# Inverse of Identity minus a stricty upper tringular matrix

Let $$S$$ be an $$n\times n$$ strictly upper triangular matrix. Show that $$(I-S)^{-1} = I+S+S^2+ \dots + S^{n-1}$$.

This seems like it should be an easy problem to do by induction, but I am having trouble justifying the last step.

We can define a sequence of matrices $$\{S_i\}$$ such that $$S_n$$ and $$S_{n+1}$$ agree in the $$n\times n$$ upper left corner of $$S_{n+1}$$, so that the only new entries being added as we increase $$n$$ appear in the furthest right column.

By induction: \begin{align*} \begin{bmatrix} 1 & -s_{12} & \dots & -s_{1,n+1}\\ 0 & \ddots & \dots & -s_{2,n+1}\\ \vdots & & \ddots & \vdots\\ 0 & \dots & \dots & 1 \end{bmatrix} (I + S_{n+1} + S_{n+1}^2 + \dots + S_{n+1}^n) = \begin{bmatrix} 1 & 0 & \dots & \text{stuff}\\ 0 & \ddots & \dots & \text{stuff}\\ \vdots & & \ddots & \vdots\\ 0 & \dots & \dots & 1 \end{bmatrix} \end{align*}

Put into words, by induction we have that the upper left corner is the $$n \times n$$ identity matrix, but I am not sure how to justify that the "stuff" rightmost column becomes $$0$$.

## 2 Answers

Strictly upper triangular matrices are nilpotent. Indeed, the characteristic polynomial function of such a matrix $$S$$ is given by $$p(\lambda)=\lambda^n$$ (since all the diagonal entries of the matrix are zero) whence by cayley-hamilton $$S^n=0$$. Hence $$(I+S+S^2+\dotsb+S^{n-1})(I-S)=I-S^n=I$$ as desired.

According to this Wolfram MathWorld entry a strictly upper triangular matrix is defined as an upper triangular matrix whose diagonal entries are all $$0$$; that is, if $$S=(s_{ij})_{1\leq i,j \leq n}$$ then $$s_{ij}=0$$ whenever $$i\geq j$$. This implies that $$S$$ is nilpotent of order $$n$$, i.e. $$S^n = 0_n$$, the $$n\times n$$ zero matrix. (To see this, try to write the entries of $$S^2,S^3,...,S^n$$ explicitly, using the definition of matrix multiplication.) Then:

$$(I-S)\left(\sum_{k=0}^{n-1} S^k\right)=\left(\sum_{k=0}^{n-1} S^k\right)-S\left(\sum_{k=0}^{n-1} S^k\right)=\left(\sum_{k=0}^{n-1} S^k\right)-\left(\sum_{k=1}^{n} S^k\right)=I-S^n=I,$$

and since the sum is a right inverse, it's also a left inverse of $$(I-S)$$ and we're done.