User Felix

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Scholar

Jul 25, 2022 at 10:15

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Inverse quadratic norms

got it - thanks :-)

Jul 25, 2022 at 10:12

accepted

Inverse quadratic norms

Jul 25, 2022 at 10:12

comment

Inverse quadratic norms

Thanks a lot. May I ask how you come to the conclusion? We have that $O$ also diagonalizes $A(I_n -A^{-1})$ and that this term is spd. Why is then $(I_n-A^{-1})$ also spd?

Jul 25, 2022 at 9:35

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Supporter

Jul 25, 2022 at 9:32

comment

Inverse quadratic norms

Thank you. This is a bit to abstract for my taste, but it probably also proves the statement.

Jul 25, 2022 at 9:29

comment

Inverse quadratic norms

spd = symmetric positive definite. Apologies for the misunderstanding.

Jul 25, 2022 at 9:26

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Student

Jul 23, 2022 at 16:27

asked

Inverse quadratic norms

Jul 23, 2022 at 11:40

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