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Let M be an invertible matrix- and let lambda be an eigenvalue of M- S.docx

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28 jan. 2023

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Let M be an invertible matrix, and let lambda be an eigenvalue of M. Show that 1/lambda is an
eigenvalue of M-1.
Solution
...

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Let M be an invertible matrix, and let lambda be an eigenvalue of M. Show that 1/lambda is an eigenvalue of M-1.
Solution
is eigenvalue so we have
Mx=x for some nonzero vector x
By multiplying both sides with M -1
we get
M -1 Mx=M -1 x
x=M -1 x
Now is nonzero because if =0, then Mx=0, so M -1 Mx=x=0. But x was nonzero
so we can divide by
1/x=M -1 x which shows 1/ is an eigenvalue of M -1
.

Let M be an invertible matrix, and let lambda be an eigenvalue of M. Show that 1/lambda is an eigenvalue of M-1.
Solution
is eigenvalue so we have
Mx=x for some nonzero vector x
By multiplying both sides with M -1
we get
M -1 Mx=M -1 x
x=M -1 x
Now is nonzero because if =0, then Mx=0, so M -1 Mx=x=0. But x was nonzero
so we can divide by
1/x=M -1 x which shows 1/ is an eigenvalue of M -1
.

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1. Let M be an invertible matrix, and let lambda be an eigenvalue of M. Show that 1/lambda is an eigenvalue of M-1. Solution is eigenvalue so we have Mx=x for some nonzero vector x By multiplying both sides with M -1 we get M -1 Mx=M -1 x x=M -1 x Now is nonzero because if =0, then Mx=0, so M -1 Mx=x=0. But x was nonzero so we can divide by 1/x=M -1 x which shows 1/ is an eigenvalue of M -1