[線代] A^TA eigenvalue的證明

[ntme (ntme)]

A的eigenvalue 為入

A^TA的eigenvalue為入^2

突然想要來證證看卻不知道怎麼證ＸＤ

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melody

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推 JohnMash    :wrong
→ Annihilator :consider ㄨ^TAㄨ=入 as ||ㄨ||=1 and Aㄨ=入,then..
→ Annihilator :[更正]                              Aㄨ=入ㄨ
→ Annihilator :                            (need A^Tㄨ=入ㄨ)
→ Annihilator :sorry, I am wrong. Don't kill me. Happy New Year!
→ ntme        :A和AT不是有相同eigenvalu??
→ ntme        :XD 我一直以為A^TA的eigenvalue是入^2,正確是什麼阿