Inverse of $3\times3$ block upper triangle matrix

Question

How to find the inverse of $3\times 3$ block upper triangular matrix

$$X = \begin{bmatrix} \mathbb{1} & \mathbb{B} & 0\\ 0 & \mathbb{1} & \mathbb{B}\\ 0 & 0 & \mathbb{1} \end{bmatrix}$$

where $\mathbb{B}$ is a $3\times 3$ matrix.

Edit:

Is there a general proof for such a problem?

Here's an approach that I have tried. I started by decomposing the matrix into kronecker product of individual matrices.

$$ X = \mathbb{1}_{3x3}\otimes\mathbb{1}_{3x3} + \mathbb{B}_{3x3}\otimes\mathbb{C}_{3x3} $$ where $$ \mathbb{C}_{3x3} = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}$$

Answer 1

First try to have a look at the case that $B$ is $1\times1$.

You should be able to check relatively easily that: $$ \begin{pmatrix} 1 & b & 0 \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}^{-1}= \begin{pmatrix} 1 &-b & b^2 \\ 0 & 1 &-b \\ 0 & 0 & 1 \end{pmatrix} $$

Now you only have to check the same pattern works for block matrices consisting of square blocks: $$ \begin{pmatrix} I & B & B^2 \\ 0 & I & B \\ 0 & 0 & I \end{pmatrix}^{-1}= \begin{pmatrix} I &-B & 0 \\ 0 & I &-B \\ 0 & 0 & I \end{pmatrix} $$

Simply check that if you multiply the two matrices, you get the identity matrix.

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As you correctly pointed out in your edit, you can rewrite your matrix as $A=I+N$, where $N^3=0$. This means that \begin{align*} (I+N)(I-N+N^2)&=I-N^3=I\\ (I+N)^{-1}&=I-N+N^2 \end{align*} So you get that $A^{-1}=I-N+N^2$.

You can get similar equality for $(I+N)^{-1}$ where $N$ is a nilpotent matrix. See also:

• Prove that $A+I$ is invertible if $A$ is nilpotent
• Proving that if $A^n = 0$, then $I - A$ is invertible and $(I - A)^{-1} = I + A + \cdots + A^{n-1}$
• For a nilpotent square matrix $A$, $I + aA$ is invertible (for scalar $a$)
• Let M be invertible and N be nilpotent. Show that M-N is invertible

Answer 2

The answer by @MartinSleziak is the correct one

$${\rm X}^{-1}=\begin{bmatrix}{\bf 1} & -\mathbb{B} & \mathbb{B}^{2}\\ 0 & {\bf 1} & -\mathbb{B}\\ 0 & 0 & {\bf 1} \end{bmatrix}$$

and the check

$$ \small \begin{bmatrix}{\bf 1} & -\mathbb{B} & \mathbb{B}^{2}\\ 0 & {\bf 1} & -\mathbb{B}\\ 0 & 0 & {\bf 1} \end{bmatrix}\begin{bmatrix}{\bf 1} & \mathbb{B} & 0\\ 0 & {\bf 1} & \mathbb{B}\\ 0 & 0 & {\bf 1} \end{bmatrix}=\begin{bmatrix}{\bf 1} & {\bf 1}\left(\mathbb{B}\right)+\left(-\mathbb{B}\right){\bf 1} & \left(-\mathbb{B}\right)\mathbb{B}+{\bf 1}\mathbb{B}^{2}\\ 0 & {\bf 1} & {\bf 1}\mathbb{B}+\left(-\mathbb{B}\right){\bf 1}\\ 0 & 0 & {\bf 1} \end{bmatrix}=\begin{bmatrix}{\bf 1}\\ & {\bf 1}\\ & & {\bf 1} \end{bmatrix} $$