Linear Algebra proof, diagonalization

[hedgie]

Yes, it is invertible, because V_1 and V_2 have to be linearly independent.
(If you're up to it... why?)

So let's define J=[(0,0) (1,0)].
Then we'd have N P = P J with P an invertible matrix.
Can you rewrite this?

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.

 

[I like Serena]

Yes.
What happens if you multiply the LHS and the RHS with the inverse of P?

 

[hedgie]

Yes.
What happens if you multiply the LHS and the RHS with the inverse of P?

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 you not do that because we do not know what N is?

 

[I like Serena]

So I get N*((1,0),(0,1))=((0,0),(0,1)) which is what we want.

Umm... no...
You need to show that N is similar to J.
But you don't and can't know what N, P, V_1, or V_2 are...
You can only find some properties of them, like to which matrix N is similar to, and that V_1 and V_2 are linearly independent.

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 you not do that because we do not know what N is?

N*(0,1)=(0,0) may or may not be true. It's not possible to say.

 

[hedgie]

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))

 

[I like Serena]

Suppose N=\begin{pmatrix}1 & 1 \\ -1 & -1 \end{pmatrix}.
Does it satisfy the problem criteria?
Are V_1 and V_2 as you're suggesting?

 

[hedgie]

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.

 

[I like Serena]

Suppose V_1=(1,1).
Does it fit?

What would V_2 be?
And what would N^2 be?

 

[hedgie]

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

 

[I like Serena]

No, it's not arbitrary. :)

Why should it be the first column?

 

[hedgie]

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).

 

[I like Serena]

That's why I adjusted post #36, to avoid confusion.

So what is:
N\cdot \begin{pmatrix}1 \\ 1\end{pmatrix}=\begin{pmatrix}1 & 1 \\ -1 & -1 \end{pmatrix} \cdot \begin{pmatrix}1 \\ 1\end{pmatrix}

And what do you get if you multiply N with the resulting vector?

What are therefore V_1 and V_2?

 

[hedgie]

(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))

 

[I like Serena]

Did you calculate \begin{pmatrix}1 & 1 \\ -1 & -1 \end{pmatrix} \cdot \begin{pmatrix}1 \\ 1\end{pmatrix}?

 

[hedgie]

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

 

[I like Serena]

yes (2,-2) column vector

Yes. And N*(2,-2)?

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

a=-d would be right, but how did you get a=d?

 

[hedgie]

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))

 

[I like Serena]

I don't, I only got a=-d, sorry typo.

Yes.

((1,-1),(1,-1))*(2,-2)=(0,0) and (V_1 V_2) = ((2,-2),(2,-2))

No: (V_1 V_2) = ((1,1),(2,-2))

 

[hedgie]

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