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## 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?

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