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seen Sep 8 at 2:50

Dec
18
comment Understanding the change-of-coordinate matrix
Thank you for all your help!
Dec
18
comment Understanding the change-of-coordinate matrix
You mean $[T]_\beta^\gamma$? What would be the interpretation of this?
Dec
18
comment Understanding the change-of-coordinate matrix
Okay, I think I see what you mean. Thank you. So would $[T]_\beta$ mean the representation of $T$ with respect to the basis $\beta$?
Dec
18
comment Understanding the change-of-coordinate matrix
So $A$ would change the scalar elements?
Dec
17
comment Exponential distribution, am I doing this correctly?
Thanks for your help!
Dec
17
comment Exponential distribution, am I doing this correctly?
Thank you for your help!
Dec
17
comment Probability question..how to set up?
Thank you, but would I need to calculate the binomial for 3 or more?
Dec
5
comment Proof for diagonalizable matrix
So you mean branching off from $Av= \lambda v$?
Dec
5
comment Proof for diagonalizable matrix
Yes, I know the process but would I need to prove that?
Dec
4
comment Leibniz Formula…determinants
Thanks for all the information! Can you please explain the $\sigma(i)$ notation a bit?
Dec
4
comment Leibniz Formula…determinants
Thank you! Can you explain how the $\sigma(i)$ works? With the permutations...
Dec
2
comment Determinant proof
@HenrySwanson Ohh okay, I get it now. Thanks!
Dec
2
comment Eigenspace definition
@Vladhagen and Any Thank you for your help
Dec
2
comment Eigenspace definition
So is the definition above accurate?
Nov
23
comment Prove positive, semi-definite
Luis Valerin- Thank you for providing a different method in approaching the problem!
Nov
21
comment Prove positive, semi-definite
Thank you. How did you know $<h,A^*Ah>=<Ah,Ah>$?
Nov
21
comment Prove positive, semi-definite
@LuisValerin I don't know what that means...
Nov
21
comment Change of coordinate matrix proof
@user99680 I forgot to tag you in the question above
Nov
21
comment Computing $[T]_\beta$
Okay, I see now, thanks!
Nov
21
comment Change of coordinate matrix proof
So I wouldn't need to show that for an $v \in V$, $[v]_\beta=Q[v]_\beta'$?