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Jun
26
comment Commutative permutations
@user159519 - I am working with $G=S_n$
Jun
26
comment Commutative permutations
@user159519 - Please try a little bit more and write down what you tried so I can better direct you. I think it will help more then writing down the answer..
Jun
21
comment $A$ matrix, $+i, -i$ are eigenvalues.
@user3628041 note that the determinant is zero iff one of the eigenvalues is zero and use Surb answer to complete the argument that indeed the implication I made holds
Jun
21
comment How on earth will anyone prove $n^3-3n^2+n-1=Θ(n^3)$
Yes you can, it follows from the definition
Jun
14
comment $T: M_{2\times 2}(R) \rightarrow M_{2 \times 2}(R)$ linear map
@IlanAizelmanWS use the standard basis (all entries are 0 except for one entry which is 1). Work with 4 dimensional coordinate vectors to represent an element of V
Jun
14
comment Matrix $A$ with characteristic polynomial
@IlanAizelmanWS - What book are you using ?
Jun
14
comment Matrix $A$ with characteristic polynomial
@IlanAizelmanWS - oh yes, your'e right
Jun
14
comment Matrix $A$ with characteristic polynomial
@IlanAizelmanWS - Good luck! Its a great course in my opinion
Jun
14
comment Matrix $A$ with characteristic polynomial
@IlanAizelmanWS - No, I deduced that because the dimension of the matrix is the same of the degree of the characteristic polynomial
Jun
14
comment Matrix $A$ with characteristic polynomial
@IlanAizelmanWS - By the way, do you study for the "Algebra aleph" exam ?
Jun
14
comment Matrix $A$ with characteristic polynomial
@IlanAizelmanWS - Yes. This would imply that the geometric multiplexing is $\leq 1$. But you also know that it is at least $1$ since it is an eigenvalue and so the geometric multiplexing is $1$
Jun
14
comment How to check whether this matrix is diagonalizable or not.
@David - I have shown a matrix $A$ with $A^3=-1$ but not the eigenvalues are not distinct. This is a counterexample for statement $1$
Jun
14
comment How to check whether this matrix is diagonalizable or not.
@Surb - a scalar $c$ means, in abuse of notation, $cI$
May
19
comment b such that Ax = b has no solution having found column space
@HenningMakholm thanks for the correction , you're right! I'll edit my answer
May
19
comment b such that Ax = b has no solution having found column space
@HenningMakholm I don't understand your example. Do you claim that my example is not a counter example to your claim? I believe that $V-W$ together with the zero vector does form a subspace, although $W$ is a proper and non trivial subspace of $V$
May
18
comment If $|X|<|Y|$ then $|Y|=|Y-X|$ (with $Y$ infinite)
I didn't understand how to deduce the desired equality. From what I understand showing that the sum of two infinite cardinals equals to one of them only gives us an inequality between them (I'm sorry that I am not writing exactly what I meant with tex, I'm using a mobile phone)
May
18
comment If $|X|<|Y|$ then $|Y|=|Y-X|$ (with $Y$ infinite)
How does this series of inequalities (which became a series of equalities) help us?
May
18
comment b such that Ax = b has no solution having found column space
@HenningMakholm what about $\mathbb{R}^2 $ and $W=sp\{(1,0)\} $?
May
17
comment $x^3+3x^2+4x+5=0$ and $x^3+2x^2+7x+3=0$, how many common roots they have?
@Macavity - Yes, I looked at Andreas answer and I saw where I made the mistake and edited to correct it
May
17
comment $x^3+3x^2+4x+5=0$ and $x^3+2x^2+7x+3=0$, how many common roots they have?
@Macavity -already edited! :)