Inverse of a triangular matrix( both upper & lower ) is triangular

How can we prove that inverse of upper(lower) triangular matrix is upper(lower) triangular...

Can anybody answer to this question.......

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Why not try a constructive proof? Better yet, look at the 2-by-2 case first, and figure out how you can generalize your observations from it. – J. M. Sep 17 '10 at 8:11
A further hint: look up "forward elimination" and "backsubstitution", and figure out how to use these to find the inverse of a triangular matrix. – J. M. Sep 17 '10 at 8:12

Personally, I prefer arguments which are more geometric to arguments rooted in matrix algebra. With that in mind, here is a proof.

First, two observations on the geometric meaning of an upper triangular invertible linear map.

1. Define $S_k = {\rm span} (e_1, \ldots, e_k)$, where $e_i$ the standard basis vectors. Clearly, the linear map $T$ is upper triangular if and only if $T S_k \subset S_k$.

2. If $T$ is in addition invertible, we must have the stronger relation $T S_k = S_k$.

Indeed, if $T S_k$ was a strict subset of $S_k$, then $Te_1, \ldots, Te_k$ are $k$ vectors in a space of dimension strictly less than $k$, so they must be dependent: $\sum_i \alpha_i Te_i=0$ for some $\alpha_i$ not all zero. This implies that $T$ sends the nonzero vector $\sum_i \alpha_i e_i$ to zero, so $T$ is not invertible.

With these two observations in place, the proof proceeds as follows. Take any $s \in S_k$. Since $TS_k=S_k$ there exists some $s' \in S_k$ with $Ts'=s$ or $T^{-1}s = s'$. In other words, $T^{-1} s$ lies in $S_k$, so $T^{-1}$ is upper triangular.

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 Your last $T$ should be $T^{-1}$, shouldn't it? – Agustí Roig Sep 18 '10 at 6:53 Yes - corrected now that I reworded the last few sentences. – alex Sep 18 '10 at 17:08 This is my preferred proof also. It explicitly exhibits the group of invertible upper triangular matrices as the group of symmetries of something, which (to my mind) is always the most natural way to define a group. – Qiaochu Yuan Sep 18 '10 at 20:23

Another method is as follows. An invertible upper triangular matrix has the form $A=D(I+N)$ where $D$ is diagonal (with the same diagonal entries as $A$) and $N$ is upper triangular with zero diagonal. Then $N^n=0$ where $A$ is $n$ by $n$. Both $D$ and $I+N$ have upper triangular inverses: $D^{-1}$ is diagonal, and $(I+N)^{-1}=I-N+N^2-\cdots +(-1)^{n-1}N^{n-1}$. So $A^{-1}=(I+N)^{-1}D^{-1}$ is upper triangular.

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Just a tiny terminology note: $N$ in your answer would be termed a "strictly upper triangular matrix"; the definition of "strictly lower triangular matrix" is similar. – J. M. Sep 18 '10 at 10:40
@Robin: How can you say that $I+N$ has upper triangular inverses? – ramanujan_dirac Feb 23 at 13:38

Remeber that you can compute the inverse of a matrix by reducing it to row echelon form and solving the simultaneous systems of linear equations $(A \vert I)$, where $A$ is the matrix you want to invert and $I$ the unit matrix. When you have finished the process, you'll get a matrix like $(I\vert A^{-1})$ and the matrix on the right, yes!, is the inverse of $A$. (Why?)

$$\begin{pmatrix} a^1_1 & a^1_2 & \cdots & a^1_{n-1} & a^1_n & 1 & 0 & \cdots & 0 & 0 \\ & a^2_2 & \cdots & a^2_{n-1} & a^2_n & & 1 & \cdots & 0 & 0 \\ & & \ddots & \vdots & \vdots & & & \ddots & \vdots & \vdots \\ & & & a^{n-1}_{n-1} & a^{n-1}_n & & & & 1 & 0 \\ & & & & a^n_n & & & & & 1 \end{pmatrix}$$
Now, what happens when you do back substitution starting with $a^n_n$ and then continuing with $a^{n-1}_{n-1}$...?
Suppose that $U$ is upper. The $i$th column $x_i$ of the inverse is given by $Ux_i=e_i$ where $e_i$ is the $i$th unit vector. By backward subsitution you can see that $(x_i)_j=0$ for $i+1\leq j\leq n$. I.e all the entries in the $i$th column of the inverse below the diagonal are zero. This is true for all $i$ and hence the inverse $U^{-1}=[x_1|\ldots|x_n]$ is upper triangular.