daniel.jackson
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 Apr16 asked Maximizing rectangle area without intersecting others 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. Feb22 accepted Depending on variable to calculate joint distribution Feb22 asked Depending on variable to calculate joint distribution Feb4 awarded Tumbleweed Nov7 awarded Yearling Oct31 awarded Nice Question Oct31 awarded Nice Answer Jul10 comment Maximal subspace that a quadratic form is non-negative on @Geoff: why must $z=0$? Jul10 accepted Maximal subspace that a quadratic form is non-negative on 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? Jul10 accepted if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$ Jul10 asked Maximal subspace that a quadratic form is non-negative on 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. Jul6 asked if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$ 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.