# 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. Commented Sep 17, 2010 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. Commented Sep 17, 2010 at 8:12
• – user53259
Commented Dec 10, 2013 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. Commented Sep 18, 2010 at 10:40
• @Robin: How can you say that $I+N$ has upper triangular inverses?
– user23238
Commented Feb 23, 2013 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! Commented Jul 18, 2013 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. Commented Jul 18, 2013 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.) Commented Jul 26, 2017 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? Commented Sep 18, 2010 at 6:53
• Yes - corrected now that I reworded the last few sentences. Commented Sep 18, 2010 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. Commented Sep 18, 2010 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) Commented Feb 7, 2017 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 $TS_k = \mathrm{span}(Te_1,Te_2,Te_3,\dots,Te_k) = \mathrm{span}(T_{11}e_1, T_{12}e_1 + T_{22}e_2,\dots,\sum_{i\le k}T_{ik}e_i) \subset S_k$. Commented Oct 16, 2018 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.

Let $$A=\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1,n-1} & a_{1,n}\\ & a_{22} & \cdots & a_{2,n-1} & a_{2,n}\\ & & \ddots & \vdots & \vdots\\ & & & a_{n-1,n-1} & a_{n-1,n}\\ & & & & a_{n,n} \end{pmatrix}.$$ Let $$i,j$$ be two integers such that $$i,j\in\{1,\dots,n\}$$ and $$i.
Let $$A_{i,j}$$ be an $$n-1\times n-1$$ matrix which is obtained by crossing out row $$i$$ and column $$j$$ of $$A$$.
Then, $$A_{i,j}$$ is
$$\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1,i-1} & a_{1,i} &a_{1,i+1}&a_{1,i+2}&\cdots&a_{1,j-1}&a_{1,j+1}&a_{1,j+2}&\cdots&a_{1n}\\ & a_{22} & \cdots & a_{2,i-1} & a_{2,i} &a_{2,i+1}&a_{2,i+2}&\cdots&a_{2,j-1}&a_{2,j+1}&a_{2,j+2}&\cdots&a_{2n}\\ & & \ddots & \vdots & \vdots &\vdots&\vdots&\cdots&\vdots&\vdots&\vdots&\cdots&\vdots\\ & & & a_{i-1,i-1} & a_{i-1,i} &a_{i-1,i+1}&a_{i-1,i+2}&\cdots&a_{i-1,j-1}&a_{i-1,j+1}&a_{i-1,j+2}&\cdots&a_{i-1,n}\\ & & & & 0 & a_{i+1,i+1}&a_{i+1,i+2}&\cdots&a_{i+1,j-1}&a_{i+1,j+1}&a_{i+1,j+2}&\cdots&a_{i+1,n}\\ & & & & & 0&a_{i+2,i+2}&\cdots&a_{i+2,j-1}&a_{i+2,j+1}&a_{i+2,j+2}&\cdots&a_{i+2,n}\\ & & & & & &0&\cdots&\vdots&\vdots&\vdots&\cdots&\vdots\\ & & & & & &&\ddots&a_{j-1,j-1}&a_{j-1,j+1}&a_{j-1,j+2}&\cdots&a_{j-1,n}\\ & & & & & &&&0&a_{j,j+1}&a_{j,j+2}&\cdots&a_{j,n}\\ & & & & & &&&&a_{j+1,j+1}&a_{j+1,j+2}&\cdots&a_{j+1,n}\\ & & & & & &&&&&a_{j+2,j+2}&\cdots&a_{j+2,n}\\ & & & & & &&&&&&\ddots&\vdots\\ & & & & & &&&&&&&a_{n,n}\\ \end{pmatrix}.$$

So, $$\det A_{i,j}=0$$ if $$i,j$$ are two integers such that $$i,j\in\{1,\dots,n\}$$ and $$i.
Let $$C_{i,j}$$ be the $$(i,j)$$-cofactor of $$A$$.
Then, $$C_{i,j}=(-1)^{i+j}\det A_{i,j}=0$$ if $$i,j$$ are two integers such that $$i,j\in\{1,\dots,n\}$$ and $$i.
So, $$A^{-1}=\frac{1}{\det A}\begin{pmatrix}C_{11}&C_{21}&\cdots&C_{n,1}\\ C_{12}&C_{22}&\cdots&C_{n,2}\\ \vdots&\vdots&&\vdots\\ C_{1n}&C_{2n}&\cdots&C_{n,n}\\ \end{pmatrix}=\frac{1}{\det A}\begin{pmatrix}C_{11}&C_{21}&\cdots&C_{n,1}\\ 0&C_{22}&\cdots&C_{n,2}\\ \vdots&\vdots&&\vdots\\ 0&0&\cdots&C_{n,n}\\ \end{pmatrix}.$$

• the product of A with its adjugate yields a diagonal matrix whose diagonal entries are the determinant det(A): $A^{-1} = det(A)^{-1} *adj(A)$
– Kuo
Commented Oct 20, 2022 at 16:30

The proof is based on cayley–hamilton theorem.

Suppose $$A$$ is upper triangular matrix. Then we know the characteristic polynomial of $$A$$ is $$P_A(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)...(\lambda-\lambda_n)$$. By cayley–hamilton theorem we know $$P_A(A)=0$$, i.e.$$(A-\lambda _1I)(A-\lambda _2I)...(A-\lambda _nI)=0$$, from where we can see that $$I=A\frac{\sum_i{c_iA^i}}{\underset{i}{\Pi}\left( -\lambda _i \right)}$$ where $$c_i$$ can be solved from $$P_A(A)=0$$. Compare $$I=A\frac{\sum_i{c_iA^i}}{\underset{i}{\Pi}\left( -\lambda _i \right)}$$ with $$I=AA^{-1}$$ we can see that $$A^{-1}$$ is also upper triangular matrix.

Just to be add a simple answer, I will add another one. Note that, by the $$QR$$ decomposition, for any invertible upper-triangular matrix $$U$$, there exist an upper-triangular matrix $$R$$ and an orthogonal matrix $$Q$$ so that $$U^{-1} = Q R.$$ Hence, $${(R U)^{-1}} = Q.$$ This implies that, inverting both sides, $$R U = Q^{T}.$$ Since the product of two upper-triangular matrices are still upper triangular, we have found an orthogonal upper-triangular matrix, which is impossible unless $$Q=D$$, for some diagonal Matrix $$D$$, see this link.. Thus the result follows by noting that $$U^{-1}=Q R = DR,$$ which is upper-triangular.

Here's an alternative approach using flags:

Definition (Upper Triangular w.r.t. Flag). Let $$T \in \mathscr{L}(V)$$, and let $$\mathfrak{F}$$ be a complete flag in $$V$$; i.e., $$\mathfrak{F}$$ is a collection of subspaces $$V_i$$ with $$\{0\} = V_0 \subset V_1 \subset \cdots \subset V_n = V,$$ such that $$\dim(V_i/V_{i-1})=1$$ for all $$i \in [n]$$. The transformation $$T$$ is "upper triangular" with respect to $$\mathfrak{F}$$ if each $$V_i$$ is $$T$$-invariant.

Note. For each $$i \in [n]$$, let $$\hat{T}_i : V_i/V_{i-1} \to V_i/V_{i-1}$$ be the quotient map. Since $$\mathfrak{F}$$ is complete, each $$\hat{T}_i$$ is determined by multiplication by a unique $$\lambda_i \in F$$ (the base field of $$V$$).

Lemma 2. $$T$$ is invertible if and only if each $$\lambda_i \neq 0$$.

Lemma 3. The matrix $$A = [T]_{\beta}$$ is upper triangular if and only if $$T$$ is upper triangular w.r.t. the flag associated to the basis $$\beta$$.

Reformulated Question. If $$T$$ is upper triangular w.r.t. $$\mathfrak{F}$$, the inverse $$T^{-1}$$ is also upper triangular w.r.t. the same flag $$\mathfrak{F}$$.

Proof. Suppose $$T$$ is invertible and upper triangular w.r.t. $$\mathfrak{F}$$. We proceed by induction. The base case $$i=1$$ is given by the fact that for $$x \in V_1$$, the inverse $$T^{-1}(x)=(1/\lambda_1)x \in V_1$$, by closure. Hence, $$V_1$$ is $$T^{-1}$$-invariant. Now, assume $$V_i$$ is preserved by $$T^{-1}$$ for all $$i$$. Consider the inverse quotient map $$[T^{-1}] : V/V_i \to V/V_i : [x] \mapsto (1/\lambda_i)[x].$$ This is an inverse for $$[T] : V/V_i \to V/V_i$$, as $$\left[T_i\left(T_i^{-1}\left([x]\right)\right)\right] = T_i \left( \left[ \frac{1}{\lambda_i}[x]\right] \right) = \lambda_i \left[ \frac{1}{\lambda_i} x \right] = [x] \in V/V_i.$$ As such, the individual quotient space $$V/V_i$$ is invariant for $$T_i^{-1}$$, as the individual inverse preserves the space. Now, recall the chained composition construction of $$T^{-1}$$, which when acting on the quotient space, yields $$\prod_{i=1}^{n+1} \left[ \frac{1}{\lambda_i} [\lambda_i x] \right]= [x] \in V_{i+1}/V_i,$$ as the singular addition of another $$T_i^{-1}$$ can be likened to another multiplication of $$\lambda_i$$ on the quotient, which, by the base case, ensures that the $$i+1$$ case holds. This chained composition of inverses gives a $$T^{-1}$$ that preserves $$V_{i+1}$$ by the coset operation falling back into the subspace $$V_i$$, completing the proof.