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danielI

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So we know that [tex]\det(A^2) = (\det A)^2 = \det 0 \Leftrightarrow \det A = 0[/tex]

We should now show that [tex]\det(I-A) \not= 0[/tex].

But I'm not sure how to do that. Could someone kick me in the right directon?

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- Thread starter danielI
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- #1

danielI

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So we know that [tex]\det(A^2) = (\det A)^2 = \det 0 \Leftrightarrow \det A = 0[/tex]

We should now show that [tex]\det(I-A) \not= 0[/tex].

But I'm not sure how to do that. Could someone kick me in the right directon?

- #2

Muzza

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- #3

matt grime

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what is (1-x)^{-1} as a series expansion when |x|<1?

This can be suitably altered to tell us what the inverse of 1-A is for any A up to some convergence questions. Since A^2=0 these convergence questions vanish entirely as they do if instead of A^2=0 we had A^r=0

obviously since you have been given r=2 i could be using something, talyor series, you've never heard of, and instead you are supposed to use something that, up to making a sign change, is what Muzza said (difference of two squares, anyone?)

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mathwonk

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danielI

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http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/series/taylor.html

[tex](I-A)^{-1}=(I+A)\text{, since }A^n=A^2\cdot A^{n-2} = 0\cdot A^{n-2}=0[/tex]

[tex](I - A)(I + A) = II + IA - AI - AA = I + A - A - 0 = I[/tex]

[tex](I + A)(I - A) = II - IA + AI - AA = I - A + A - 0 = I[/tex]

Now when I've solved this problem I would like to see your solutions (if it's different). It's allways good to learn more ways to solve a problem

mathwonk: exactly what do you mean by 'kernel'?

[tex](I-A)^{-1}=(I+A)\text{, since }A^n=A^2\cdot A^{n-2} = 0\cdot A^{n-2}=0[/tex]

[tex](I - A)(I + A) = II + IA - AI - AA = I + A - A - 0 = I[/tex]

[tex](I + A)(I - A) = II - IA + AI - AA = I - A + A - 0 = I[/tex]

Now when I've solved this problem I would like to see your solutions (if it's different). It's allways good to learn more ways to solve a problem

mathwonk: exactly what do you mean by 'kernel'?

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- #6

HallsofIvy

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matt grime

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suppose that (1-A)x=0, that is x is in the kernel, then that is the same as x=Ax, but now apply A to both sides, what happens? Now, if U and V are zero what is U+V?

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