Results 1 to 2 of 2

Thread: Linear Algebra Question - Linear Operators

  1. #1
    Banned
    Joined
    Jan 2009
    Posts
    20

    Linear Algebra Question - Linear Operators

    Let u and v be vectors in $\displaystyle \Re^n$ and let T be a linear operator on $\displaystyle \Re^n$. Prove that T(n).T(v) = n.v if and ony if $\displaystyle A^t = A^{-1}$ where A is the standard matrix for T
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Quote Originally Posted by r2dee6 View Post
    Let u and v be vectors in $\displaystyle \Re^n$ and let T be a linear operator on $\displaystyle \Re^n$. Prove that T(n).T(v) = n.v if and ony if $\displaystyle A^t = A^{-1}$ where A is the standard matrix for T
    That is called an orthogonal operator. (i.e. a linear transformation -over the scalar field $\displaystyle \mathbb{R}$- $\displaystyle
    T:V \to V
    $ - V a vector space- such that $\displaystyle
    \left\langle {T\left( \bold{v} \right),T\left( \bold{u} \right)} \right\rangle = \left\langle {\bold{v},\bold{u}} \right\rangle
    $ , $\displaystyle
    \forall\bold{u},\bold{v} \in V
    $ )

    All this would be much shorter if you knew what the traspose of a Linear Transformation is, and its properties.


    The proof would go as follows. Let $\displaystyle
    B = \left\{ {\bold{v}_1 ,...,\bold{v}_n } \right\}
    $ be the standard basis for $\displaystyle \mathbb{R}^n$ (i.e. $\displaystyle
    \bold{v}_1 = \left( {1,0,0...,0} \right);\bold{v}_2 = \left( {0,1,0...,0} \right)...
    $ ). This is an orthonormal basis, so $\displaystyle
    \bold{v} = \sum\limits_{k = 1}^n {\left\langle {\bold{v},\bold{v}_k } \right\rangle \cdot \bold{v}_k }
    $ for each $\displaystyle \bold{v} \in \mathbb{R}^n$ (1)

    Let us compute $\displaystyle
    \left[ T \right]_B
    $. By (1) we have: $\displaystyle
    \left[ T \right]_B = \left( {\begin{array}{*{20}c}
    {\left\langle {T(\bold{v}_1) ,\bold{v}_1 } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_n) ,\bold{v}_1 } \right\rangle } \\
    \vdots & \ddots & \vdots \\
    {\left\langle {T(\bold{v}_1) ,\bold{v}_n } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_n) ,\bold{v}_n } \right\rangle } \\

    \end{array} } \right)
    $ thus: $\displaystyle
    \left( {\left[ T \right]_B } \right)^t = \left( {\begin{array}{*{20}c}
    {\left\langle {T(\bold{v}_1) ,\bold{v}_1 } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_1) ,\bold{v}_n } \right\rangle } \\
    \vdots & \ddots & \vdots \\
    {\left\langle {T(\bold{v}_n) ,\bold{v}_1 } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_n) ,\bold{v}_n } \right\rangle } \\

    \end{array} } \right)
    $

    We will prove that $\displaystyle
    \left\langle {T\left( \bold{v} \right),T\left( \bold{u} \right)} \right\rangle = \left\langle {\bold{v},\bold{u}} \right\rangle
    $ holds if and only if $\displaystyle
    \left( {\left[ T \right]_B } \right)^{ - 1} = \left( {\left[ T \right]_B } \right)^t
    $

    Direct: $\displaystyle
    \left\langle {T\left( \bold{v} \right),T\left( \bold{u} \right)} \right\rangle = \left\langle {\bold{v},\bold{u}} \right\rangle ;
    \forall\bold{u},\bold{v} \in \mathbb{R}^n
    \Rightarrow \left( {\left[ T \right]_B } \right)^{ - 1} = \left( {\left[ T \right]_B } \right)^t
    $


    Now: $\displaystyle
    \left[ {\left( {\left[ T \right]_B } \right) \cdot \left( {\left[ T \right]_B } \right)^t } \right]_{i,j} = \sum\limits_{k = 1}^n {\left\langle {T\left( {v_k } \right),v_i } \right\rangle \cdot \left\langle {T\left( {v_k } \right),v_j } \right\rangle }
    $

    We have: $\displaystyle
    \sum\limits_{k = 1}^n {\left\langle {T\left( {v_k } \right),v_i } \right\rangle \cdot \left\langle {T\left( {v_k } \right),v_j } \right\rangle } = \sum\limits_{k = 1}^n {\left\langle {v_i ,T\left( {v_k } \right)} \right\rangle \cdot \left\langle {v_j ,T\left( {v_k } \right)} \right\rangle }
    $

    Since $\displaystyle
    \left\langle {T\left( v \right),T\left( u \right)} \right\rangle = \left\langle {u,v} \right\rangle
    $ you can check it maps an orthonormal basis into an orthonormal basis. Thus $\displaystyle
    \left\{ {T\left( {\bold{v}_1 } \right);...;T\left( {\bold{v}_n } \right)} \right\}
    $ is an orthonormal basis.

    Now: $\displaystyle
    \bold{v}_i = \sum\limits_{k = 1}^n {\left\langle {\bold{v}_i ,T\left( {\bold{v}_k } \right)} \right\rangle \cdot T\left( {\bold{v}_k } \right)}
    $ and hence $\displaystyle
    \left\langle {\bold{v}_i ,\bold{v}_j } \right\rangle = \sum\limits_{k = 1}^n {\left\langle {\bold{v}_i ,T\left( {\bold{v}_k } \right)} \right\rangle \cdot \left\langle {\bold{v}_j ,T\left( {\bold{v}_k } \right)} \right\rangle }
    $

    Thus it follows that: $\displaystyle
    \left[ {\left( {\left[ T \right]_B } \right) \cdot \left( {\left[ T \right]_B } \right)^t } \right]_{i,j} = \left\langle {v_i ,v_j } \right\rangle = \left\{ \begin{gathered}
    1{\text{ if }}i = j \hfill \\
    0{\text{ otherwise}} \hfill \\
    \end{gathered} \right.
    $ and therefore: $\displaystyle
    \left( {\left[ T \right]_B } \right) \cdot \left( {\left[ T \right]_B } \right)^t = {\text{Id}}_{{\text{n}} \times {\text{n}}}
    $ and the proof of the direct is complete

    Converse: $\displaystyle \left( {\left[ T \right]_B } \right)^{ - 1} = \left( {\left[ T \right]_B } \right)^t
    \Rightarrow
    \left\langle {T\left( \bold{v} \right),T\left( \bold{u} \right)} \right\rangle = \left\langle {\bold{v},\bold{u}} \right\rangle ;
    \forall\bold{u},\bold{v} \in \mathbb{R}^n
    $

    Step 1: Since $\displaystyle \left( {\left[ T \right]_B } \right)^{ - 1}$ then $\displaystyle
    T^{ - 1} :\mathbb{R}^n \to \mathbb{R}^n
    $ exists and $\displaystyle
    \left[ {T^{ - 1} } \right]_B = \left( {\left[ T \right]_B } \right)^t
    $

    Step 2: $\displaystyle
    \left[ T^{-1}\right]_B = \left( {\begin{array}{*{20}c}
    {\left\langle {T^{-1}(\bold{v}_1) ,\bold{v}_1 } \right\rangle } & \cdots & {\left\langle {T^{-1}(\bold{v}_n) ,\bold{v}_1 } \right\rangle } \\
    \vdots & \ddots & \vdots \\
    {\left\langle {T^{-1}(\bold{v}_1) ,\bold{v}_n } \right\rangle } & \cdots & {\left\langle {T^{-1}(\bold{v}_n) ,\bold{v}_n } \right\rangle } \\

    \end{array} } \right)
    $, by the symmetry of the inner-products over the filed of the real numbers : $\displaystyle
    \left[ T^{-1}\right]_B = \left( {\begin{array}{*{20}c}
    {\left\langle { \bold{v}_1 ,T^{-1}(\bold{v}_1)} \right\rangle } & \cdots & {\left\langle {\bold{v}_1 ,T^{-1}(\bold{v}_n) } \right\rangle } \\
    \vdots & \ddots & \vdots \\
    {\left\langle { \bold{v}_n, T^{-1}(\bold{v}_1)} \right\rangle } & \cdots & {\left\langle { \bold{v}_n, T^{-1}(\bold{v}_n)} \right\rangle } \\

    \end{array} } \right)
    $ and step 1 implies $\displaystyle
    \left( {\begin{array}{*{20}c}
    {\left\langle { \bold{v}_1 ,T^{-1}(\bold{v}_1)} \right\rangle } & \cdots & {\left\langle {\bold{v}_1 ,T^{-1}(\bold{v}_n) } \right\rangle } \\
    \vdots & \ddots & \vdots \\
    {\left\langle { \bold{v}_n, T^{-1}(\bold{v}_1)} \right\rangle } & \cdots & {\left\langle { \bold{v}_n, T^{-1}(\bold{v}_n)} \right\rangle } \\

    \end{array} } \right)
    $$\displaystyle
    = \left( {\begin{array}{*{20}c}
    {\left\langle {T(\bold{v}_1) ,\bold{v}_1 } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_1) ,\bold{v}_n } \right\rangle } \\
    \vdots & \ddots & \vdots \\
    {\left\langle {T(\bold{v}_n) ,\bold{v}_1 } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_n) ,\bold{v}_n } \right\rangle } \\

    \end{array} } \right)
    $ hence: $\displaystyle
    \left\langle {T\left( {v_i } \right),v_j } \right\rangle = \left\langle {v_i ,T^{ - 1} \left( {v_j } \right)} \right\rangle
    $ for all $\displaystyle
    1 \leqslant i,j \leqslant n
    $ (2)

    Step 3. Prove that (2) implies $\displaystyle
    \left\langle {T\left( v \right),u} \right\rangle = \left\langle {v,T^{ - 1} \left( u \right)} \right\rangle
    $ for all $\displaystyle u,v\in \mathbb{R}^n$

    Step 4: $\displaystyle
    \left\langle {T\left( v \right),T\left( u \right)} \right\rangle = \left\langle {v,T^{ - 1} \left( {T\left( u \right)} \right)} \right\rangle = \left\langle {v,u} \right\rangle
    $ for all $\displaystyle u,v\in \mathbb{R}^n$ and the proof is complete
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Algebra: Linear Independence question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 3rd 2011, 06:28 AM
  2. Replies: 3
    Last Post: Oct 17th 2010, 06:13 PM
  3. another question about eigenvalues and linear operators
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jul 29th 2009, 03:55 PM
  4. boundedness and linear operators question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jan 22nd 2009, 01:27 AM
  5. Linear Algebra- Normal and Self-Adjoint Operators
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Dec 5th 2007, 04:13 AM

Search Tags


/mathhelpforum @mathhelpforum