# Take a matrix with normalized rows; no column of the inverse has too large a norm

For $n>2$, let $X$ be an $n$-by-$n$ invertible matrix where every row has Euclidean norm $1$. Let $Y=X^{-1}$. Let $\Vert y_i \Vert$ be the Euclidean norm of column $i$ of $Y$.

The following conjecture seems to be confirmed by numerical evidence: for every $i$ $$\frac{\Vert y_i \Vert}{\sum_{j=1}^n \Vert y_j \Vert} < \frac{1}{2}$$

How can we prove this?

Letting $z_i$ denote row $i$ of $X$, we know that $z_i \cdot y_i = 1$, so by Cauchy-Schwarz, $\Vert z_i \Vert \cdot \Vert y_i \Vert \geq 1$, implying that each column of $Y$ has norm at least $1$.

• For $n=3$ it seems to be true as it says something about the areas of the sides of a parallelepiped. – Wilberd van der Kallen Feb 8 '17 at 9:13

Indeed, if you think of it, it can be restated in geometric terms. What we need to prove is that the inverse altitudes of a parallelepiped spanned by $n+1$ unit vectors in $\mathbb R^{n+1}$ satisfy the "triangle inequality" (each inverse altitude does not exceed the sum of the rest). Let $v$ be one of the given unit vectors and $v_j$ be the rest. If $h$ is the altitude vector for $v$, then $v-r=h$ where $r=\sum_j a_j v_j$ and $r$ is orthogonal to $h$. Notice that we can use the same linear combination $v-r$ divided by $a_j$ to estimate the altitude $H_j$ for the vector $v_j$ from above by $\|h\|/|a_j|$ and that we can improve that estimate to $\|h-\langle h,v\rangle v\|/|a_j|$ if we modify the coefficient at $v$. Thus it would suffice to show that $$\frac{\|h-\langle h,v\rangle v\|}{\|h\|}\le \sum_j|a_j|\,.$$ However the left hand side is the sine of the angle between $v$ and $h$ in the right triangle with sides $v,h,r$, so it equals $\|r\|$. It remains to combine the definition of $r$ with the triangle inequality.