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The exercise is to show that $\sum_{i=0}^n \|u_i\|^2 \ge 1$, knowing that the set of $(u_i+e_i)$ in $\mathbb R^n$ is LD, where $e_i$ is the canonical basis.

I know that $e_i$ are orthogonal vectors of modulus $1$, and that being LD means that at least one of the vectors $(u_i+e_i)$ can be expressed by a linear combination of the others. From this I tried isolating $e_i$, and thinking of inner product to use something like Cauchy-Schwarz to get an inequality, but was unable to develop the solution.

Alexandre Tourinho
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2 Answers2

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By contraposition: Suppose that $\sum\limits_{i=1}^n \|u_i\|^2<1$. If $A$ is the matrix whose columns are $u_1,\ldots,u_n$, this means that the Frobenius norm of $A$ is less than $1$, which implies that $A+I$ is invertible, with inverse $\sum\limits_{k=0}^\infty (-1)^kA^k$. The columns of $A+I$ are $u_1+e_1,\ldots,u_n+e_n$, and they are linearly independent because $A+I$ is invertible.

Related: Prove that $v_1, \dots v_n$ is a basis of V.

Jonas Meyer
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  • This Frobenius norm looks like a powerful tool, but I am just starting to learn Linear Algebra. The answear seems very succint, but maybe too much in terms of the fact that haven't studied matrices yet, so I think this can be proved only with basic introductory Linear Algebra concepts, what I am unable to do. I also made a mistake and the terms of the sum are squared. – Alexandre Tourinho Mar 15 '19 at 21:18
  • @AlexandreTourinho: I think mechanodroid's answer looks better for you. – Jonas Meyer Apr 16 '19 at 01:14
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Let $A$ be the matrix with columns $u_1, \ldots, u_n$. The assumption that $\{u_1+e_1, \ldots, u_n+e_n\}$ is linearly dependent means that $A+I$ is not invertible.

Hence $A^* + I = (A+I)^*$ is also not invertible so pick an eigenvector $x_1 \in \mathbb{R}^n, \|x_1\| = 1$ such that $A^*x_1 = -x_1$. Extend $\{x_1\}$ to an orthonormal basis $\{x_1, \ldots, x_n\}$ for $\mathbb{R}^n$.

Notice that $u_i = Ae_i$ so Parseval's identity yields \begin{align} \sum_{i=1}^n \|u_i\|^2 &= \sum_{i=1}^n \|Ae_i\|^2 \\ &= \sum_{i=1}^n \sum_{j=1}^n \left|\langle Ae_i, x_j\rangle\right|^2 \\ &= \sum_{j=1}^n \sum_{i=1}^n |\langle A^*x_j, e_i\rangle|^2 \\ &= \sum_{j=1}^n \|A^*x_j\|^2 \\ &\ge \|A^*x_1\|^2 \\ &= \|-x_1\|^2 \\ &= 1 \end{align}

mechanodroid
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