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Problem: Consider two matrices $A, B \in \mathbb{R}^{3 \times 3}$. Suppose $A$ has three distinct real eigenvalues $\lambda_1, \lambda_2$ and $\lambda_3$ with respective eigenspaces $E_{\lambda_1}, E_{\lambda_2}$ and $E_{\lambda_3}$. Suppose furthermore that $B$ has two distinct real eigenvalues $\mu_1$ and $\mu_2$ with respective eigenspaces $E_{\mu_1} = \text{span}(E_{\lambda_1}, E_{\lambda_2})$ (the space spanned by $E_{\lambda_1}$ and $E_{\lambda_2}$) and $E_{\mu_2} = E_{\lambda_3}$.

1) Determine the eigenvalues and corresponding eigenspaces of $AB$.

2) Show that $AB = BA$.

Attempt at solution: I have no idea how to do this. I tried writing $\det(A - x \mathbb{I}_3) = (-1)^3 (x- \lambda_1) (x- \lambda_2) (x- \lambda_3)$ and then using the fact that $\det(AB) = \det(A) \det(B)$. But then I figured that the characteristic equation doesn't necessarily have to split like that ?

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  • $\begingroup$ You mean that $A$ has those three distinct eigenvalues counting with their multiplicities and no others (and same with $B$)? $\endgroup$ – A.Γ. Jul 17 '15 at 19:56
  • $\begingroup$ $A$ has the three distinct eigenvalues (all with multiplicity $1$), and $B$ has two real distinct eigenvalues (both multiplicity $1$). $\endgroup$ – Kamil Jul 17 '15 at 20:07
  • $\begingroup$ Then $A$ would be $3\times 3$ and $B$ $2\times 2$. There must be either multiplicity involved or we just do not know anything about other eigenvalues (the latter will hardly give commutation). $\endgroup$ – A.Γ. Jul 17 '15 at 20:32
  • $\begingroup$ isn't it that one of B's eigenspaces has multiplicity $2$? Considering Mike F's answer, perhaps it's implied that both A and B are $3×3$... $\endgroup$ – Dleep Jul 17 '15 at 20:39
  • $\begingroup$ I guess we don't know anything about the third eigenvalue of $B$ (if there is one). But it has two distinct real eigenvalues. The matrices $A$ and $B$ are $(3 \times 3)$-matrices. $\endgroup$ – Kamil Jul 17 '15 at 20:44
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There is a basis $\{ e_1,e_2,e_3 \}$ of eigenvectors of $A$ because there are three distinct eigenvalues for the $3\times 3$ matrix $A$, and non-zero eigenvectors corresponding to distinct eigenvalues form a linearly-independent set. These eigenvectors are also eigenvectors of $B$. So both $A$ and $B$ are diagonalizable with respect to this basis. In this basis, $$ A = \left[\begin{array}{ccc}\lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3\end{array}\right] \;\;\; B = \left[\begin{array}{ccc}\mu_1 & 0 & 0 \\ 0 & \mu_1 & 0 \\ 0 & 0 & \mu_2\end{array}\right] $$

Therefore, $A$ and $B$ commute in this basis system, which means they commute, i.e., $AB=BA$.

Let $e_j$ be a non-zero vector in $E_{\lambda_j}$ for $j=1,2,3$. Then $\{ e_1,e_2,e_3 \}$ satisfy $$ \begin{array}{cc} Ae_1 = \lambda_1 e_1 & Be_1 = \mu_1 e_1,\\ Ae_2 = \lambda_2 e_2 & Be_2 = \mu_1 e_2, \\ Ae_3 = \lambda_3 e_3 & Be_3 = \mu_2 e_3. \end{array} $$ That's enough to work out $AB$ acting on this basis $\{e_1,e_2,e_3\}$: $$ ABe_1 = A(\mu_1 e_1) = \mu_1 Ae_1 = \mu_1 \lambda_1 e_1 \\ ABe_2 = A(\mu_1 e_2) = \mu_1 Ae_2 = \mu_1 \lambda_2 e_2 \\ ABe_3 = A(\mu_2 e_3) = \mu_2 Ae_3 = \mu_2\lambda_3 e_3 $$ This also follows by multiplying the diagonal matrix representations of $A$ and $B$.

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  • $\begingroup$ How do you know that $\left\{ e_1, e_2, e_3\right\}$ forms also a basis of eigenvectors for $B$? Because $E_{\mu_2} = E_{\lambda_3}$ and $E_{\mu_1} = \text{span}(E_{\lambda_1}, E_{\lambda_2})$ ? $\endgroup$ – Kamil Jul 17 '15 at 21:22
  • $\begingroup$ Right. Since $e_1 \in E_{\lambda_1}$ we have $e_1 \in E_{\mu_1}$, so $Be_1 = \mu_1e_1$. Since $e_2 \in E_{\lambda_2}$, $e_2 \in E_{\mu_1}$ and so $Be_2 = \mu_1e_2$. And lastly $e_3 \in E_{\lambda_3}$, so $e_3 \in E_{\mu_2}$, and $Be_3 = \mu_2e_3$. $\endgroup$ – Alex Zorn Jul 17 '15 at 21:28
  • $\begingroup$ The vectors are a basis because they are eigenvectors of $A$ with distinct eigenvalues. So, if $\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3 = 0$, you can apply $(A-\lambda_1)(A-\lambda_2)(\alpha_1e_1+\alpha_2 e_2+\alpha_3 e_3)=(\lambda_3-\lambda_1)(\lambda_3-\lambda_2)\alpha_3 e_3$ which forces $\alpha_3 = 0$. You can conclude $\alpha_j = 0$ for all $j$; hence the (non-zero) eigenvectors are linearly independent. $\endgroup$ – DisintegratingByParts Jul 17 '15 at 21:30
  • $\begingroup$ I see, thanks. Now that we've found the eigenvalues corresponding to $AB$, what about the eigenspaces? Would the eigenspace of $E_{\mu_1 \lambda_1}$ just be the union of $E_{\mu_1}$ and $E_{\lambda_1}$ ? $\endgroup$ – Kamil Jul 17 '15 at 22:13
  • $\begingroup$ @Kamil : $E_{\mu_1\lambda_1}$ is spanned by $e_1$. You have to intersect eigenspaces in order to divide it into spaces where both $A$ and $B$ are constant multiplies of the identity, thereby guaranteeing that $AB$ is constant. In this case $E_{\mu_1\lambda_1}=E_{\lambda_1}\cap E_{\mu_1}$. Some of the spaces may recombine because more than one product of eigenvalues can be equal, e.g., $\frac{1}{2}\times 2, 1\times 1$, etc. $\endgroup$ – DisintegratingByParts Jul 17 '15 at 22:20

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