Tag Archives: inner-product-space

Algebra

Non linear map preserving Euclidean norm

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Let \(V\) be a real vector space endowed with an Euclidean norm \(\Vert \cdot \Vert\).

A bijective map \( T : V \to V\) that preserves inner product \(\langle \cdot, \cdot \rangle\) is linear. Also, Mazur-Ulam theorem states that an onto map \( T : V \to V\) which is an isometry (\( \Vert T(x)-T(y) \Vert = \Vert x-y \Vert \) for all \(x,y \in V\)) and fixes the origin (\(T(0) = 0\)) is linear.

What about an application that preserves the norm (\(\Vert T(x) \Vert = \Vert x \Vert\) for all \(x \in V\))? \(T\) might not be linear as we show with following example:\[
\begin{array}{l|rcll}
T : & V & \longrightarrow & V \\
& x & \longmapsto & x & \text{if } \Vert x \Vert \neq 1\\
& x & \longmapsto & -x & \text{if } \Vert x \Vert = 1\end{array}\]

It is clear that \(T\) preserves the norm. However \(T\) is not linear as soon as \(V\) is not the zero vector space. In that case, consider \(x_0\) such that \(\Vert x_0 \Vert = 1\). We have:\[
\begin{cases}
T(2 x_0) &= 2 x_0 \text{ as } \Vert 2 x_0 \Vert = 2\\
\text{while}\\
T(x_0) + T(x_0) = -x_0 + (-x_0) &= – 2 x_0
\end{cases}\]

Algebra

Non linear map preserving orthogonality

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Let \(V\) be a real vector space endowed with an inner product \(\langle \cdot, \cdot \rangle\).

It is known that a bijective map \( T : V \to V\) that preserves the inner product \(\langle \cdot, \cdot \rangle\) is linear.

That might not be the case if \(T\) is supposed to only preserve orthogonality. Let’s consider for \(V\) the real plane \(\mathbb R^2\) and the map \[
\begin{array}{l|rcll}
T : & \mathbb R^2 & \longrightarrow & \mathbb R^2 \\
& (x,y) & \longmapsto & (x,y) & \text{for } xy \neq 0\\
& (x,0) & \longmapsto & (0,x)\\
& (0,y) & \longmapsto & (y,0) \end{array}\]

The restriction of \(T\) to the plane less the x-axis and the y-axis is the identity and therefore is bijective on this set. Moreover \(T\) is a bijection from the x-axis onto the y-axis, and a bijection from the y-axis onto the x-axis. This proves that \(T\) is bijective on the real plane.

\(T\) preserves the orthogonality on the plane less x-axis and y-axis as it is the identity there. As \(T\) swaps the x-axis and the y-axis, it also preserves orthogonality of the coordinate axes. However, \(T\) is not linear as for non zero \(x \neq y\) we have: \[
\begin{cases}
T[(x,0) + (0,y)] = T[(x,y)] &= (x,y)\\
\text{while}\\
T[(x,0)] + T[(0,y)] = (0,x) + (y,0) &= (y,x)
\end{cases}\]

Algebra

A proper subspace without an orthogonal complement

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We consider an inner product space \(V\) over the field of real numbers \(\mathbb R\). The inner product is denoted by \(\langle \cdot , \cdot \rangle\).

When \(V\) is a finite dimensional space, every proper subspace \(F \subset V\) has an orthogonal complement \(F^\perp\) with \(V = F \oplus F^\perp\). This is no more true for infinite dimensional spaces and we present here an example.

Consider the space \(V=\mathcal C([0,1],\mathbb R)\) of the continuous real functions defined on the segment \([0,1]\). The bilinear map
\[\begin{array}{l|rcl}
\langle \cdot , \cdot \rangle : & V \times V & \longrightarrow & \mathbb R \\
& (f,g) & \longmapsto & \langle f , g \rangle = \displaystyle \int_0^1 f(t)g(t) \, dt \end{array}
\] is an inner product on \(V\).

Let’s consider the proper subspace \(H = \{f \in V \, ; \, f(0)=0\}\). \(H\) is an hyperplane of \(V\) as \(H\) is the kernel of the linear form \(\varphi : f \mapsto f(0)\) defined on \(V\). \(H\) is a proper subspace as \(\varphi\) is not always vanishing. Let’s prove that \(H^\perp = \{0\}\).

Take \(g \in H^\perp\). By definition of \(H^\perp\) we have \(\int_0^1 f(t) g(t) \, dt = 0\) for all \(f \in H\). In particular the function \(h : t \mapsto t g(t)\) belongs to \(H\). Hence
\[0 = \langle h , g \rangle = \displaystyle \int_0^1 t g(t)g(t) \, dt\] The map \(t \mapsto t g^2(t)\) is continuous, non-negative on \([0,1]\) and its integral on this segment vanishes. Hence \(t g^2(t)\) is always vanishing on \([0,1]\), and \(g\) is always vanishing on \((0,1]\). As \(g\) is continuous, we finally get that \(g = 0\).

\(H\) doesn’t have an orthogonal complement.

Moreover we have
\[(H^\perp)^\perp = \{0\}^\perp = V \neq H\]