diagonal matrix and invertible matrix proof

Question

I am given the following proof question:

Let $A \in {\mathbb R}^{n\times n} $.` Show that there exist invertible matrices $B$, $C$ such that $A=B+C$.

I believe it has something to do with diagonal matrix, but maybe I am wrong. thank you for the help

Answer 1

I presume you mean there exist $B$ and $C$...

Hint: let $B$ be a multiple of the identity. Think about eigenvalues.

Answer 2

Hint : Try to construct $B$ upper triangular, and $C$ lower triangular.

Answer 3

One doesn't need much information about eigenvalues to solve this problem, for if $\mu \in \Bbb R$ is sufficiently large, then both $A - \mu I$ and $\mu I$ are invertible, and

$A = (A - \mu I) + \mu I; \tag 1$

it is easy to see $\mu I$ is invertible if $\mu \ne 0$; what about $A - \mu I$? well, if

$\mu > \Vert A \Vert, \tag 2$

where $\Vert A \Vert$ is the standard operator norm of $A$ (actually, any norm on $\Bbb R^{n \times n}$ will do), then

$\Vert \mu^{-1} A \Vert = \mu^{-1} \Vert A \Vert < 1, \tag 3$

which in the usual, standard way implies the series representation of $(I - \mu^{-1}A)^{-1}$,

$(I - \mu^{-1}A)^{-1} = \displaystyle \sum_0^\infty (\mu^{-1}A)^j \tag 4$

converges, since it is dominated by

$\displaystyle \sum_0^\infty \Vert \mu^{-1} A \Vert^j = \dfrac{\Vert \mu^{-1}A \Vert}{1 - \Vert \mu^{-1}A \Vert}; \tag 5$

thus

$\mu^{-1}(\mu I - A)^{-1} = (I - \mu^{-1}A)^{-1} \tag 6$

exists, as does

$(\mu I - A)^{-1} = \mu (I - \mu^{-1}A)^{-1}; \tag 7$

thus

$A - \mu I = -(\mu I - A) \tag 8$

has an inverse, and therefore (1) expresses $A$ as the sum of the two invertible matrices $A - \mu I$ and $\mu I$.