0
$\begingroup$

$n \times n$ matrix with distinct eigenvalues $\lambda_1, \lambda_2, ... \lambda_k$ and corresponding eigenspace $E_1, E_2, ..., E_k$.

Have $\vec v_{1,1}, \vec v_{1,2}, ... , \vec v_{1,m_1}$ be a basis for $E_1$, and $\vec v_{k,1}, \vec v_{k,2}, ... , \vec v_{k,m_k} $ be a basis for $E_k$, etc.

How to show that all of the basis vectors (i.e., $\vec v_{1,1}, \vec v_{1,2}, ... , \vec v_{1,m_1}, ..., \vec v_{k,1}, \vec v_{k,2}, ... , \vec v_{k,m_k}$) are linearly independent?

I know that a linear combination of any eigenvectors with the same eigenvalue will result in another eigenvector for that same eigenvalue. I also know that eigenvectors corresponding to different eigenvalues are linearly independent.

$\endgroup$
  • $\begingroup$ Then you have it, except perhaps for a matter of notation. When you take a linear combination of all the vectors, group the terms with $\vec v_{i,j}$ into one vector $\vec w_i$. How can $\vec w_1+\dots+\vec w_k=\vec 0$? $\endgroup$ – Ted Shifrin Mar 9 '16 at 5:16
  • $\begingroup$ @TedShifrin Ok, I see now that would mean $c_1 \vec w_1 + \cdots + c_k \vec w_k = \vec 0$ only when $c_1 = \cdots = c_k = 0$. Then expanding the $\vec w_i$ vectors back out to linear combinations of $\vec v_{i,j}$ vectors would mean all coefficients must be $0$, indicating linear independence. Thank you. $\endgroup$ – Tim Mar 9 '16 at 5:28
  • $\begingroup$ Well, actually, the $\vec w_i$ will all be $\vec 0$ themselves in this case. $\endgroup$ – Ted Shifrin Mar 9 '16 at 5:30
  • $\begingroup$ @TedShifrin Oh. I see how the $\vec w_i$ would all be $\vec 0$. But I am having trouble going from this to proving that every basis vector of all the eigenspaces are linearly independent. $\endgroup$ – Tim Mar 9 '16 at 5:45
  • $\begingroup$ Remember that $\vec w_1=c_{11}\vec v_{1,1}+\dots+c_{1m_1}\vec v_{1,m_1}$, etc. If this is $\vec 0$, what do you conclude? $\endgroup$ – Ted Shifrin Mar 9 '16 at 5:49
1
$\begingroup$

Suppose $A$ is the matrix and let $v_i \in E_i$ and suppose $v=\sum_i \alpha_i v_i = 0$.

Choose some index $j_0$. Then $\prod_{j \neq j_0} (A-\lambda_{j} I) v = \prod_{j \neq j_0} (\lambda_{j_0}-\lambda_{j} ) \alpha_{j_0}v_{j_0} = 0$ and so $\alpha_{j_0} = 0$. Since $j_0$ was arbitrary, we see that $\alpha_i = 0$ for all $i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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