# Linear Algebra Question - Linear Operators

• March 2nd 2009, 09:04 PM
r2dee6
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
• March 3rd 2009, 04:33 AM
PaulRS
Quote:

Originally Posted by r2dee6
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 $\mathbb{R}$- $
T:V \to V
$
- V a vector space- such that $
\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 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 $
B = \left\{ {\bold{v}_1 ,...,\bold{v}_n } \right\}
$
be the standard basis for $\mathbb{R}^n$ (i.e. $
\bold{v}_1 = \left( {1,0,0...,0} \right);\bold{v}_2 = \left( {0,1,0...,0} \right)...
$
). This is an orthonormal basis, so $
\bold{v} = \sum\limits_{k = 1}^n {\left\langle {\bold{v},\bold{v}_k } \right\rangle \cdot \bold{v}_k }
$
for each $\bold{v} \in \mathbb{R}^n$ (1)

Let us compute $
\left[ T \right]_B
$
. By (1) we have: $
\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: $
\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 $
\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 $
\left( {\left[ T \right]_B } \right)^{ - 1} = \left( {\left[ T \right]_B } \right)^t
$

Direct: $
\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: $
\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: $
\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 $
\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 $
\left\{ {T\left( {\bold{v}_1 } \right);...;T\left( {\bold{v}_n } \right)} \right\}
$
is an orthonormal basis.

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

Step 2: $
\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 : $
\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 $
\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)
$
$
= \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: $
\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 $
1 \leqslant i,j \leqslant n
$
(2)

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

Step 4: $
\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 $u,v\in \mathbb{R}^n$ and the proof is complete