# Need Linear Algebra Proofs Proof-read

## 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?

Last edited:

Mark44
Mentor

## 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.

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

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?

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...

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 $$u^TAu=(u^TAu)^T$$...

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.

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

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

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?

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.

Okay. Thank you so much, micromass. 