Question

Show that if L is invertible and lower triangular, then so is L^(-1).

Answer 1

Suppose $L$ is invertible and lower triangular matrix. We need to prove that $L^{-1}$ is also invertible and low triangular.

Since $L$ is invertible there exists inverse matrix $L^{-1}$ satisfying

$$L L ^ {- 1} = L ^ {- 1} L = I$$

where $I$ is identity matrix. Changing $L$ into $L^{-1}$ in the equalities we get:

$$L ^ {- 1} (L ^ {- 1}) ^ {- 1} = (L ^ {- 1}) ^ {- 1} L ^ {- 1} = I$$

The following property holds for an invertible matrix $L: (L^{-1})^{-1} = L$. Thus the inverse matrix to $L^{-1}$ exists, so $L^{-1}$ is invertible matrix.

Let's prove that $L^{-1}$ is lower triangular matrix.

Suppose $L^{-1} = [y_{1}y_{2}\dots y_{n}]$, so $y_{i}$ is the $i - th$ column of the matrix $L^{-1}$.

Then, by definition,

$$L L ^ {- 1} = I$$ $$L L ^ {- 1} = L \left[ y _ {1} y _ {2} \dots y _ {n} \right] = \left[ L y _ {1} \dots L y _ {n} \right] = \left[ e _ {1} e _ {2} \dots e _ {n} \right]$$

So

$$L y _ {k} = e _ {k}, k = \overline {{1 , n}}$$

Denoting elements of the matrix $L$ by $l_{ij}$ and elements of the vector $y_j$ by $y_j^i$ we get:

$$l _ {1 1} y _ {k} ^ {1} = 0$$ $$l _ {2 1} y _ {k} ^ {1} + l _ {2 2} y _ {k} ^ {2} = 0$$

...

$$l _ {k - 1 1} y _ {k} ^ {1} + \dots + l _ {k - 1 k - 1} y _ {k} ^ {k - 1} = 0$$ $$l _ {k 1} y _ {k} ^ {1} + \dots + l _ {k k} y _ {k} ^ {k} = 1$$

Since $L$ is lower triangular matrix then $l_{11} \neq 0$.

The first equation implies $y_{k}^{1} = 0$, then we obtain that the second one implies

$y_{k}^{2} = 0$, ..., the k-1-th equation implies $y_{k}^{k - 1} = 0$. Thus all the elements of the matrix $L^{-1}$ that are located above the main diagonal are equal to 0, so $L^{-1}$ is lower triangular matrix.