Matrix multiplication and the dot product

[EV33]

Homework Statement

Suppose that A is an n x n matrix such that A(Transpose)A=I. Let x be any vector in R^n. Show that llAxll=llxll; that is, multiplication of x by A produces a vector Ax having the same length as x.

Homework Equations

Sqrt(x(transpose)x)=llxll

The Attempt at a Solution

So I started setting up the equation wth an n x n matrix...

l a b l =A
l c d l

l a c l= A (transpose)
l b d l

A(transpose)A= I=
l a^(2) +c^(2) ab+cd l
l ba+dc b^(2)+d^(2)l

then I let the vector x=

l x1 l
l x2 l


then from there I was going to use the distance formula on Ax

but I wasn't sure if doing the dot product of Ax is the same thing as matrix multiplication of Ax, and my second problem is that I am not sure how to take the square root of a 2x2 matrix if they are the same thing.

If someone can answer those to questions for me I think I can finish the problem.

Thank you.

 

[Dick]

You don't have to work with any explicit matrices. If you are writing expressions like Ax then you should be thinking of x as a column vector. So ||x||=sqrt(transpose(x)x). So ||Ax||=sqrt(transpose(Ax)Ax). Can you express transpose(Ax) in terms of transpose(A) and transpose(x)?

 

[EV33]

I'm not too sure how to do that.

Would this argument work?

If A(transpose)A=I, then A=A(transpose)=I.

If A=I, then Ax=Ix=x

therefore llAxll=llIxll=llxl

 

[vela]

I'm not too sure how to do that.

Would this argument work?

If A(transpose)A=I, then A=A(transpose)=I.

If A=I, then Ax=Ix=x

therefore llAxll=llIxll=llxl

No, because your first step is wrong. There are matrices other than I where $A^TA=I$.

Try looking up the properties of transpose, in particular, the transpose of a product.

 

[EV33]

ok I looked up all I know about transpose, and I also looked up all I know about I...
so here is what I know, but I ended up using other theorums to solve it.

Thrm 10: If A and B are m x n matrices and c is an n x p matrix, then:

1. (A+B)^T=A^T +B^T
2.(AC)^T=C^T * A^T
3. (A^T)^T=A


What I know about I...
.I is a matrix whose diagnal is all 1's and the rest are 0's.
.AI=A
.I is nonsingular (the only soultion is the trivial solution)
. (A^(-1)*A)^T=I


Here is my second attempt. I ended up using mainly other thrms then the ones above but I am pretty sure I have it this time. Thank you to eveyone who has helped me thus far, and anyone who spends the time to read my proof.


If I is non singular then A and A transpose must be nonsingular,because if either A or A transpose was singular then I would have to be singular and I happens to be nonsingular. The lemma states, "Let P,Q, and R be n x n matrices such that PQ=R. IF either P or Q is singular then so is R.

Then I also know that all non singular matrices can row reduce to I. If all non singular matrices can reduce to I, and A is non singular then I must be equal to A, by thrm 1.

thrm 1 states, " if one of the 3 row operations is applied to a system of linear equations, then the resulting system is equivalent to the original system.

So because any matrix time I is equal to that matrix and I is equal to A then...

A=I,

so Ax=Ix=x, therefore

llAxll=llxll

 

[vela]

Then I also know that all non singular matrices can row reduce to I. If all non singular matrices can reduce to I, and A is non singular then I must be equal to A, by thrm 1.

This is completely wrong. Just because a matrix is non-singular doesn't mean it equals I. It can be reduced to I, but the reduced matrix isn't equal to the original matrix.

An obvious counterexample to your logic: Take A=2I. It's obviously non-singular, but |Ax| = 2|x|. "Aha!" you say, "but $A^TA \ne I$!" Yes, that's true, but your proof doesn't rely on that assumption either, so according to your proof's logic, A=2I should preserve the length of vectors.

So look at property 2 of theorem 10 and use the hint Dick gave above.

 

[EV33]

oh wow that made things much easier but I am still a little confused.

so I have (Ax)^T=x^(T)*A^(T)

llxll=llAxll

sqrt(x^(T)*x)=sqrt((x^(T)*A^(T))(Ax))


x^(T)*x=(x^(T)*A^(T))(Ax)

My problem starts here now. I thought the order of multiplication mattered, and for me to get this to simplifiy I would have to break that order up I feel like because I would multiply the A^T by the A first rather than the B^T by the A^T. Does order matter here? And if it does, does it disable me from being able to do A^T by A first?

 

[vela]

What you're thinking of is the fact that matrix multiplication is not commutative, that is, $AB\ne BA$ generally, but it is associative, so $(AB)C = A(BC)$.

 

[Dick]

You've got (x^(T)*A^(T))*(A*x). You are confusing commutativity with associativity. It's true you can't change the order of matrices but you can regroup them. I.e. (AB)(CD)=A(BC)D. That's just associativity. And it is true.

 

[EV33]

ahhhhhhhh got ya. Thank you so much for all the help everyone.