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 $1$.



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


Certainly this all should be very well known since 18.., so if someone can carry out the corresponding literature search and point out the reference, it would be great :-).


