This post is about how to invert a lower triangular matrix by forward substitution.

Consider a matrix which is the matrix inverse of another sized triangular matrix . We take the definition of matrix inversion as a starting point:

Here is the matrix multiplication of times and is the identity matrix, kind of the equivalent of the number 1 of the natural numbers 1,2,3… but as a matrix.

By writing out elements this could for example look like:

However it is a fact that the inverse of a triangular matrix is also a triangular matrix and therefore we can write the equation as follows:

We then proceed by solving n different equation systems for each column of . Here denotes the n:th column of and the somewhat unorthodox denotes the n:th column of the identity matrix .

Equation system 1: :

Equation system 2: :

Equation system 3: :

Equation system 4: :

Equation system 5: :

Equation system 6: :

Now recall matrix multiplication with this example:

Which has the intermediary step:

That is also how we will rewrite the equations systems:

Equation system 1: :

Equation system 2: :

Equation system 3: :

Equation system 4: :

Equation system 5: :

Equation system 6: :

These equation systems are then solved by forward substitution. See wikipedia Triangular matrix, forward substitution . If the matrix to be inverted would have been upper triangular then we would have arrived at similar equations to be solved by back substitution.

Mats Granvik mats.granvik(AT)abo.fi

Thanks, it helped a lot