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Is the inverse of an invertible and lower triangular matrix still both lower triangular and invertible?

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marked as duplicate by Marc van Leeuwen, Davide Giraudo, user61527, Nick Peterson, Thomas Andrews Nov 29 '13 at 19:33

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Yes. The lower triangular matrix $L$ is invertible if and only if its determinant is nonzero, and since its determinant is the product of its diagonal entries, if and only if its diagonal entries are nonzero.

If we show the sum and product of lower triangular matrices is again lower triangular, then the proof goes quickly from there.

The sum of two lower triangular matrices is pretty evident, so let's jump into the product of two $n\times n$ lower triangular matrices $L$ and $M$. Consider an entry of the product above the main diagonal, i.e. row $i$ and column $j$ with $i \lt j$, obtained by multiplying the $i$th row of $L$ and the $j$th column of $M$:

$$ \sum_{k=1}^n L(i,k) M(k,j) $$

Notice that for $k \gt i$ the factor $L(i,k)=0$, while for $k \lt j$, the factor $M(k,j)=0$. Use the fact that $i \lt j$ and it turns out all the summands in the above expression are zero.

Now one way to finish the proof is by appealing to Cayley-Hamilton for an expression of $L^{-1}$ as a polynomial in $L$. From what we just showed, any polynomial in lower triangular $L$ is again lower triangular, and Cayley-Hamilton for invertible $L$ gives:

$$ L (L^{n-1} + \ldots ) = \pm \det(L) $$

$$ L^{-1} = \frac{1}{\pm \det(L)} (L^{n-1} + \ldots ) $$

showing the inverse of $L$ is indeed a polynomial in $L$.

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yes. Because the set of invertible lower triangular matrices forms a subgroup of $GL(n)$.

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    $\begingroup$ How do you prove that, without proving what the OP asked? $\endgroup$ – Vedran Šego Nov 29 '13 at 15:14
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Let $A$ be a lower triangular matrix with invertible diagonal entries. Then for any given column vector $b$ the equation $A\cdot x=b$, for an unknown vector $x$, gives a lower triangular system of linear equations for the coordinates of$~x$. It can be solved in a straightforward way from top to bottom.

For instance for $$A=\begin{pmatrix}a_{1,1}&0&0\\ a_{2,1}&a_{2,2}&0\\ a_{3,1}&a_{3,2}&a_{3,3}\\\end{pmatrix}$$ one gets the system $$ \begin{align} &a_{1,1}x_1&=b_1\\ &a_{2,1}x_1{}+a_{2,2}x_2&=b_2\\ &a_{3,1}x_1+a_{3,2}x_2+a_{3,3}x_3&=b_3\\ \end{align} $$ that can be solved as $$ \begin{align} x_1&=a_{1,1}^{-1}b_1,\\ x_2&=a_{2,2}^{-1}(-a_{2,1}x_1+b_2),\\ x_3&=a_{3,3}^{-1}(-a_{3,1}x_1-a_{3,2}x_2+b_3).\\ \end{align} $$ Clearly if $b_1=b_2=\cdots=b_k=0$ for some $k$, one will also find $x_1=x_2=\cdots=x_k=0$. Now applying this repeatedly, where the column vector $b$ runs through the standard basis, one finds as the solutions$~x$ of $A\cdot x=b$ the successive columns of $A^{-1}$; what I have said implies that these columns form a lower triangular matrix.

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