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?

  • $\begingroup$ 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 $\alpha$. $\endgroup$ – Arturo Magidin Feb 27 '12 at 18:12
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    $\begingroup$ 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 $2$). $\endgroup$ – Joel Cohen Feb 27 '12 at 18:13
  • $\begingroup$ so for eigenspace we can say that there is a whole line of eigenvectors? $\endgroup$ – Binarylife Feb 27 '12 at 18:35
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    $\begingroup$ @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 $\lambda$. $\endgroup$ – Arturo Magidin Feb 27 '12 at 19:08
  • $\begingroup$ 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. $\endgroup$ – Gerry Myerson Feb 28 '12 at 0:02

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.

  • $\begingroup$ 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? $\endgroup$ – Binarylife Feb 27 '12 at 19:34
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    $\begingroup$ @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.... $\endgroup$ – zapkm Feb 28 '12 at 2:29

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