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The inverse of a non-singular lower triangular matrix is lower triangular.

Construct a proof of this fact as follows. Suppose that $L$ is a non-singular lower triangular matrix. If $b \in \mathbb{R^n}$ is such that $b_i = 0$ for $i = 1, . . . , k \leq n$, and $y$ solves $Ly = b$, then $y_i = 0$ for $i = 1, . . . , k \leq n$.

Hint: partition $L$ by the first $k$ rows and columns.

Can someone tell me what exactly we are showing here and why it will prove that the inverse of any non-singular lower triangular matrix is lower triangular?

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  • $\begingroup$ In order to have an inverse, a matrix must be non-singular. The question as stated doesn't quite make sense: obviously what was meant was "The inverse of a non-singular lower triangular matrix is lower triangular". $\endgroup$ Nov 27, 2012 at 19:25
  • $\begingroup$ Thanks, I've edited that in. $\endgroup$
    – sonicboom
    Nov 27, 2012 at 19:26
  • $\begingroup$ why is this tagged numerical methods? $\endgroup$
    – user31280
    Nov 27, 2012 at 19:42
  • $\begingroup$ @F'OlaYinka: The OP is probably in a Numerical Methods class. $\endgroup$ Nov 27, 2012 at 19:43
  • $\begingroup$ Related. $\endgroup$ Nov 12, 2015 at 1:47

4 Answers 4

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Let's write $$L^{-1}=[y_1\:\cdots\:y_n],$$ where each $y_k$ is an $n\times 1$ matrix.

Now, by definition, $$LL^{-1}=I=[e_1\:\cdots\:e_n],$$ where $e_k$ is the $n\times 1$ matrix with a $1$ in the $k$th row and $0$s everywhere else. Observe, though, that $$LL^{-1}=L[y_1\:\cdots\:y_n]=[Ly_1\:\cdots\: Ly_n],$$ so $$Ly_k=e_k\qquad(1\leq k\leq n)$$

By the proposition, since $e_k$ has only $0$s above the $k$th row and $L$ is lower triangular and $Ly_k=e_k$, then $y_k$ has only $0$s above the $k$th row. This is true for all $1\leq k\leq n$, so since $$L^{-1}=[y_1\:\cdots\:y_n],$$ then $L^{-1}$ is lower triangular, too.

$$********$$

Here's an alternative (but related) approach.

Observe that a lower triangular matrix is nonsingular if and only if it has all nonzero entries on the diagonal. Let's proceed by induction on $n$. The base case ($n=1$) is simple, as all scalars are trivially "lower triangular". Now, let's suppose that all nonsingular $n\times n$ lower triangular matrices have lower triangular inverses, and let $A$ be any nonsingular $(n+1)\times(n+1)$ lower triangular matrix. In block form, then, we have $$A=\left[\begin{array}{c|c}L & 0_n\\\hline x^T & \alpha\end{array}\right],$$ where $L$ is a nonsingular $n\times n$ lower triangular matrix, $0_n$ is the $n\times 1$ matrix of $0$s, $x$ is some $n\times 1$ matrix, and $\alpha$ is some nonzero scalar. (Can you see why this is true?) Now, in compatible block form, we have $$A^{-1}=\left[\begin{array}{c|c}M & b\\\hline y^T & \beta\end{array}\right],$$ where $M$ is an $n\times n$ matrix, $b,y$ are $n\times 1$ matrices, and $\beta$ some scalar. Letting $I_n$ and $I_{n+1}$ denote the $n\times n$ and $(n+1)\times(n+1)$ identity matrices, respectively, we have $$I_{n+1}=\left[\begin{array}{c|c}I_n & 0_n\\\hline 0_n^T & 1\end{array}\right].$$ Hence, $$\left[\begin{array}{c|c}I_n & 0_n\\\hline 0_n^T & 1\end{array}\right]=I_{n+1}=A^{-1}A=\left[\begin{array}{c|c}ML+bx^T & M0_n+b\alpha\\\hline y^TL+\alpha x^T & y^T0_n+\beta\alpha\end{array}\right]=\left[\begin{array}{c|c}ML+bx^T & \alpha b\\\hline y^TL+\alpha x^T & \beta\alpha\end{array}\right].$$ Since $\alpha$ is a nonzero scalar and $\alpha b=0_n$, then we must have $b=0_n$. Thus, $$A^{-1}=\left[\begin{array}{c|c}M & 0_n\\\hline y^T & \beta\end{array}\right],$$ and $$\left[\begin{array}{c|c}I_n & 0_n\\\hline 0_n^T & 1\end{array}\right]=\left[\begin{array}{c|c}ML & 0_n\\\hline y^TL+\alpha x^T & \beta\alpha\end{array}\right].$$ Since $ML=I_n$, then $M=L^{-1}$, and by inductive hypothesis, we have that $M$ is then lower triangular. Therefore, $$A^{-1}=\left[\begin{array}{c|c}M & 0_n\\\hline y^T & \beta\end{array}\right]$$ is lower triangular, too, as desired.

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  • $\begingroup$ Cheers but I am aware of that case already, it is the specific problem in the question that I have an issue with, i.e. why the proposition implies that any lower triangular matrix will be lower triangular. $\endgroup$
    – sonicboom
    Nov 27, 2012 at 20:48
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    $\begingroup$ I'm not sure you're saying what you intend to say. "...[A]ny lower triangular matrix will be lower triangular"? Of course it will. Did you mean "...implies that the inverse of any nonsingular lower triangular matrix will be lower triangular"? $\endgroup$ Nov 27, 2012 at 21:00
  • $\begingroup$ Ok, I have edited the original question to fix that mistake. $\endgroup$
    – sonicboom
    Nov 27, 2012 at 23:47
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    $\begingroup$ Does my edited answer do the trick for you? $\endgroup$ Nov 27, 2012 at 23:49
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    $\begingroup$ Thanks alot mate, I understand the process now. $\endgroup$
    – sonicboom
    Nov 28, 2012 at 13:47
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Suppose you have an invertible lower-triangular matrix $L$. To find its inverse, you must solve the matrix equation $LX = I$, where $I$ denotes the $n$-by-$n$ identity matrix.

Based on how matrix multiplication works, the $i^{\text{th}}$ column of $LX$ is equal to $L$ times the $i^{\text{th}}$ column of $X$. In order for $LX = I$, it must be that the first $i-1$ entries in the $i^{\text{th}}$ column of $LX$ are all zero. The hint is that you can prove that this implies that the first $i-1$ entries in the $i^{\text{th}}$ column of $X$ must all be zero. To do this, you can explicitly write out your calculation, using your assumption that $L$ is lower-triangular. You'll get a fairly easy system of linear equations to analyze.

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  • $\begingroup$ I understand that and can prove that case. It is the specific problem in the question that I have an issue with, i.e. why the proposition implies that any lower triangular matrix will be lower triangular $\endgroup$
    – sonicboom
    Nov 27, 2012 at 20:48
  • $\begingroup$ Take $y$ to be a column of the matrix $X$ and $b$ to be the corresponding column of the identity matrix. $\endgroup$ Nov 27, 2012 at 22:28
  • $\begingroup$ Ok, cheers, I'll have to think about it for a bit. $\endgroup$
    – sonicboom
    Nov 27, 2012 at 23:51
  • $\begingroup$ It may help to point out that when you are considering the $j^{\text{th}}$ column, you should take $k = j-1$. $\endgroup$ Nov 27, 2012 at 23:53
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In simple form, we can write A = D*(I+L); where A is lower triangular matrix, D is diagonal matrix, I is identity matrix and L is lower triangular with all zeros in diagonal. Since $A^{-1} = (I+L)^{-1}*D^{-1}$ and inverse of D is simply inverse of diagonal element. And for very large n $L^{-n} = 0$ since it is having only lower triangular elements. And we can write $(I+L)^{-1} = I - L + L^2 - L^3 + .... (-1)^n*L^n$ which itself is lower triangular matrix.

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I was thinking about this same question and have an explanation from an informal perspective:

With invertible matrix $A$ and $$LA = B$$

We know that the first row of $B$ is a multiple of the first row of $A$, and the second row of $B$ is a linear combination of the first two rows of $A$, ..., the $k$th row of $B$ is a linear combination of the first $k$ rows of $A$,...

It follows that for any $k$, the first $k$ rows of $A$ and the first $k$ rows of $B$ span the same subspace. Therefore the $k$th row of $A$ is in the subspace spanned by the first $k$ rows of $B$. Furthermore, the $k$th row of $A$ cannot be in the subspace spanned by the first $k-1$ rows of $B$. Otherwise it is in the subspace spanned by the first $k-1$ rows of $A$, which contradicts the assumption that $A$ is invertible (rows are linearly independent). Because for any $k$ the $k$th row of $A$ is a linear combination of the first $k$ rows of $B$, for $$L^{-1}B=A$$ $L^{-1}$ must be lower triangular.

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