daniel.jackson
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 Jul 6 comment if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$ I see. Could you be so kind to provide a proof (a general idea will be fine too)? Jul 6 comment if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$ Why must it have a nonreal eigenvalue? Jul 6 comment if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$ So if $A^2$ had a negative eigenvalue, say $-1$, the same argument would not work? Because then it's possible that $A$ characteristic polynomial has $i$ as one of its roots, which is not an eigenvalue of $A$. Jul 6 comment if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$ @Chandru, not at all, thank you. Jul 6 asked if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$ Jul 5 comment Confused about quadratic forms I don't understand why this is true. By orthogonal diagonalization you can bring $q$ to the form $\sum \lambda_i x_i^2$, how did you conclude that $\lambda_i=1$ in this case? I know that there is a basis where $q(x)=\sum x_i^2$ but why is it necessarily orthogonal? Wikipedia says something similar too. Jul 5 comment Confused about quadratic forms You say the norm defined by q, but I'm not sure I understand that. Does that mean that if $f$ is the associated symmetric bilinear form then $f$ also defines an inner product in $R^n$ such that $f(u,v)=\langle u,v \rangle$ and $q(v)=\langle v,v \rangle = \lVert v \lVert^2$? Jul 4 accepted Confused about quadratic forms Jul 4 comment Confused about quadratic forms @wild: Care to elaborate? Jul 4 asked Confused about quadratic forms Jun 23 accepted Finding $a_n$ using a given matrix Jun 23 accepted If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator Jun 23 asked Finding $a_n$ using a given matrix Jun 18 comment If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator Proof for lemma: let $u \in U^\perp$, then $=0,\ \forall v \in U$. Since U is invariant, $Tv \in U$ as well, hence $==0$, therefore $T^*u \in U^\perp$. But how does this help? Jun 17 comment If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator Ok, thanks for trying. So according to your proof, $T$ and $T^*$ can be represented by these diagonal matrices: $\text{diag}(a_{11},\ldots,a_{nn}),\ \text{diag}(\overline{a_{11}},\ldots,\overline{a_{nn}})$, and since diagonal matrices commute, so does the transformations? Jun 17 comment If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator That's a very nice proof, thanks. Out of curiosity, can you think of one that uses Jordan forms? I'm trying to think how this relates to the material I just read. Jun 17 comment If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator @Jonas: Yes to the last question. Never heard the term 'unitarily triangularized' though. Jun 17 asked If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator Jun 4 comment Non diagonalizable matrix I think I understand the example but could you also add a small intuitive explanation as to what's going on here? Jun 4 accepted Non diagonalizable matrix