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\begin{pmatrix} 1 & .. &.. &.. \\ .. & 1 &.. &.. \\ .. &l_{k+1,k} &.. &..\\ .. &.. .. &.. &.. \\ .. &l_{n,k} &.. &1 \end{pmatrix}

the inverse is $L_i*$

\begin{pmatrix} 1 &.. &.. &.. \\ .. &.. 1 &.. &.. \\ .. &-l_{k+1,k} &.. &..\\ .. &.. .. &.. &.. \\ .. &-l_{n,k} &.. &1.. \end{pmatrix}

how can I show in a nice formal way that indeed $L_i * L_i* = I$?

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  • $\begingroup$ Is multiplying the matrices formal enough? You should add $0$'s where they belong. As they are, they don't look like atromic matrices. $\endgroup$
    – Git Gud
    Mar 11, 2013 at 22:16

3 Answers 3

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Write $L_i$ as $I + A$, where $A$ is the same as $L_i$ but with all $0$ on the diagonal, and do the same with $L_i^*$. Now it's quite easy:

$L_i*L_{i}^* = (I+A)(I+A^*) = I + ( A + A^* ) + AA^*$

now, the sum in the brackets is obviously $0$, and the product $AA^* = 0 $ too ( should be quite easy to prove)

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Let $a = (0,..,0,l_{i+1,i},..,l_{n,i})$.

Than $$(I + e_i a^T)(I-e_ia^T) = I+e_ia^T e_ia^T$$

But $a^Te_i=0$.

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More generally, for any block lower triangular matrix of the form $$ M=\begin{pmatrix} I_k&0\\ F&I_{n-k} \end{pmatrix}, $$ we have $$ M^{-1}= \begin{pmatrix} I_k&0\\ -F&I_{n-k} \end{pmatrix}. $$ In your particular case, $$ \quad F=\begin{pmatrix} 0&\ldots&0&l_{k+1,k} \\ 0&\ldots&0&0 \\ \vdots&&\vdots&\vdots \\ 0&\ldots&0&0 \\ 0&\ldots&0&l_{n,k} \end{pmatrix}. $$

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