MHB Is A Invertible When Each Diagonal Element is Nonzero?

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SUMMARY

The matrix \( A = \textit{diag}(a_1, a_2, \ldots, a_n) \) is invertible if and only if each diagonal element \( a_i \) is non-zero. This conclusion is established by noting that a matrix is invertible when its determinant is non-zero. For the diagonal matrix \( A \), the determinant is the product of its diagonal elements, \( \prod_{i=1}^{n} a_i \). The inverse of \( A \) is the diagonal matrix \( \textit{diag}(1/a_1, 1/a_2, \ldots, 1/a_n) \), confirming the invertibility condition.

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karush
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$\textsf{Let }$
$A=\textit{diag} (a_1,a_2,...,a_n)$.
$\textsf{Show that A is invertible iff each}$
$a_i\ne 0.$

$\textsf{Ok I didn't know formally how to answer this.}$
$\textsf{Except i can see that an $a=0$ would mess things up}$
 
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I don't know if this helps but a matrix is invertible iff its determinant is non-zero.
 
The inverse matrix for the diagonal matrix with a_1, a_2, … , a_n on the diagonal, is, rather trivially, the diagonal matrix with 1/a_1, 1/a_2, …, 1/a_n on the diagonal. Show that and you are finished.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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