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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
16
asked Maximizing rectangle area without intersecting others
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 $A$