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    Upper Triangular matrix

    How do I show that the inverse of an upper triangular matrix is also upper triangular? Thanks
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Upper Triangular matrix

    Quote Originally Posted by jzellt View Post
    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 .
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    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 sameŚnonzeroŚdeterminant); 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.
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    MHF Contributor FernandoRevilla's Avatar
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    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<I_1,\ldots I_i> that is, A^{-1} is an upper triangular matrix.
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