$A$ upper triangular $n\times n$ over $\mathbb{R}$. Show that $I-A$ is invertible and express inverse of $I-A$ as a function of $A$.

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Let $A$ be a strictly upper triangular $n\times n$ matrix with real entries. Show that $I-A$ is invertible and express the inverse of $I-A$ as a function of $A$.

$A$ is upper triangular, so the diagonal is all $0$, everything below the diagonal is all $0$, and then we have entries in the upper part, say, $a_{1,2}$, as the entry in row $1$ column $2$. If we consider $I-A$, then $I-A$ has $1$ in every entry of the diagonal, only $0$ below the diagonal, and negative whatever the entry of $A$ was in the upper part. So, det$(I-A)=1n=1$, so $I-A$ is invertible. Consider

$$X=I+A+A^2+ \cdots +A^{n-1}$$

Then, I want to show that $(I-A)X=X(I-A)=I$. I can see why this "might" work, since we would have $I$ and then add all the powers of $A$, then subtract all the powers of $A$… but are there any unexpected issues I should be careful of here? Thank you much!

Best Answer

  1. You don't need to argue about $\det\left(I-A\right)$. Once you have shown that $\left(I-A\right)X = X\left(I-A\right) = I$, the invertibility of $I-A$ follows automatically. The determinant is thus unnecessary (although the argument is correct).

  2. As @nejimban already noticed, you need to show that $A^n = 0$. I would be wary of using the fact that any nilpotent $n\times n$-matrix $B$ satisfies $B^n = 0$, since some proofs of this fact involve reducing it to the case of strictly upper-triangular matrices and then applying the very fact you are trying to prove. That would be circular reasoning. Instead, you can argue as follows: Say that an $n\times n$-matrix $C$ is $k$-upper-triangular (for some integer $k$) if all its entries $C_{i,j}$ are zero when $j < i+k$. Thus, a matrix is strictly upper-triangular if and only if it is $1$-upper-triangular. Now, show that the product of a $p$-upper-triangular matrix with a $q$-upper-triangular matrix is $p+q$-upper-triangular (for any integers $p$ and $q$). Use this to conclude that $A^k$ is $k$-upper-triangular for any $k \geq 0$. Apply this to $k = n$, and observe that the only $n$-upper-triangular $n\times n$-matrix is the zero matrix. You can find this proof in full detail in §3.12 of my (abandoned, but this section is complete) notes on linear algebra (although I am working with lower-triangular matrices instead of upper-triangular ones there).

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