(adsbygoogle = window.adsbygoogle || []).push({}); Show that every eigenvalue of A is zero iff A is nilpotent (A^k = 0 for k>=1)

i m having trouble with going from right to left (left to right i got)

we know that det A = product of the eignevalues = 0

when we solve for the eigenvalues and put hte characteristic polynomial = 0

then

[tex] det (\lambda I - A) = det (-A) = 0 [/tex]

but i have a feeling taht i am not allowed to do that last step because the characteristic polynomial need not be equal to zero.

(Also i do not know the Cayley Hamilton theorem.)

If A is symmetric show taht every eigenvalue of A is nonnegaitve iff A = B^2 for some symmetric matrix B

im not even quite sure where to start with this one! Please help

thank you for any help!

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# Linear Algebra

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