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I is the identity matrix.

Let D ∈ $R^{n\times m}$ and E ∈ $R^{m\times n}$

Show that if I − ED is nonsingular, then I − DE is nonsingular and

$(I − DE)^{−1} = I + D(I − ED)^{−1}E$

i don't get how to this first of all i don't get how i could show the first proof. Because DE and ED are different dimensions i don't get how to you could ever compare them.

Second i dont get how to show that the equation is true because the I-ED is inbetween the E and D and with matrices you can't just multiple in any order i thought so i don't get how to get rid of it.

Danielvanheuven
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  • This is a consequence of $\det(I-DE)=\det(I-ED)$ which is known as Sylvester's identity, https://math.stackexchange.com/q/17831 – Jean Marie Nov 23 '18 at 15:14

2 Answers2

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Compute the product of the matrix $I-DE$ and its wannabe inverse $I+D(I-ED)^{-1}E$: $$ (I+D(I-ED)^{-1}E)(I-DE)=I-DE + D(I-ED)^{-1}E(I-DE)\\=I-DE + D(I-ED)^{-1}(I-ED)E=I-DE+DE=I. $$ The product is the identity, which proves the claim.

daw
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We can show that $I-DE$ is nonsingular by contradiction. If it is singular, there is a vector $v\in\mathbb{R}^n$ such that $$ (I-DE) v = 0 .$$ Consider $w = E v\in\mathbb{R}^m$. $$(I-ED)w = E v - EDE v = E (I-DE) v = 0$$ so $I-ED$ is also singular (contradiction). Therefore, $I-DE$ is not singular.

The form of the inverse is verified with simple matrix algebra, see daw's answer.

fqq
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