Need Linear Algebra Proofs Proof-read

[Screwdriver]

Homework Statement

Let A be a skew-symmetric n x n matrix with entries in R.

a) Prove that

u^{T}Au=0 for every u E R^{n}

b) Prove that

I_{n} + A is an invertible matrix.

Homework Equations

A^{T} = -A

The Attempt at a Solution

a)

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

As required.

b)

Let x = I_{n} + A

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

xx^{-1} = I_{n}

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?

 

[Mark44]

Homework Statement

Let A be a skew-symmetric n x n matrix with entries in R.

a) Prove that

u^{T}Au=0 for every u E R^{n}

b) Prove that

I_{n} + A is an invertible matrix.

Homework Equations

A^{T} = -A

The Attempt at a Solution

a)

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

As required.

b)

Let x = I_{n} + A

If the inverse of x exists

Then you're done, but you have assumed what needed to be proved.

, there will be a matrix x^{-1} such that:

xx^{-1} = I_{n}

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.

\left[ \begin{array}{cc}0 & 1\\ 0 & 0 \end{array}\right]

 

[micromass]

Your solution to 1 doesn't make very much sense. You start from u^TAu=0, 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?

 

[Screwdriver]

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:
A^{T} = -A
Multiply both sides by u:
uA^{T} = -uA
Take the transpose of both sides:
(uA^{T})^{T} = (-uA)^{T}
(A^{T})^{T}(u^{T}) = -(A^{T})(u^{T})
(A)(u^{T}) = -(A^{T})(u^{T})
(A)(u^{T}) + (A^{T})(u^{T})= 0
Multiply both sides by u...again
u(A)(u^{T}) + u(A^{T})(u^{T})= 0

Now that first part kinda looks like what I'm trying to get to, but it's still not quite it...

 

[micromass]

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: u^TAu equals its transposition, since it is a 1x1-matrix. Thus u^TAu=(u^TAu)^T...

 

[Screwdriver]

m.k III (Also, thank you for helping)

u^TAu=(u^TAu)^T
u^TAu=(u^T)(A^T)(u^T)^T
u^TAu=-(u^T)(A)(u)
u^TAu + (u^T)(A)(u)= 0
2u^TAu = 0
u^TAu = 0

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:

(u^T)_{bxn}A_{nxn}u_{nxb}

A'_{bxn}u_{nxb}

A''_{bxb}

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.

 

[micromass]

Well, in this case, it IS the unique solution. Use 1 to prove this...

 

[Screwdriver]

(I_{n}+A)u = 0
u + Au = 0
(u^{T})(u) + (u^{T})(Au) = 0
(u^{T})(u) = 0

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

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

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?

 

[micromass]

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.

 

[Screwdriver]

Okay. Thank you so much, micromass.