Sum of Elements of Inverse of Matrix with Column of Ones
This article needs to be linked to other articles. In particular: throughout You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
Theorem
Let $\mathbf B = \sqbrk b_n$ denote the inverse of a square matrix $\mathbf A$ of order $n$.
Let $\mathbf A$ be such that it has a row or column of all ones.
Then the sum of elements in $\mathbf B$ is one:
- $\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{ij} = 1$
Proof
If ones appear in a row of $\mathbf A$, then replace $\mathbf A$ by $\mathbf A^T$ and $\mathbf B$ by $\mathbf B^T$.
Assume $\mathbf A$ has a column of ones.
Apply Sum of Elements of Invertible Matrix to the inverse $\mathbf B = \mathbf A^{-1}$:
- $\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{i j} = 1 - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}$
where $\mathbf J_n$ denotes the square ones matrix of order $n$.
If $\mathbf A = \mathbf B^{-1}$ has a column of ones, then $\mathbf B^{-1} - \mathbf J_n$ has a column of zeros, implying determinant zero.
Substitute $\map \det {\mathbf B^{-1} - \mathbf J_n} = 0$ in Sum of Elements of Invertible Matrix:
- $\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{i j} = 1 - 0$
which implies the statement.
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $43$ (solution)