# Transformations...

• May 4th 2010, 06:55 AM
Also sprach Zarathustra
Transformations...
Let V be an inner product space(real ao complex), when dim(V)=n,
T:V-->V is linear transformation.

1. Prove that there exist orthonormal basis {v_1,...,v_n} to V, so that for every i != j : <Tv_i,Tv_j>=0.

2. Prove that there exist U,D:V-->V, so that U is an unitary transformation and D is an adjoint transformation.
• May 4th 2010, 08:13 AM
tonio
Quote:

Originally Posted by Also sprach Zarathustra
Let V be an inner product space(real ao complex), when dim(V)=n,
T:V-->V is linear transformation.

1. Prove that there exist orthonormal basis {v_1,...,v_n} to V, so that for every i != j : <Tv_i,Tv_j>=0.

This cannot be true: if T is an invertible transformation, then it is an orthogonal (unitary) transformation iff it maps orthonormal basis to orthonormal basis, so take any invertible T which is not orthogonal (unitary) and you'll get a contradiction.

2. Prove that there exist U,D:V-->V, so that U is an unitary transformation and D is an adjoint transformation.

This question is most probably incorrect or, at least, incomplete: take U=D=I = the identity transformation and we're done . (Giggle)

Tonio
• May 4th 2010, 08:48 AM
Tikoloshe
Quote:

Originally Posted by tonio
This cannot be true: if T is an invertible transformation, then it is an orthogonal (unitary) transformation iff it maps orthonormal basis to orthonormal basis, so take any invertible T which is not orthogonal (unitary) and you'll get a contradiction.

I do not believe the OP is looking for orthonormality of the images of the original basis vectors, only orthogonality, so the result can be (in fact is) true.

This comes down to finding the singular value decomposition. Basically, if $\displaystyle s_1,\ldots,s_n$ are the singular values of $\displaystyle T$, then there exist orthonormal bases $\displaystyle \{v_1,\ldots,v_n\}$ and $\displaystyle \{u_1,\ldots,u_n\}$ such that for an arbitrary $\displaystyle x\in V$, we can write $\displaystyle Tx=\sum_{i=1}^n s_i\langle x,v_i\rangle u_i$. One can obtain the singular value decomposition by taking the polar decomposition of $\displaystyle T$ and then applying the finite-dimensional spectral theorem to $\displaystyle \sqrt{T^\ast T}$.
Quote:

Originally Posted by tonio
This question is most probably incorrect or, at least, incomplete

Yes, I have a feeling the purpose of part 2 was to show the existence of a polar decomposition.
• May 4th 2010, 09:13 AM
Also sprach Zarathustra
Quote:

Originally Posted by Tikoloshe
I do not believe the OP is looking for orthonormality of the images of the original basis vectors, only orthogonality, so the result can be (in fact is) true.

This comes down to finding the singular value decomposition. Basically, if $\displaystyle s_1,\ldots,s_n$ are the singular values of $\displaystyle T$, then there exist orthonormal bases $\displaystyle \{v_1,\ldots,v_n\}$ and $\displaystyle \{u_1,\ldots,u_n\}$ such that for an arbitrary $\displaystyle x\in V$, we can write $\displaystyle Tx=\sum_{i=1}^n s_i\langle x,v_i\rangle u_i$. One can obtain the singular value decomposition by taking the polar decomposition of $\displaystyle T$ and then applying the finite-dimensional spectral theorem to $\displaystyle \sqrt{T^\ast T}$.

Yes, I have a feeling the purpose of part 2 was to show the existence of a polar decomposition.

hmmm... oops (for 2)
2. Prove that there exist U,D:V-->V so that U is an unitary, D is self adjoint and T=DU
• May 4th 2010, 09:28 AM
Tikoloshe
You should know that every positive (semi-definite) operator has a positive square root. So for 2, let $\displaystyle D=\sqrt{T^\ast T}$. U should be easy to find then.

For 1, I would proceed as I have already explained (using the result of 2).