1. ## Upper Triangular matrix

How do I show that the inverse of an upper triangular matrix is also upper triangular? Thanks

2. ## Re: Upper Triangular matrix

Originally Posted by jzellt
How do I show that the inverse of an upper triangular matrix is also upper triangular? Thanks
One way: if $A=(a_{ij})$ is an invertible upper triangular matrix then $A^{-1}=\dfrac{1}{\det A}\textrm{Adj}\;(A)$ . If $A^{-1}=(b_{ij})$ , you can easily verify that $b_{ij}=0$ for $i>j$ .

3. ## Re: Upper Triangular matrix

Another way: write $A = D + N$, where D is the diagonal part of A and N is the strictly upper-diagonal part of A. If A is invertible then so is D (because A and D have the same—nonzero—determinant); and $D^{-1}$ is diagonal and hence upper-triangular. The strictly upper-diagonal matrix N is nilpotent, with $N^n=0$ if A is an $n\times n$ matrix. You can check that

$A^{-1} = D^{-1} - D^{-1}ND^{-1} + D^{-1}N^2D^{-1} -\ldots + (-1)^{n-1}D^{-1}N^{n-1}D^{-1}$

(just multiply that by $D+N$ and check that you get the identity).

The result then follows from the fact that sums and products of upper-diagonal matrices are upper-diagonal.

4. ## Re: Upper Triangular matrix

Very nice Opalg. Another way, denoting

$A=(a_{ij})\;,\;A^{-1}=\begin{bmatrix}C_1,C_2,\hdots,C_n\end{bmatrix}, \;I=\begin{bmatrix}I_1,I_2,\hdots,I_n\end{bmatrix}$

Then,

$\begin{bmatrix} C_1,C_2,\hdots,C_n\end{bmatrix}\begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n}\\ 0 &a_{22} & \ldots & a_{2n} \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & a_{nn}\end{bmatrix}=\begin{bmatrix}I_1,I_2,\hdots, I_n\end{bmatrix}\Leftrightarrow \ldots$

We get a triangular system on the unknowns $C_i$ and easily proved: $C_i\in$ that is, $A^{-1}$ is an upper triangular matrix.