Let $${A_{n \times n}} = \left[ {\begin{array}{*{20}{c}} {-2}&{1}&{}&{}&{}\\ {1}&{-2}&{1}&{}&{}\\ {}&{1}&{\ddots}&{\ddots}&{}\\ {}&{}&{\ddots}&{\ddots}&{1}\\ {}&{}&{}&{1}&{-2} \end{array}} \right]$$ be a tridiagonal matrix. How we can prove that its inverse is the matrix $B=(b_{ij})$ where $$b_{ij}=-\frac{i(n+1-j)}{n+1} \; ,\quad i\leq j.$$

  • 6
    $\begingroup$ Compute each entry of $AB$ explicitly? $\endgroup$ – hmakholm left over Monica Jan 2 '15 at 16:27
  • 1
    $\begingroup$ And $b_{ij} = 0$ for $i>j$? $\endgroup$ – user14717 Jan 2 '15 at 16:43
  • 2
    $\begingroup$ @NickThompson $A$ is symmetric so $B$ is as well ($b_{ij}=b_{ji}$ for $i>j$). $\endgroup$ – Algebraic Pavel Jan 2 '15 at 16:54
  • 1
    $\begingroup$ @MathMan: By the rule for matrix multiplication, each entry of $AB$ is the sum of no more than three terms. You can write down explicit formulas for them and simplify. $\endgroup$ – hmakholm left over Monica Jan 2 '15 at 17:32
  • 1
    $\begingroup$ It is quite straightforward (but rather tedious) to derive this formula using the fact that $\det A_{n\times n}=(-1)^n(n+1)$ (which follows from this recursion) and the Cramer's rule by expanding the minors by the suitable rows/columns. $\endgroup$ – Algebraic Pavel Jan 2 '15 at 17:46

As Henning points out, checking that two matrices are inverses is much easier than computing an inverse; all we need to do is find the product of the matrices. Also, note that your definition of $B$ is incomplete, but we can deduce the rest since $B$ must be symmetric.

By definition, the $i,j$ of the matrix product $AB$ is given by $$ [AB](i,j) = \sum_{k=1}^n a_{ik}b_{kj} $$ in the case of $i=1$, we have $$ [AB](1,j) = \sum_{k=1}^n a_{1k}b_{kj} = \\ \begin{cases} -2\cdot\frac{1(n+1 - j)}{n+1} + 1 \cdot \frac{2(n + 1 - j)}{n+1} & j \geq 2\\ -2 \cdot \frac{1(n+1-j)}{n+1} + 1 \cdot \frac{j(n + 1 - 2)}{n+1} & j=1 \end{cases} = \delta_{ij} $$ We can repeat a similar computation for the cases $2 \leq i \leq n-1$ and $i = n$.

Alternatively: we can find the characteristic polynomial of $A$ using induction (Pavel's method). From there, we could find the inverse using the Cayley Hamilton theorem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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