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seen Jul 23 '13 at 15:47

Apr
17
comment Maximizing rectangle area without intersecting others
Is it possible for you to add a picture illustrating this?
Apr
17
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.
Apr
4
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?
Feb
22
comment Depending on variable to calculate joint distribution
Now that you mention it, of course they're not! Thanks.
Jul
10
comment Maximal subspace that a quadratic form is non-negative on
@Geoff: why must $z=0$?
Jul
10
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?
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
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
comment Confused about quadratic forms
@wild: Care to elaborate?
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 $<u, v>=0,\ \forall v \in U$. Since U is invariant, $Tv \in U$ as well, hence $<u, Tv>=<T^*u, v>=0$, 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
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
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.
Jun
4
comment Non diagonalizable matrix
@Mark: that makes sense, but where do I go from there?