1
$\begingroup$

This question already has an answer here:

Let $A,B\in \mathbf{R}^{3\times 3}$. Let $A$ have 3 distinct eigenvalues $\lambda_1$, $\lambda_2$, $\lambda_3$ with corresponding eigenspaces $E_{\lambda_1}$, $E_{\lambda_2}$, $E_{\lambda_3}$. Let $B$ have 2 distinct eigenvalues $\mu_1$, $\mu_2$ with corresponding eigenspaces $E_{\mu_1}=\operatorname{span}(E_{\lambda_1}+E_{\lambda_2})$ and $E_{\mu_2}=E_{\lambda_3}$.

(a) Determine the eigenvalues and eigenvectors of $AB$.

(b) Prove that $AB=BA$.

I don't understand what my book means by "$\operatorname{span}(E_{\lambda_1}+E_{\lambda_2})$", is this simply the sum $E_{\lambda_1}+E_{\lambda_2}$ or something more complicated?

For part (a). I took a vector $v\in E_{\mu_2}$, then $v\in E_{\lambda_3}$ so $(AB)v=A(Bv)=A(\mu_2 v)=\mu_2 AV=\mu_2 \lambda_3 v$ so $\lambda_3 \mu_2$ is an eigenvalue (corresponding to eigenvector $v\in E_{\mu_2}=E_{\lambda_3}$?).

Similarly, take a vector $E_{\lambda_1}$, then (about this step I am not sure) $w\in E_{\mu_1}$. Therefore $(AB)w=A(Bw)=A(\mu_1 w)=\mu_1 (Aw)=\mu_1\lambda_1 w$ so $\lambda_1 \mu_1$ is an eigenvalue (corresponding to what eigenvector? Since $w$ must lie in $E_{\lambda_1}$ and $E_{\mu_1}$ I would say $E_{\lambda_1} \cap E_{\mu_1}$, but is this correct?

For part (b), I know that simultaneously diagonalizable matrices commute, so I need to show that $A$ and $B$ have the same basis of eigenvectors, which again is trivial if I know what my book means by the $\operatorname{span}$-thing.

$\endgroup$

marked as duplicate by Rolf Hoyer, Mark Bennet, Siong Thye Goh, Paramanand Singh, mechanodroid Nov 10 '17 at 21:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

The span a vector space is the set of every linear combination of its vectors. Basically, it is the set containing every vector in the space. So span$(E_{\lambda_1} + E_{\lambda_2})$ is basically the formal way of referring to all the vectors that are linear combinations of vectors in $E_{\lambda_1} + E_{\lambda_2}$.

Hence, span$(E_{\lambda_1} + E_{\lambda_2}) = \{ax + by \space | \space x \in E_{\lambda_1} \space \space y \in E_{\lambda_2} \space \space a,b \in \mathbf{R} \} $

This question has already been asked: if you want the full proof, check out Eigenvalues and eigenspaces of AB.

$\endgroup$
  • $\begingroup$ Thanks for your answer! So $\operatorname{span}(E_{\lambda_1}+E_{\lambda_2})$ contains all vectors $x+y$ with $x\in E_{\lambda_1}$ and $y\in E_{\lambda_2}$? $\endgroup$ – Heinz Doofenschmirtz Nov 9 '17 at 16:25
  • $\begingroup$ Yes. More precisely, it contains every vector $ax + by$ with $x \in E_{\lambda_1}$ and $y \in E_{\lambda_2}$ and $ a, b \in \mathbf{R} $. (This is the same set you as the one you described (since vector spaces are closed.) $\endgroup$ – helper Nov 9 '17 at 21:08

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