# Linear Algebra Question - Linear Operators

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

Originally Posted by r2dee6
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