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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 $$ \mathbb{C}_{3x3} = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}$$

Fracton
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2 Answers2

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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.

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:

Martin Sleziak
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  • Actually, I tried this trick, but is there a more general way to prove this? – Fracton Nov 14 '23 at 01:49
  • Thanks for the update, does this server as a general proof for any nxn upper triangular block matrix then? – Fracton Nov 15 '23 at 02:30
  • This works for any matrix of the form you specified, i.e., $\begin{pmatrix} I & B & 0 \ 0 & I & B \ 0 & 0 & I \end{pmatrix}$ where all blocks $0$, $I$, $B$ are square matrices of the same size. I am not sure whether this is what you meant when you wrote: " general proof for any nxn upper triangular block matrix". – Martin Sleziak Nov 15 '23 at 06:07
  • Yes, this is what I meant. I did know the solution using taylor series expansion, but wasn't convinced if it would be a general solution. I was stuck with the point, that C was a singular matrix, and so would be N in your notation. – Fracton Nov 17 '23 at 01:51
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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

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

John Alexiou
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  • Yes, even I have tried this, I there a more general way to find the inverse? or a way to prove that the limit where B is 1x1 holds for all nxn? – Fracton Nov 14 '23 at 01:51
  • If your elements are matrices, just carry on like they are scalars, except you do not have the commutative property with multiplication. If you want to prove the inverse, instead of plug and try, you can find the reduced echelon matrix with an identity matrix augmented. – John Alexiou Nov 14 '23 at 13:53
  • Each submatrix $\mathbb{B}$ can be n × n above. there is no restriction, as all you need is matrix multiplication to make the point. – John Alexiou Nov 15 '23 at 17:46