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Does anybody have a solution to the given word problem below?

Let A be a lower triangular n x n matrix with nonzero entries on the diagonal. Show that A is invertible and and that A-inverse is lower triangular. [HINT: Explain why A can be changed into I using only row replacements and scaling. (Where are the pivots?) Also, explain why the row operations that reduce A to I change I into a lower triangular matrix.]

Big Hint: Think about row reducing [A I]

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Here's a somewhat more abstract way to do it which doesn't use row reduction technology:

First of all, observe that if $A_0$ is strictly lower triangular, and $D$ is diagonal, then both $A_0D$ and $DA_0$ are strictly lower triangular. This is very easy demonstrate: note that if $B$ and $C$ are any two square matrices, and $c_1, c_2, . . . , c_n$ are the columns of $C$, so that we may write $C$ in columnar form

$C = [c_1 c_2 . . . c_n], \tag{1}$

then

$BC = [Bc_1 Bc_2 . . . Bc_n], \tag{2}$

that is, $BC$ is given by simply multiplying each column of $C$ by $B$; likewise, if $B$ is presented in terms of its rows $b_1, b_2, . . . , b_n$:

$B = \begin{bmatrix} b_1 \\ b_2 \\ . \\ . \\ . \\ b_n \end{bmatrix}, \tag{3}$

then we have

$BC = \begin{bmatrix} b_1C \\ b_2C \\ . \\ . \\ . \\ b_nC \end{bmatrix}, \tag{4}$

so we see that $BC$ may be had by right multiplying each row of $B$ by $C$. If we apply these concepts to the lower triangular matrix $A_0$ and the diagonal matrix $D$, it is easy to see by inspection that both $DA_0$ and $A_0D$ are in fact strictly lower triangular. Indeed column $j$ of $A_0$, $a_{0j}$, has the property that $a_{0jk} =0$ for $1 \le k \le j$; from this observation it is easy to see that $Da_{0j}$ shares this property for diagonal $D$, that is $(Da_{0j})_k = 0$ for the same range of $k$. An effectively identical argument applies to the rows under right multiplication by $D$; and indeed, this demonstration can easily be extended to cover the case $A$ lower triangular though not strictly so, and with no essential modification to the classes of upper triangular matrices, strict and non-strict, as well. It is also possible to establish these facts for either upper or lower triangular matrices and then use matrix transposition to argue the other case, e.g. $A$ upper triangular implies $A^T$ lower triangular implies $DA^T$, $A^TD$ lower triangular implies $(DA^T)^T = AD^T = AD$ etc. upper triangular. In any event, the point is that these sets of triangular matrices are each closed under left or right multiplication by diagonal matrices.

Having first established that multiplication by diagonal matrices preserves triangularity, the second thing that should be mentioned is that these results can be generalized to affirm that, for any two lower triangular matrices $A$ and $B$, both $AB$ and $BA$ are lower triangular. That is, the set of lower triangular matrices is closed under multiplication; and of course the corresponding result holds for upper triangular matrices. This may easily be seen by a straightforward extension of the technique used in the previous paragraph to show $DA$ etc. are lower triangular; that is, we can consider the lower triangular matrix $A$ in columnar form

$A = [a_1 a_2 . . . a_n], \tag{5}$

and use

$BA = [Ba_1 Ba_2 . . . Ba_n]; \tag{6}$

from (6), it is easy to see that if $B$ is also lower triangular, then the column vectors $Ba_i$ have all zero entries above the diagonal of $BA$; due to their triangularity, the zeroes of $A$ and $B$ are arranged in such a manner that all products $Ba_j$ have components satisfying $(Ba_j)_k = 0$ for $1 \le k \le j$. The analogous assertions hold for $A$, $B$ strictly lower triangular and upper and strictly upper triangular; all these classes of matrices are closed under multiplication. And in fact, the transposition trick employed above works in this case as well: if $A$ and $B$ are upper triangular, then $A^T$ and $B^T$ are lower triangular, hence is $B^TA^T$; thus $(B^TA^T)^T = AB$ is upper triangular etc.

Returning then to the problem at hand, we see that $A$ may be expressed in the form

$A = D + A_0, \tag{5}$

with $A_0$ strictly lower triangular and $D$ diagonal. Then we may write

$A= D(I + D^{-1}A_0), \tag{6}$

where as we have seen $D^{-1}A_0$ is strictly lower triangular. As such, it is nilpotent of degree at most $n = \text{size}(A)$; $(D^{-1}A_0)^n = 0$. But then $I + D^{-1}A_0$ is invertible, since we have the identity

$(I + B)\sum_0^m (-1)^k B^k = I + (-1)^mB^{m + 1}, \tag{7}$

which holds for any matrix $B$; since $D^{-1}A_0$ is nilpotent with $(D^{-1}A_0)^n = 0$, if we take $B = D^{-1}A_0$ with $m = n$ in (7) we obtain

$(I + D^{-1}A_0) \sum_0^n (-1)^k(D^{-1}A_0)^k = I, \tag{8}$

which shows that $I + D^{-1}A_0$ is invertible with inverse $\sum_0^n(-1)^k (D^{-1}A_0)^k$. Since every term in (8) with $k \ge 1$ is strictly lower triangular, by virtue of the fact that the product of such matrices is again such a matrix, and the term $(D^{-1}A_0)^0 = I$, we see that $(I + D^{-1}A_0)^{-1}$ is itself lower triangular. Finally, note that $(I + D^{-1}A_0)^{-1}D^{-1}$ is also lower triangular, by what we have seen, and is in fact the inverse of $A$:

$(I + D^{-1}A_0)^{-1}D^{-1}A = (I + D^{-1}A_0)^{-1}D^{-1}(D(I + D^{-1}A_0)) = (I + D^{-1}A_0)^{-1}(I + D^{-1}A_0) = I. \tag{9}$

We have thus shown that $A = D(I + D^{-1}A_0)$ has a lower triangular inverse $A^{-1} = (I + D^{-1}A_0)^{-1}D^{-1}$, and in so doing we have seen how this result generalizes to include upper triangular matrices having no zeroes on the diagonal as well. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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Do you believe that the product of two lower triangular matrices is again a lower triangular matrix? Row reducing $[A\;I]$ amounts to multiplying $[A\;I]$ by certain matrices which correspond to elementary row operations. The only matrices you need for this particular row reduction are all lower triangular...

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