$n \times n$ 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.

  • $\begingroup$ 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

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 $i$.


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