Recent content by hedgie

Why would it be the first row and the column that i am multiplying it against transposed to a row?

I don't, I only got a=-d, sorry typo. ((1,-1),(1,-1))*(2,-2)=(0,0) and (V_1 V_2) = ((2,-2),(2,-2))

yes (2,-2) column vector...also tried to solve the problem using the matrix ((a,b),(c,d)) and got a=+-d

(a_11+a_12,a_21+a_22) as a 2x1 matrix if we do not know what v_1 and v_2 are.. V_1 and V_2 are both (1,1) or the 2x2 matrix of ((1,1),(1,1))

We had always used columns in class and from the book...and the basis are all columns and everything we have been writing or I at least have been writing is in columns. Which is why I would've thought (1,-1) is V1 and why I asked if it was arbitrary...if I was using rows i'd expect (1,1).

Are you just picking arbitrary V_1? shouldn't it be the first column?

Do they satisfy the transformation or the ((1,1),(-1,-1)) * V_1= V_2...no for the second one if V_1 = (0,1)...completely lost now.

Why do we not know V_1 and V_2 are they not E_1 and E_2 for R^2 and then transformed to get ((0,1),(0,0))

So I get N*((1,0),(0,1))=((0,0),(1,0)) which is what we want. Because N times the identity equals the similar matrix. So we can just use anything that generates R^2... Isn't it a lot simpler to say N*(0,1)=(0,0) where (0,1) is V_1 as I noted in my original attempt at the problem or cannot...

So we don't know N and when I write that I get ((0,0),(0,1)) on the RHS and N*(V_2,V_1) on the LHS.

Since [V_2 V_1] is a matrix that would make [(0,1),(1,0] the change of coordinate matrix but why can we switch the order of V_1 and V_2. I think it is invertible.

So do I just guess to find the basis...say N*(0,1) then you'd find V2=(0,0)...but why does just finding a basis for R^2 show that its similar?

[V_2 V_1] is the basis for R^2 such that N*V_1=V_2 and N*V_2=0 is true

We cannot use any of that. Thanks though.

What I was asking here is can you swap the order of the vectors/basis. The equalities would change to the matrix you wrote early((1,0),(0,0))