daniel.jackson
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 Apr17 comment Maximizing rectangle area without intersecting others Is it possible for you to add a picture illustrating this? Apr17 comment Maximizing rectangle area without intersecting others Looking at the example above, if there's a rectangle E that is above R and to the left of it then it will never intersect with R but you are still taking it in account when finding M. Apr4 comment Given an infinite bounded set A in $R^n$,$2\leq n$, show there are infinite boundary points Why is there a nonzero vector that is orthogonal to L, and why do you need it to be orthogonal? Feb22 comment Depending on variable to calculate joint distribution Now that you mention it, of course they're not! Thanks. Jul10 comment Maximal subspace that a quadratic form is non-negative on @Geoff: why must $z=0$? Jul10 comment Maximal subspace that a quadratic form is non-negative on I must say I didn't completely understand your proof there. Could you point out how to finish the proof I started? Jul6 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)? Jul6 comment if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$ Why must it have a nonreal eigenvalue? Jul6 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 $A$. Jul6 comment if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$ @Chandru, not at all, thank you. Jul5 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. Jul5 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 $q(v)=\langle v,v \rangle = \lVert v \lVert^2$? Jul4 comment Confused about quadratic forms @wild: Care to elaborate? Jun18 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? Jun17 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? Jun17 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. Jun17 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. Jun4 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? Jun4 comment Non diagonalizable matrix @Aaron: by Cayley-Hamilton, $Q$ is the characteristic polynomial of $A$, but I'm not sure I followed the rest of your suggestion. Jun4 comment Non diagonalizable matrix @Mark: that makes sense, but where do I go from there?