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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 $0$.

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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.

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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.

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