# invertible matrices

• Feb 5th 2010, 06:41 PM
deniselim17
invertible matrices
I have 2 questions here.

Let $\displaystyle A \in M _{n} ( \mathbb{R})$.

1. If $\displaystyle A$ is an invertible upper triangular matrix, show that $\displaystyle A^{-1}$ also an upper triangular matrix.

2. If $\displaystyle A^{m} =0$ for some positive integer $\displaystyle m$, show that $\displaystyle A-I_{n}$ is invertible and find $\displaystyle (A - I_{n})^{-1}$.

I know how to show (1) by using cofactor. Is there any other method without using determinant and adjoin?

For (2), I dont even know where to start.
• Feb 5th 2010, 11:48 PM
Opalg
Quote:

Originally Posted by deniselim17
I have 2 questions here.

Let $\displaystyle A \in M _{n} ( \mathbb{R})$.

1. If $\displaystyle A$ is an invertible upper triangular matrix, show that $\displaystyle A^{-1}$ also an upper triangular matrix.

2. If $\displaystyle A^{m} =0$ for some positive integer $\displaystyle m$, show that $\displaystyle A-I_{n}$ is invertible and find $\displaystyle (A - I_{n})^{-1}$.

I know how to show (1) by using cofactor. Is there any other method without using determinant and adjoin?

For (2), I dont even know where to start.

1. For $\displaystyle 1\leqslant k\leqslant n$, let $\displaystyle E_k$ be the subspace of $\displaystyle \mathbb{R}^n$ spanned by the first k vectors in the standard basis. The condition for A to be upper triangular is that $\displaystyle AE_k\subseteq E_k$ for each k. If A is invertible then its kernel consists only of the zero vector. So $\displaystyle AE_k$ has the same dimension as $\displaystyle E_k$ and therefore $\displaystyle AE_k=E_k$ for each k. Thus $\displaystyle E_k = A^{-1}AE_k = A^{-1}E_k$, which says that $\displaystyle A^{-1}$ is upper triangular.

2. If some power of A is 0 then the "binomial series" $\displaystyle I_n + A + A^2 + A^3 +\ldots$ becomes a finite series series which gives you a formula for $\displaystyle (A - I_{n})^{-1}$.