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Apr
28
asked Essentially bounded function which is continuously bounded Riemann integrable?
Apr
28
comment If f*g is Riemann integrable, g continuous, nonzero and bounded, show that f is Riemann integrable
@Arthur Upvoted, you beat me to the answer!
Apr
28
answered If f*g is Riemann integrable, g continuous, nonzero and bounded, show that f is Riemann integrable
Apr
28
revised Matrices Eqvialence Relation
edited body
Apr
28
comment Matrices Eqvialence Relation
Yes, thanks. Will change it
Apr
28
reviewed Approve For what values of $a$ will $y=ax$ be a tangent to $x^2+y^2+20x-10y+100=0$
Apr
28
reviewed Approve $\inf A = -\sup(-A)$
Apr
28
reviewed Approve From Cosine formula between two vectors to Schwarz inequality and Triangle inequality?
Apr
28
answered Matrices Eqvialence Relation
Apr
28
comment Logic - Binomial Theorem
I believe there is an answer here: math.stackexchange.com/questions/11601/…
Apr
28
revised Why do we study real numbers?
added 485 characters in body
Apr
28
answered Why do we study real numbers?
Apr
28
comment Proof that there are infinitely many primes (Euclid)
When you learn topology, try reading Furstenberg's topological proof of infinite primes!
Apr
28
accepted Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable
Apr
27
comment Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable
@Omnomnomnom Shouldn't it be every eigenvalue is a root of the minimal polynomial and thus $p$, but not every root of $p$ is an eigenvalue?
Apr
27
comment Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable
Update: $p$ isn't necessarily the characteristic polynomial
Apr
27
revised Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable
added 56 characters in body
Apr
27
comment Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable
Wait... How do we know the eigenvalues are precisely the roots of $p$? ($p$ is not necessarily the characteristic polynomial here, sorry for the confusing notation)
Apr
27
comment Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable
Ok I think I understand.
Apr
27
asked Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable