1
$\begingroup$

Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation such that all it's eigenvalues are $\lambda_1$ and $\lambda_2$ and the eigenvectors are $v_1, v_2$ (corresponding with $\lambda_1$) and $v_3$ (corresponding with $\lambda_2$) . Find all the T invariant subspaces of $\mathbb{R}^3$.

This question helped me to understand invariant subspace if every eigenvalue are associated with only one eigenvector. But if one eigenvalue are related with two vectors?

I think that the answer is related with my other question: Are eigenvectors related with same eigenvalue always linearly independent?

$\endgroup$
0
$\begingroup$

Luckily, here we have as many eigenvectors as the dimension of the space. (I presume that, as is common, when someone says that "the eigenvectors are $v_1$, $v_2$", they mean that these are linearly independent eigenvectors.) Since we have enough eigenvectors, they form a basis for the entire space. And in particular, each invariant subspace is the span of the corresponding eigenvectors.

As to your second question, the answer is NO. But… As I pointed out above, people often slightly abuse the language.

  1. If $v$ is an eigenvector, then e.g. $2v$ is an eigenvector corresponding to the same eigenvalue, and so is any scalar multiple of $v$. So if you know that several vectors are eigenvectors corresponding to the same eigenvalue, you can't conclude that they are independent.

  2. But when describing eigenvalues and eigenvectors, it is a common figure of speach to just say something like "Let $v_1,v_2$ be the eigenvectors corresponding to an eigenvalue $\lambda$." This is an inaccurate (but quite common!) statement, where it is silently assumed that these eigenvectors are linearly independent. The purpose of such a phrase is literally to state that the invariant subspace is two dimensional, and these two vectors form one of its possible bases.

$\endgroup$

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.