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seen Mar 14 '13 at 17:46

Apr
24
awarded  Notable Question
Nov
18
awarded  Popular Question
Mar
3
accepted Matrix Exponential using JNF and transition Matrix
Mar
2
asked Matrix Exponential using JNF and transition Matrix
Feb
20
awarded  Scholar
Feb
20
accepted Algebraic and geometric multiplicity, eigenspace and Transition Matrix
Feb
19
comment Algebraic and geometric multiplicity, eigenspace and Transition Matrix
ohhh okay i understand it now :) thankyouuuuu :D
Feb
19
comment Algebraic and geometric multiplicity, eigenspace and Transition Matrix
so in my question, using my eigenvalues $2$ and $3$ what would my eigenspaces be?
Feb
19
comment Algebraic and geometric multiplicity, eigenspace and Transition Matrix
@GitGud one last quick question, what is the eigenspace?
Feb
19
awarded  Commentator
Feb
19
awarded  Editor
Feb
19
comment Algebraic and geometric multiplicity, eigenspace and Transition Matrix
i think i understand, the geometric multiplicity is 1 for both values of $\lamda$ as they both produce only one eigenvector each?
Feb
19
revised Algebraic and geometric multiplicity, eigenspace and Transition Matrix
added 23 characters in body
Feb
19
comment Algebraic and geometric multiplicity, eigenspace and Transition Matrix
yes this is clearer, this is the same for the geometric multiplicity? so they'd both be 1?? as in are they relitive to each eigenvector?
Feb
19
comment Algebraic and geometric multiplicity, eigenspace and Transition Matrix
so in this case, the algebraic multiplicity would be 2? as it is the largest $k$ value? .... i've memorised the method of computing the JNF so i dont exactly understand it, however now im currently attempting to understand it
Feb
19
awarded  Student
Feb
19
asked Algebraic and geometric multiplicity, eigenspace and Transition Matrix
Feb
6
comment Normal Distribution Stats
ohhh sorry musta been a typo, thankyou i understand it now. And that other explanation is AMAZING :)
Feb
6
comment Normal Distribution Stats
im not 100% sure where i've gone wrong with my working out?
Feb
6
comment Normal Distribution Stats
@StefanHansen i've gotten to $\frac {x-65}{2(5)^{1/2}} =0.1764$ and hence $x = 67.789$ ??