# The inverse of a lower triangular matrix is lower triangular

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?

• 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". Nov 27, 2012 at 19:25
• Thanks, I've edited that in. Nov 27, 2012 at 19:26
• why is this tagged numerical methods?
– user31280
Nov 27, 2012 at 19:42
• @F'OlaYinka: The OP is probably in a Numerical Methods class. Nov 27, 2012 at 19:43
• Nov 12, 2015 at 1:47

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.

• 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. Nov 27, 2012 at 20:48
• 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"? Nov 27, 2012 at 21:00
• Ok, I have edited the original question to fix that mistake. Nov 27, 2012 at 23:47
• Does my edited answer do the trick for you? Nov 27, 2012 at 23:49
• Thanks alot mate, I understand the process now. Nov 28, 2012 at 13:47

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.

• 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 Nov 27, 2012 at 20:48
• Take $y$ to be a column of the matrix $X$ and $b$ to be the corresponding column of the identity matrix. Nov 27, 2012 at 22:28
• Ok, cheers, I'll have to think about it for a bit. Nov 27, 2012 at 23:51
• It may help to point out that when you are considering the $j^{\text{th}}$ column, you should take $k = j-1$. Nov 27, 2012 at 23:53

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.

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.