Inverse of an invertible triangular matrix (either upper or lower) is triangular of the same kind

How can we prove that the inverse of an upper (lower) triangular matrix is upper (lower) triangular?

• 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. is a poor mathematician 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. is a poor mathematician Sep 17 '10 at 8:12
• – Greek - Area 51 Proposal Dec 10 '13 at 13:39

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.

• 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. is a poor mathematician Sep 18 '10 at 10:40
• @Robin: How can you say that $I+N$ has upper triangular inverses? – user23238 Feb 23 '13 at 13:38
• I know that $(1+x)^{-1}$ has the power series expansion. Why is this true for matrices? and I would never think that inverse is equivalent to $-1$ power. I'll believe it but I would like to know why. Cool proof though! – CodeKingPlusPlus Jul 18 '13 at 1:52
• @ramanujan_dirac If you believe you can write $(I+N)^{-1}$ as a power series expansion, then $N^k$ is nilpotent and so the series is finite. $N^k$ is always upper triangular. Thus the power series is upper triangular. – CodeKingPlusPlus Jul 18 '13 at 2:09
• @Travis, I do not follow your objection. $(\mathbf I+\mathbf N)^{-1}$ is the formal result of plugging in the strict upper triangular matrix $\mathbf N$ in the geometric series, and an infinite sum of triangular matrices remains triangular. (You may or may not have to prove this in the course of using this proof tho.) – J. M. is a poor mathematician Jul 26 '17 at 4:56

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.

• 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. – morgan 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
• This was 7 years ago.... but I'm sorry I do not understand why T must be upper triangular in order for there to be the inclusion from (1) – jgcello Feb 7 '17 at 13:56
• @jgcello: If $T$ is upper triangular with respect to the chosen basis, then by reading the columns of the matrix representation of $T$, then we see that the first column tells us that $T$ maps $e_1\mapsto T_{11}e_1$, $e_2\mapsto T_{12}e_1 + T_{22}e_2$, $e_3\mapsto T_{13}e_1+T_{23}e_2+T_{33}e_3$, and so on. Hence $TS_k = \mathrm{span}(Te_1,Te_2,Te_3,\dots,Te_k) = \mathrm{span}(T_{11}e_1, T_{12}e_1 + T_{22}e_2,\dots,\sum_{i\le k}T_{ik}e_i) \subset S_k$. – Alex Ortiz Oct 16 '18 at 23:03

I'll add nothing to alext87 answer, or J.M. comments. Just "display" them. :-)

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?)

In your case, half of the work is already done:

$$\begin{pmatrix} a^1_1 & a^1_2 & \cdots & a^1_{n-1} & a^1_n & 1 & 0 & \cdots & 0 & 0 \\\ & a^2_2 & \cdots & a^2_{n-1} & a^2_n & & 1 & \cdots & 0 & 0 \\\ & & \ddots & \vdots & \vdots & & & \ddots & \vdots & \vdots \\\ & & & a^{n-1}_{n-1} & a^{n-1}_n & & & & 1 & 0 \\\ & & & & a^n_n & & & & & 1 \end{pmatrix}$$

Now, what happens when you do back substitution starting with $a^n_n$ and then continuing with $a^{n-1}_{n-1}$...?

You can prove by induction.

Suppose $A$ is upper triangular. It is easy to show that this holds for any $2\times 2$ matrix. (In fact, $A^{-1}=\left[\begin{array}{cc} a & b\\ 0 & d \end{array}\right]^{-1} =\frac{1}{ad}\left[\begin{array}{cc} d & -b\\ 0 & a \end{array}\right]$. )

Suppose the result holds for any $n\times n$ upper triangular matrix. Let $A=\left[\begin{array}{cc} A_{1} & a_{2}\\ 0 & x \end{array}\right]$, $B=\left[\begin{array}{cc} B_{1} & b_{2}\\ b_{3}^{T} & y \end{array}\right]$ be any $(n+1)\times (n+1)$ upper triangular matrix and its inverse. (Mind that $a_2$, $b_2$, $b_3$ are $n\times 1$ vectors, $x$, $y$ are scalars.) Then $AB=BA=I_{n+1}$ implies that $$\left[\begin{array}{cc} A_{1} & a_{2}\\ 0 & x \end{array}\right] \left[\begin{array}{cc} B_{1} & b_{2}\\ b_{3}^{T} & y \end{array}\right]= \left[\begin{array}{cc} B_{1} & b_{2}\\ b_{3}^{T} & y \end{array}\right] \left[\begin{array}{cc} A_{1} & a_{2}\\ 0 & x \end{array}\right] =I_{n+1},$$

From the upper left corner of the second multiplication, we have $B_1 A_1 = I_n$. Hence $B_1$ is upper triangular from our hypothesis. From the lower left block of the multiplication , we know that $b_3=0$. ($x\ne 0$ since $A$ is invertible.) Therefore $B=\left[\begin{array}{cc} B_{1} & b_{2}\\ 0 & y \end{array}\right]$ is also upper triangular.

Another proof is by contradiction. Let $A = [a_{ij}]$ be an upper triangular matrix of size $N$. Assume $B = A^{-1} = [b_{ij}]$ is not upper triangular. Thus there exists an entry $b_{ij} \neq 0$ for $i > j$. Let $b_{ik}$ be the element with the smallest $k$ in row $i$ such that $b_{ik} \neq 0$ and $i > k$. Consider the product $C = B A$. The element $c_{ik}$ of matrix C is off-diagonal ($i > k$) and computed as $$c_{ik} = \sum b_{ij}a_{jk} = b_{i1}a_{1k} + b_{i2}a_{2k} + \dots + b_{ik}a_{kk} + \dots + b_{iN}a_{Nk}$$

Since $b_{ik}$ is the first non-zero element in its row, all the terms to the left of $b_{ik}a_{kk}$ vanish. Since A is upper triangular (given), all the terms to the right of $b_{ik}a_{kk}$ vanish. Since $A$ is invertible, all its diagonal elements are non-zero. Thus $c_{ik} = b_{ik}a_{kk} \neq 0$. However, since $C$ is the identity matrix and $c_{ik}$ is off diagonal, this is a contradiction! Thus, $B = A^{-1}$ is upper triangular.

Same applies to lower triangular matrix by noticing that $(A^T)^{-1} = (A^{-1})^T$

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.

The same thing works for lower triangular using forward subsitution.