Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$A$ is an invertible $n \times n$ matrix. Interpret each row of $A$ as a point in $\mathbb{R}^n$; then these $n$ points define a unique hyperplane in $\mathbb{R}^n$ that passes through each point (this hyperplane does not intersect the origin).

Under this geometric interpretation, $A^{-1}$ has an interesting property: the normal vector to the hyperplane is given by the row sums of $A^{-1}$ (i.e. $A^{-1} * 1$, where $1 = \langle 1, \dots, 1 \rangle^T$).

Within this same geometric interpretation of $A$, what other interesting properties does $A^{-1}$ have? Do the individual entries of $A^{-1}$ have geometric meaning? How about the column sums (besides the obvious row sums of $A^T$ intepretation)?

share|improve this question
add comment

1 Answer

Here is one simple geometric relationship: if $X_1,\ldots,X_n$ are the rows of $A$, and $C_1,\ldots,C_n$ are the columns of $A^{-1}$, then the length of the column $C_i$ is equal to the reciprocal of the distance between $X_i$ and the hyperplane $V_i$ spanned by the other $n-1$ rows $X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n$ of $A$. This is because $C_i$ is orthogonal to $V_i$ and has an inner product of $1$ with $X_i$. If one then square sums in $i$, we obtain the negative second moment identity

$$ \sum_{i=1}^n \sigma_i(A)^{-2} = \sum_{j=1}^n \hbox{dist}(X_j,V_j)^{-2}$$

where $\sigma_1(A),\ldots,\sigma_n(A)$ are the singular values of $A$, which turns out to be a useful identity in random matrix theory (see e.g. this blog post of mine). In particular, it highlights the importance of understanding the distance between a row and the hyperplane spanned by the other rows if one is to get some control on the small singular values.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.