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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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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". –  Robert Israel Nov 27 '12 at 19:25
    
Thanks, I've edited that in. –  sonicboom Nov 27 '12 at 19:26
    
why is this tagged numerical methods? –  user31280 Nov 27 '12 at 19:42
    
@F'OlaYinka: The OP is probably in a Numerical Methods class. –  Cameron Buie Nov 27 '12 at 19:43

2 Answers 2

up vote 3 down vote accepted

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, $$\left[\begin{array}{c|c}I_n & 0_n\\\hline 0_n^T & 1\end{array}\right]=I_{n+1}=A^{-1}A=\left[\begin{array}{c|c}ML+by^T & M0_n+b\alpha\\\hline x^TM+\alpha y^T & y^T0_n+\beta\alpha\end{array}\right]=\left[\begin{array}{c|c}ML+by^T & \alpha b\\\hline x^TM+\alpha y^T & \beta\alpha\end{array}\right].$$ 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 $$\left[\begin{array}{c|c}I_n & 0_n\\\hline 0_n^T & 1\end{array}\right]=\left[\begin{array}{c|c}ML & 0_n\\\hline x^TM+\alpha y^T & \beta\alpha\end{array}\right].$$ 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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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. –  sonicboom Nov 27 '12 at 20:48
2  
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"? –  Cameron Buie Nov 27 '12 at 21:00
    
Ok, I have edited the original question to fix that mistake. –  sonicboom Nov 27 '12 at 23:47
1  
Does my edited answer do the trick for you? –  Cameron Buie Nov 27 '12 at 23:49
    
Thanks alot mate, I understand the process now. –  sonicboom Nov 28 '12 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.

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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 –  sonicboom Nov 27 '12 at 20:48
    
Take $y$ to be a column of the matrix $X$ and $b$ to be the corresponding column of the identity matrix. –  Michael Joyce Nov 27 '12 at 22:28
    
Ok, cheers, I'll have to think about it for a bit. –  sonicboom Nov 27 '12 at 23:51
    
It may help to point out that when you are considering the $j^{\text{th}}$ column, you should take $k = j-1$. –  Michael Joyce Nov 27 '12 at 23:53

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