Results 1 to 2 of 2

Math Help - 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 \Re^n and let T be a linear operator on \Re^n. Prove that T(n).T(v) = n.v if and ony if 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 \Re^n and let T be a linear operator on \Re^n. Prove that T(n).T(v) = n.v if and ony if 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 \mathbb{R}- <br />
T:V \to V<br />
- V a vector space- such that <br />
\left\langle {T\left( \bold{v} \right),T\left( \bold{u} \right)} \right\rangle  = \left\langle {\bold{v},\bold{u}} \right\rangle <br />
, <br />
\forall\bold{u},\bold{v} \in V<br />
)

    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 <br />
B = \left\{ {\bold{v}_1 ,...,\bold{v}_n } \right\}<br />
be the standard basis for \mathbb{R}^n (i.e. <br />
\bold{v}_1  = \left( {1,0,0...,0} \right);\bold{v}_2  = \left( {0,1,0...,0} \right)...<br />
). This is an orthonormal basis, so <br />
\bold{v} = \sum\limits_{k = 1}^n {\left\langle {\bold{v},\bold{v}_k } \right\rangle  \cdot \bold{v}_k } <br />
for each \bold{v} \in \mathbb{R}^n (1)

    Let us compute <br />
\left[ T \right]_B <br />
. By (1) we have: <br />
\left[ T \right]_B  = \left( {\begin{array}{*{20}c}<br />
{\left\langle {T(\bold{v}_1) ,\bold{v}_1 } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_n) ,\bold{v}_1 } \right\rangle } \\<br />
    \vdots  &  \ddots  &  \vdots   \\<br />
{\left\langle {T(\bold{v}_1) ,\bold{v}_n } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_n) ,\bold{v}_n } \right\rangle } \\<br /> <br />
 \end{array} } \right)<br />
thus: <br />
\left( {\left[ T \right]_B } \right)^t  = \left( {\begin{array}{*{20}c}<br />
{\left\langle {T(\bold{v}_1) ,\bold{v}_1 } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_1) ,\bold{v}_n } \right\rangle } \\<br />
    \vdots  &  \ddots  &  \vdots   \\<br />
{\left\langle {T(\bold{v}_n) ,\bold{v}_1 } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_n) ,\bold{v}_n } \right\rangle } \\<br /> <br />
 \end{array} } \right)<br />

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

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


    Now: <br />
\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 } <br />

    We have: <br />
\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 } <br />

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

    Now: <br />
\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)} <br />
and hence <br />
\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 } <br />

    Thus it follows that: <br />
\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}<br />
  1{\text{ if }}i = j \hfill \\<br />
  0{\text{ otherwise}} \hfill \\ <br />
\end{gathered}  \right.<br />
and therefore: <br />
\left( {\left[ T \right]_B } \right) \cdot \left( {\left[ T \right]_B } \right)^t  = {\text{Id}}_{{\text{n}} \times {\text{n}}} <br />
and the proof of the direct is complete

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

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

    Step 2: <br />
\left[ T^{-1}\right]_B  = \left( {\begin{array}{*{20}c}<br />
{\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 } \\<br />
    \vdots  &  \ddots  &  \vdots   \\<br />
{\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 } \\<br /> <br />
 \end{array} } \right)<br />
, by the symmetry of the inner-products over the filed of the real numbers : <br />
\left[ T^{-1}\right]_B  = \left( {\begin{array}{*{20}c}<br />
{\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 } \\<br />
    \vdots  &  \ddots  &  \vdots   \\<br />
{\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 } \\<br /> <br />
 \end{array} } \right)<br />
and step 1 implies <br />
  \left( {\begin{array}{*{20}c}<br />
{\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 } \\<br />
    \vdots  &  \ddots  &  \vdots   \\<br />
{\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 } \\<br /> <br />
 \end{array} } \right)<br />
<br />
= \left( {\begin{array}{*{20}c}<br />
{\left\langle {T(\bold{v}_1) ,\bold{v}_1 } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_1) ,\bold{v}_n } \right\rangle } \\<br />
    \vdots  &  \ddots  &  \vdots   \\<br />
{\left\langle {T(\bold{v}_n) ,\bold{v}_1 } \right\rangle } & \cdots & {\left\langle {T(\bold{v}_n) ,\bold{v}_n } \right\rangle } \\<br /> <br />
 \end{array} } \right)<br />
hence: <br />
\left\langle {T\left( {v_i } \right),v_j } \right\rangle  = \left\langle {v_i ,T^{ - 1} \left( {v_j } \right)} \right\rangle <br />
for all <br />
1 \leqslant i,j \leqslant n<br />
(2)

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

    Step 4: <br />
\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 <br />
for all 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, 05:28 AM
  2. Replies: 3
    Last Post: October 17th 2010, 05:13 PM
  3. another question about eigenvalues and linear operators
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: July 29th 2009, 02:55 PM
  4. boundedness and linear operators question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 22nd 2009, 12:27 AM
  5. Linear Algebra- Normal and Self-Adjoint Operators
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: December 5th 2007, 03:13 AM

Search Tags


/mathhelpforum @mathhelpforum