I am trying to find a simple proof that $\|v \otimes u\| = 1 $ if $\|v\|=1$ and $\|u\|=1$.

I have a proof by induction, where I can fix the length of $u$ and show by induction on the length of $v$ that the previous statement is true. The base case is for length of $v=2$.

Let $u = [\alpha_1, \alpha_2]$ and $v=[v_1, \dots, v_n]$, we get a formula of the kind

$$ \sum_{i=1}^n (\alpha_1v_i)^2 + \sum_{i=1}^n (\alpha_2v_i)^2 = \sum_{i=1}^n v_i^2(\alpha_1+\alpha_2)^2 = 1$$

Do you have a another idea, maybe using properties of the tensor product or properties of the norm?


1 Answer 1


Let $\mathbf{u}$ and $\mathbf{v}$ be two vectors, which we write as: $$\mathbf{u}=\sum_{j=1}^nu_j\mathbf{e}_j$$ and: $$\mathbf{v}=\sum_{j=1}^nv_j\mathbf{e}_j$$ with $\left\{\mathbf{e}_j\right\}_j$ being an orthonormal basis of the Hilbert space we're working in. We thus have: $$\begin{align}\mathbf{u}\otimes\mathbf{v}&=\left[\sum_{j=1}^nu_j\mathbf{e}_j\right]\otimes\left[\sum_{k=1}^nv_k\mathbf{e}_k\right]\\&=\sum_{j=1}^nu_j\left[\mathbf{e}_j\otimes\sum_{k=1}^nv_k\mathbf{e}_k\right]\\&=\sum_{j=1}^n\sum_{k=1}^nu_jv_k\left[\mathbf{e}_j\otimes\mathbf{e}_k\right]\end{align}$$ We know want to compute the squared norm of this vector. Note that $\left\{\mathbf{e}_j\otimes\mathbf{e}_k\right\}_{j,k}$ is an orthonormal basis of the new Hilbert space we're working in. $$\begin{align}\|\mathbf{u}\otimes\mathbf{v}\|^2&=\sum_{j=1}^n\sum_{k=1}^n\overline{u_jv_k}u_jv_k\\&=\underbrace{\left(\sum_{j=1}^n\overline{u_j}u_j\right)}_{\|\mathbf{u}\|^2}\underbrace{\left(\sum_{k=1}^n\overline{v_k}v_k\right)}_{\|\mathbf{v}\|^2}\end{align}$$ Thus, since $\sqrt{\|\mathbf{u}\|^2\times\|\mathbf{v}\|^2}=\sqrt{\|\mathbf{u}\|^2}\times\sqrt{\|\mathbf{v}\|^2}=\|\mathbf{u}\|\times\|\mathbf{v}\|$, it follows that if $\mathbf{u}$ and $\mathbf{v}$ are unit vectors, then so is $\mathbf{u}\otimes\mathbf{v}$.


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