 # In my syllabus we have the alternative definition of the condition of a matrix: kappa(A)=(text(max)_(norm(vec(y))=1) norm(A vec(y)))/(text(min)_(norm(vec(y))=1) norm(A vec(y))) In it, it also says that by definition of the condition of a matrix it follows that kappa(A^(-1))=kappa(A). So there is no explanation for this. Therefore, my question is: Why is kappa(A^(-1))=kappa(A)? Zack Chase 2022-09-16 Answered
In my syllabus we have the alternative definition of the condition of a matrix:
$\kappa \left(A\right)=\frac{{\text{max}}_{‖\stackrel{\to }{y}‖=1}‖A\stackrel{\to }{y}‖}{{\text{min}}_{‖\stackrel{\to }{y}‖=1}‖A\stackrel{\to }{y}‖}$
In it, it also says that by definition of the condition of a matrix it follows that $\kappa \left({A}^{-1}\right)=\kappa \left(A\right)$. So there is no explanation for this. Therefore, my question is: Why is $\kappa \left({A}^{-1}\right)=\kappa \left(A\right)$
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First you must accept that if ${A}^{-1}$ doesn't exist then by convention $\kappa \left(A\right)=\mathrm{\infty }$
Once you've postulated that ${A}^{-1}$ exists, you want to show
$\underset{‖x‖=1}{max}‖{A}^{-1}x‖={\left(\underset{‖x‖=1}{min}‖Ax‖\right)}^{-1}.$
To see that, take a minimizer ${x}^{\ast }$ on the right side. Let $b=\frac{A{x}^{\ast }}{‖A{x}^{\ast }‖}$; then look at ${A}^{-1}b=\frac{{x}^{\ast }}{‖A{x}^{\ast }‖}$. You picked ${x}^{\ast }$ so that among unit vectors, $‖A{x}^{\ast }‖$ is as small as possible, so $‖{A}^{-1}b‖=\frac{1}{‖A{x}^{\ast }‖}$ is as large as possible among unit vectors b
Intuitively, if A maps ${x}^{\ast }$ to some much smaller vector b, then ${A}^{-1}$ maps b back to ${x}^{\ast }$, which is much bigger than b. Juggling the normalizations obfuscates this intuition a little bit, I think.

We have step-by-step solutions for your answer! ghulamu51
For an invertable matrix, your numerator and denominator are the absolute values of the largest and smallest eigenvalues. The eigenvalues of the inverse matrix are the reciprocals of its eigenvalues.

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