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seen Jul 23 '13 at 15:47

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.
Feb
22
accepted Depending on variable to calculate joint distribution
Feb
22
asked Depending on variable to calculate joint distribution
Feb
4
awarded  Tumbleweed
Nov
7
awarded  Yearling
Oct
31
awarded  Nice Question
Oct
31
awarded  Nice Answer
Jul
10
comment Maximal subspace that a quadratic form is non-negative on
@Geoff: why must $z=0$?
Jul
10
accepted Maximal subspace that a quadratic form is non-negative on
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
10
accepted if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$
Jul
10
asked Maximal subspace that a quadratic form is non-negative on
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 $q(v)=\langle v,v \rangle = \lVert v \lVert^2$?