(2,2)= 2(1,1)$ will also be an eigen vector....2012-02-28