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comment How do I prove that for $\|T(x)\|=\|x\|$ for all $x$ in a vector space iff $\left<T(x),T(y)\right>=\left<x,y\right>$?
I did both directions above...
59m
answered How do I prove that for $\|T(x)\|=\|x\|$ for all $x$ in a vector space iff $\left<T(x),T(y)\right>=\left<x,y\right>$?
9h
comment Can $f''(x)$ exist if $f'(x)$ is undefined?
@GEdgar: It was an answer to the title question, but have removed my comment since it wasn't very helpful.
12h
revised Multivariable Calculus, Parametrization and extreme values
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1d
answered Minimize Energy Function
1d
answered Proving $\frac{n}{n+1} < \frac{n+1}{n+2}$ by induction?
1d
answered Proving Lipschitz continuity of a piecewise function
1d
comment Recalculate Standard Deviation
Correct. If you knew the number of samples you could compute $\sum_k x_k $ and $\sum_k x_k^2$, but without that you are missing some constraint...
1d
comment Recalculate Standard Deviation
Then you can't do it. If there were a huge number of samples, one more would have little impact where as if there were just two samples, the impact would be more significant.
1d
comment Recalculate Standard Deviation
Do you know how many samples you have?
1d
comment Recalculate Standard Deviation
In general, adding a different value will change the mean and the variance. I don't understand how you concluded that adding the above value won't affect the variance?
1d
comment distance on a normed space
@user308560: Yes, it looks correct. In general, for these sorts of things, since $d_Y$ is defined in terms of $\|\cdot\|$, I think is is easier to work from the corresponding relation for $\|\cdot\|$ and then use that to get the relation for $N$.
1d
comment distance on a normed space
If $y_k$ is a sequence in $Y$ such that $ \|a-y_k\| \to N(\bar{a})$ then $y'_k=(b-a)+y_k$ is a sequence in $Y$ such that $\|b-y'_k \| = \|a-y_k\| \to N(\bar{a})$, from which it follows that $N(\bar{b}) \le N(\bar{a})$. Switching the roles of $a,b$ gives the desired result.
1d
comment how to determine a matrix has a single eigenvalue
@MooS: The OP mentioned nothing about computation effort. It is trivial to verify that $A^4 = (A^2)^2 = 0$ (which is tantamount to what you have done in your answer) from which it follows that (i) the characteristic polynomial is $x \mapsto x^4$ and (ii) all eigenvalues are zero. This sort of pettiness is why my interest in helping out at MSE has significantly waned.
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revised how to determine a matrix has a single eigenvalue
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1d
comment distance on a normed space
Glad to help! ${}{}$
1d
comment how to determine a matrix has a single eigenvalue
Why the downvote? Really, please tell me.
1d
comment distance on a normed space
Your solution is correct, but I prefer proofs that follow directly from the relevant inequality directly.
1d
revised distance on a normed space
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answered distance on a normed space