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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

Brian61354270
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ga as
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3 Answers3

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

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

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

TheSilverDoe
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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 $\mu I$.

Robert Lewis
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    Thank you, you think there is a way involoving diagnoel matrix in a different approach? – ga as Apr 03 '19 at 16:59
  • Well, $\mu I$ is a diagonal matrix, as simple as they come. So though I'm pretty sure there are other approaches involving diagonal matrices, they will likely be more complicated. The point is the $\mu I$ "swamps" $A$ when $\mu$ is big enough. .. – Robert Lewis Apr 03 '19 at 17:06
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    My professor said he want me to this kind of approach – ga as Apr 03 '19 at 17:06
  • All I can show is what I know! Cheers! – Robert Lewis Apr 03 '19 at 17:07
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    This is more complicated than necessary. All you need is $\mu \ne 0$ and not an eigenvalue of $A$. There are at most $n$ eigenvalues of $A$, so ... – Robert Israel Apr 03 '19 at 20:36
  • @RobertIsrael: i guess my response to your comment would be the equivocal "yes and no" . . . if one considers the material necessary to establish the machinery of eigenvalues, then I think my answer is relaitively simple, especially considering the fact that I probably said more than is necessary, i.e. presented the power series for $(I - \mu^{-1}A)^{-1}$ etc. A similar approach using eigenvalues would have to establish the the existence of same, which needs the fundamental theorem of algebra: – Robert Lewis Apr 03 '19 at 21:01
  • It seems to me if one figures in the distance one must traverse from the starting gate to attain this result, my approach is relatively simple. Anyway, thanks for the input. By the way, I am a self-professed admirer of you work here. You seem to have mastered the art of the short and sweet answer, unlike myself. Hell's bells, what long comments I do post! :) – Robert Lewis Apr 03 '19 at 21:01