Linear Algebra proof, diagonalization

In summary: But is that matrix B?I am sorry for the confusion. I was just trying to say that the matrix given is A and I am trying to use the formula for a similar matrix written in terms of A. I was just trying to use the formula to find B or N in the problem.In summary, the problem states that N is a 2x2 matrix such that N^2 = 0. The task is to prove that either N = 0 or N is similar to the matrix ((0,0), (1,0)). This can be done by finding a basis for R^2 and showing that N is similar to a matrix with this basis. The basis can be found by setting N(v1
  • #36
Suppose [itex]N=\begin{pmatrix}1 & 1 \\ -1 & -1 \end{pmatrix}[/itex].
Does it satisfy the problem criteria?
Are V_1 and V_2 as you're suggesting?
 
Physics news on Phys.org
  • #37
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.
 
  • #38
Suppose V_1=(1,1).
Does it fit?

What would V_2 be?
And what would N^2 be?
 
  • #39
Are you just picking arbitrary V_1? shouldn't it be the first column?
 
  • #40
No, it's not arbitrary. :)

Why should it be the first column?
 
  • #41
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).
 
  • #42
That's why I adjusted post #36, to avoid confusion.

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

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

What are therefore V_1 and V_2?
 
  • #43
(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))
 
  • #44
Did you calculate [itex]\begin{pmatrix}1 & 1 \\ -1 & -1 \end{pmatrix} \cdot \begin{pmatrix}1 \\ 1\end{pmatrix}[/itex]?
 
  • #45
yes (2,-2) column vector...also tried to solve the problem using the matrix ((a,b),(c,d)) and got a=+-d
 
  • #46
hedgie said:
yes (2,-2) column vector

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


hedgie said:
...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?
 
  • #47
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))
 
  • #48
hedgie said:
I don't, I only got a=-d, sorry typo.

Yes.


hedgie said:
((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))
 
  • #49
Why would it be the first row and the column that i am multiplying it against transposed to a row?
 

Similar threads

  • Calculus and Beyond Homework Help
Replies
14
Views
603
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
24
Views
807
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
301
  • Calculus and Beyond Homework Help
Replies
7
Views
419
  • Calculus and Beyond Homework Help
Replies
3
Views
897
  • Calculus and Beyond Homework Help
Replies
15
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
Back
Top