# Proving that T(u).T(V)=u.v if an only if A^T =A^-1

1. Nov 20, 2014

### Myr73

Let u and v be vectors in R^n , and let T be a linear operator on R^n. Prove that T(u).T(v)= u.v if and only if A^T=A^-1 where A is the standard matrix for T.

What I have so far is --> If T is a linear operator on R^n, then ; T:R^n --> R^n
and if u and v be the vectors then ; T(u).T(V) = Au.Av
Where A is the standard matrix of T
{ and so we want Au.Av=u.v, which needs to be if and only if A^T=A^-1}

Therefore first we find what it means if A^T=A^-1
----> if A^T=A^-1 then multiplying both by A --> AA^T= I
Therefore the matrix A is and orthogonal matrix/
And so from the last statement , if Au.Av= u.v if and only if A^T=A^-1
Then --> Au.Av=u.v if and only if A is an orthogonal matrix.
Therefore we must prove that Au.Av= u.v means that this is only true if the matrix A is orthogonal.

This is where I am stuck I'm not sure how to prove this.
I'm not sure if this next part is necessary but I found that

u.v = u1v1+u2v2...UnVn

and that T(u)=A(u) such that a(11)u(1), a(12)u(2),..,a(1n)u(1)
a(21)u(1), a(22)u(2)...,a(2n)u(2)
.................................................
...............................................
a(n1)u(1), a(n2)u(2)....a(nn)u(n)

and same for T(v)=A(v) such that
a(11)v(1), a(12)v(2),..,a(1n)v(1)
a(21)v(1), a(22)v(2)...,a(2n)v(2)
.................................................
...............................................
a(n1)v(1), a(n2)v(2)....a(nn)v(n)

Im not sure if I need to develop T(u).T(v) further then A(u).A(v)..

I would really appreciate some help on this one, thanks

2. Nov 20, 2014

### Fredrik

Staff Emeritus
Are you sure that you're allowed to use $Tu\cdot Tv=Au\cdot Av$ without proof? If you use this, you don't even have to use the definition of "standard matrix for T". The simplest way to proceed is to use that $x\cdot y=x^Ty$.

This is a valid approach even if you're not allowed to use $Tu\cdot Tv=Au\cdot Av$ without proof. You just have to do the proof as well. This is pretty straightforward if you expand u and v in the standard basis, and use the definitions of dot product, matrix multiplication, and "standard matrix for T".

Moderators: I moved the thread here from the linear algebra forum.

3. Nov 20, 2014

### Myr73

I only meant for example; Tu in matrix form would be [A][x] like -->a(11)u(1), a(12)u(2),..,a(1n)u(1)
a(21)u(1), a(22)u(2)...,a(2n)u(2)
.................................................
...............................................
a(n1)u(1), a(n2)u(2)....a(nn)u(n) ....Because A is the standard matrix of T. is that not the right way?

Also, I don't really understand what I am suppose to do with x.y=x^Ty

4. Nov 21, 2014

### Fredrik

Staff Emeritus
OK, you have a point. The observation you just made makes it obvious that $Tu\cdot Tv=Au\cdot Av$. So I guess it's not necessary to do the proof I had in mind:
\begin{align}
&Tu\cdot Tv =\sum_{i,j} u_i v_j Te_i\cdot T e_j =\sum_{i,j,k,l} u_i v_j (Te_i)_k (Te_j)_l e_k\cdot e_l\\
&=\sum_{i,j,k,l} u_i v_j A_{ki} A_{lj}\delta_{kl} =\sum_{i,j,k} u_i v_j A_{ki} A_{kj}=\sum_k(Au)_k (Av)_k =Au\cdot Av
\end{align}

Use it to rewrite both sides of $u\cdot v=Au\cdot Av$. You know how to deal with $(Au)^T$, right?

Last edited: Nov 21, 2014
5. Nov 24, 2014

### Myr73

no sorry, I'm unsure where to go from "Therefore we must prove that Au.Av= u.v means that this is only true if the matrix A is orthogonal."

6. Nov 24, 2014

### Fredrik

Staff Emeritus
If you rewrite $u\cdot v=Au\cdot Av$ using $x\cdot y=x^Ty$, you should almost immediately see that the equality holds if and only if $A$ is orthogonal.

The general formula for the transpose of a product is $(AB)^T=B^TA^T$. This follows almost immediately from the definitions:
$$((AB)^T)_{ij}=(AB)_{ji}=\sum_k A_{jk}B_{ki} =\sum_k (A^T)_{kj} (B^T)_{ik} =\sum_k (B^T)_{ik} (A^T)_{kj}=(B^TA^T)_{ij}.$$

7. Nov 24, 2014

### Myr73

ok, so are you meaning for vector u to be x and vector v to be y?

8. Nov 24, 2014

### Fredrik

Staff Emeritus
When you rewrite $u\cdot v$, yes. Of course, that's not the only thing you need to rewrite.

Last edited: Nov 24, 2014
9. Nov 26, 2014

hmm.. ok