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Math and science::Algebra

Inverse of triangular matrix

The inverse of an upper triangular matrix is upper triangular (assuming it's invertible).

The reverse side has an intuitive justification.

Can you generalize the result to not refer to the triangular nature of the matrix? What does it mean to be triangular?


Proof idea

Consider the upper triangular matrix:

\[ A = \begin{bmatrix} a_{1} & a_{2} & a_{3} \\ 0 & b_{2} & b_{3} \\ 0 & 0 & c_{3} \end{bmatrix} \]

The 3rd input has an exclusive route route to the third output via \(c_{3}\). And so, the 3rd input is retrievable by \( \frac{1}{c_{3}} \) times the third output. This lets us partially fill in the inverse matrix:

\[ A^{-1} = \begin{bmatrix} \ast & \ast & \ast \\ \ast & \ast & \ast \\ \ast & \ast & \frac{1}{c_{3}} \end{bmatrix} \]

Now, the 2nd output takes input from the 2nd and 3rd inputs. In this way, the 2nd row of the inverse matrix will have non-zero entries in the 2nd and 3rd columns. Continuing this process, we see that the inverse matrix will be upper triangular—the same as the original.

Generalization

The triangular-ness is not a necessary condition for the result. The triangular-ness is a result of sorted inputs and outputs. If instead we have a matrix like:

\[ A = \begin{bmatrix} a_{1} & a_{2} & a_{3} \\ 0 & 0 & b_{3} \\ 0 & c_{2} & c_{3} \end{bmatrix} \]

Then we again have the situation that one output is exclusively determined by one input, and another output is determined by two inputs, one of which is the input from the first output. This situation also gives rise to the iterative process of filling in rows using the work from the previous row. An inverse won't necessarily be the same shape as the original matrix, but it will still have a 1 element row, a 2 element row, etc.