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.

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?

share|improve this question
    
There are at least as many solutions as there are nonzero elements in your field: if $\mathbf{v}$ is an eigenvector corresponding to $\lambda_1$, then so is $\alpha\mathbf{v}$ for every nonzero scalar $\alpha$. –  Arturo Magidin Feb 27 '12 at 18:12
1  
For a given eigenvalue, the set of possible eigenvectors is a vector space (technically, a vector space minus $\{0\}$) called the eigenspace. So if you're working on real or complex vector spaces (or over any infinite field), there's an infinite number of possible eigenvectors. What might be a better measure of the size of the eigenspace is its dimension. We know it's at least $1$, and that the sum of the dimensions of all eigenspaces is at most the size of the matrix (in your case $2$). –  Joel Cohen Feb 27 '12 at 18:13
    
so for eigenspace we can say that there is a whole line of eigenvectors? –  Binarylife Feb 27 '12 at 18:35
1  
@Binarylife: Not quite; the zero vector is not an eigenvector; and eigenspaces may have dimension greater than $1$, and so not be lines. But if $\lambda$ is an eigenvalue, then there is at least "a whole line, with the origin removed] of eigenvectors" corresponding to $\lambda$. –  Arturo Magidin Feb 27 '12 at 19:08
    
aha I see :), how about answering the question so i can accept it. –  Binarylife Feb 27 '12 at 19:35
show 1 more comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
    
So, If $\lambda_1$ = -1 and $\lambda_2$ = 2 , then I have only two corresponding eigenvectors which are {1,1} {5,2} , or there are other values maybe {2,2} instead of {1,1} and so on for the other eigenvalue? –  Binarylife Feb 27 '12 at 19:34
1  
@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.... –  zapkm Feb 28 '12 at 2:29
add comment

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.