# Inner product space

• May 4th 2010, 06:34 AM
Also sprach Zarathustra
Inner product space
Let V be Inner product space(real or complex).
For this exercise we define "twin transformations" S,T:V-->V if for all u,w in V : <Tu,Sw>=<u,w>.

1. Prove that if T is singular,so T not have "twin transformations".
2. Prove that if T nonsingular, so there is only one "twin transformations".
• May 4th 2010, 09:24 AM
Failure
Quote:

Originally Posted by Also sprach Zarathustra
Let V be Inner product space(real or complex).
For this exercise we define "twin transformations" S,T:V-->V if for all u,w in V : <Tu,Sw>=<u,w>.

1. Prove that if T is singular,so T not have "twin transformations".

Indirect proof: Let $\displaystyle T$ be singular and $\displaystyle S$ its twin transformation. Since $\displaystyle T$ is singular, there exists an $\displaystyle x\in V, x\neq 0$ with $\displaystyle Tx=0$. Thus we have $\displaystyle \langle Tx, Sx\rangle = \langle 0,Sx\rangle = 0$, and we also have that $\displaystyle \langle Tx, Sx\rangle = \langle x, x\rangle\neq 0$, a contradiction.

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2. Prove that if T nonsingular, so there is only one "twin transformations".
Since $\displaystyle T$ is non-singular we have that for every basis vector $\displaystyle e_i$ there exists a uniquely determined $\displaystyle x_i\in V$, such that $\displaystyle T x_i = e_i$.
Therefore the component of the image $\displaystyle S e_j$ of the basis vector $\displaystyle e_j$ under $\displaystyle S$ in the direction of $\displaystyle e_i$ is given by $\displaystyle \langle e_i, S e_j\rangle = \langle T x_i, S e_j\rangle = \langle x_i, S e_j\rangle$, and thus is uniquely determined by $\displaystyle T$.