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seen Jul 23 '13 at 15:47

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 $q(v)=\langle v,v \rangle = \lVert v \lVert^2$?
Jul
4
accepted Confused about quadratic forms
Jul
4
comment Confused about quadratic forms
@wild: Care to elaborate?