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one-to-one linear transformations |
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| Feb24-13, 08:37 AM | #1 |
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one-to-one linear transformations
Why is a linear transformation T(x)=Ax one-to-one if and only if the columns of A are linearly independent?
I don't get it... |
| Feb24-13, 09:14 AM | #2 |
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| Feb24-13, 11:54 AM | #3 |
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Is there no alternative to insanely difficult wikipedia proofs?
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| Feb24-13, 11:55 AM | #4 |
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one-to-one linear transformations
What does your textbook say? What is your textbook?
Do they prove the rank-nullity theorem?? |
| Feb24-13, 02:33 PM | #5 |
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T is one-to-one if and only if T(x) = T(y) implies x = y, if and only if T(x-y) = 0 implies x - y = 0, if and only if T(v) = 0 implies v = 0. But T(v) is a linear combination of the columns of A, so this says the only way to combine the columns of A to get zero is if the vector of coefficients (v) is zero. In other words, the columns of A are linearly independent.
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| Feb24-13, 05:05 PM | #6 |
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Ahh, thanks Jbunny. It makes perfect sense now!
Micromass: Yes, it was, but the proofs in my book are written similarly to wikipedia - very tiresomely. |
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