Need Linear Algebra Proofs Proof-read

  • #1
129
0

Homework Statement



Let [tex]A[/tex] be a skew-symmetric n x n matrix with entries in [tex]R[/tex].

a) Prove that

[tex]u^{T}Au=0[/tex] for every u E [tex]R^{n}[/tex]

b) Prove that

[tex]I_{n} + A[/tex] is an invertible matrix.


Homework Equations



[tex]A^{T} = -A[/tex]

The Attempt at a Solution



a)

[tex]u^{T}Au=0[/tex]
Transpose both sides:
[tex](u^{T}Au)^{T} = 0^{T}[/tex]
The transpose of the product is the reverse-order product of transposes and the transpose of 0 is 0:
[tex]u^{T}A^{T}(u^{T})^{T} = 0[/tex]
The transpose of a transpose is the original thing:
[tex]u^{T}A^{T}u = 0[/tex]
Sub in (-A) for (A transpose):
[tex]-u^{T}Au = 0[/tex]
Divide by -1:
[tex]u^{T}Au = 0[/tex]

As required.

b)

Let [tex]x = I_{n} + A[/tex]

If the inverse of x exists, there will be a matrix [tex]x^{-1}[/tex] such that:

[tex]xx^{-1} = I_{n}[/tex]

And quite frankly, that's where I'm at. I don't even know what it means to prove that a matrix is invertible. A is square, so how could it not be invertible?
 
Last edited:

Answers and Replies

  • #2
35,236
7,056

Homework Statement



Let [tex]A[/tex] be a skew-symmetric n x n matrix with entries in [tex]R[/tex].

a) Prove that

[tex]u^{T}Au=0[/tex] for every u E [tex]R^{n}[/tex]

b) Prove that

[tex]I_{n} + A[/tex] is an invertible matrix.


Homework Equations



[tex]A^{T} = -A[/tex]

The Attempt at a Solution



a)

[tex]u^{T}Au=0[/tex]
Transpose both sides:
[tex](u^{T}Au)^{T} = 0^{T}[/tex]
The transpose of the product is the reverse-order product of transposes and the transpose of 0 is 0:
[tex]u^{T}A^{T}(u^{T})^{T} = 0[/tex]
The transpose of a transpose is the original thing:
[tex]u^{T}A^{T}u = 0[/tex]
Sub in (-A) for (A transpose):
[tex]-u^{T}Au = 0[/tex]
Divide by -1:
[tex]u^{T}Au = 0[/tex]

As required.

b)

Let [tex]x = I_{n} + A[/tex]

If the inverse of x exists
Then you're done, but you have assumed what needed to be proved.
, there will be a matrix [tex]x^{-1}[/tex] such that:

[tex]xx^{-1} = I_{n}[/tex]

And quite frankly, that's where I'm at. I don't even know what it means to prove that a matrix is invertible. A is square, so how could it not be invertible?

Just being a square matrix is not enough to be able to say that its inverse exists. For example, the following matrix is square, but doesn't have an inverse.

[tex]\left[ \begin{array}{cc}0 & 1\\ 0 & 0 \end{array}\right][/tex]
 
  • #3
22,129
3,297
Your solution to 1 doesn't make very much sense. You start from [tex]u^TAu=0[/tex], while that was what you were trying to show... You're using the right techniques to solving this though...

For 2, you will need to use 1. A hint: how many solutions does the system (I+A)x have?
 
  • #4
129
0
you have assumed what needed to be proved.

...while that was what you were trying to show

Yes, I was afraid of that xD.

how many solutions does the system (I+A)x have?

0, 1 or an infinite number? If you meant x to be the same thing that I let it be in OP then it would be (I + A)(I + A) = I^2 + 2A + A^2...

Okay, here's proof mk.II

Consider the definition of a skew-symmetric matrix:
[tex]A^{T} = -A[/tex]
Multiply both sides by u:
[tex]uA^{T} = -uA[/tex]
Take the transpose of both sides:
[tex](uA^{T})^{T} = (-uA)^{T}[/tex]
[tex](A^{T})^{T}(u^{T}) = -(A^{T})(u^{T})[/tex]
[tex](A)(u^{T}) = -(A^{T})(u^{T})[/tex]
[tex](A)(u^{T}) + (A^{T})(u^{T})= 0[/tex]
Multiply both sides by u...again
[tex]u(A)(u^{T}) + u(A^{T})(u^{T})= 0[/tex]

Now that first part kinda looks like what I'm trying to get to, but it's still not quite it...
 
  • #5
22,129
3,297
No, I'm sorry, I meant x as a vector. But I'll call it u instead.
I want to know how many u exist such that (I+A)u=0...

As for the proof of 1: [tex]u^TAu[/tex] equals its transposition, since it is a 1x1-matrix. Thus [tex]u^TAu=(u^TAu)^T[/tex]...
 
  • #6
129
0
m.k III (Also, thank you for helping)

[tex]u^TAu=(u^TAu)^T[/tex]
[tex]u^TAu=(u^T)(A^T)(u^T)^T[/tex]
[tex]u^TAu=-(u^T)(A)(u)[/tex]
[tex]u^TAu + (u^T)(A)(u)= 0[/tex]
[tex]2u^TAu = 0[/tex]
[tex]u^TAu = 0[/tex]

The fact that a 1x1 matrix equals its transpose is obvious, but how did you know that it was a 1x1 matrix? Is the reason something like:

[tex](u^T)_{bxn}A_{nxn}u_{nxb}[/tex]

[tex]A'_{bxn}u_{nxb}[/tex]

[tex]A''_{bxb}[/tex]

And since u was a vector, b must be 1?

I want to know how many u exist such that (I+A)u=0...

Well u = 0 is clearly a solution - that doesn't mean it's the only solution though, one can multiply two non-zero vectors and get the zero vector.
 
  • #7
22,129
3,297
Well, in this case, it IS the unique solution. Use 1 to prove this...
 
  • #8
129
0
[tex](I_{n}+A)u = 0[/tex]
[tex]u + Au = 0[/tex]
[tex](u^{T})(u) + (u^{T})(Au) = 0[/tex]
[tex](u^{T})(u) = 0[/tex]

Which implies that u = 0 because, say u = n x 1:

[tex](u^{T})(u) = 0[/tex]
[tex](u^{T}_{1xn})(u_{nx1}) = 0[/tex]
[tex](u'_{1x1}) = 0[/tex]

And the only way for a 1x1 matrix to be 0 is if the matrix itself, u, is 0.

Okay so now I know (assuming that what I wrote is correct) that the only way to get the matrix (I + A) to be 0 is to multiply it by 0. Does that imply that its determinant can't be 0?
 
  • #9
22,129
3,297
Yes, that is correct. Now you must use some linear algebra. You know that the kernel I+A is trivial, this implies that I+A is invertible.
 
  • #10
129
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Okay. Thank you so much, micromass. :smile:
 

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