- #1

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Show that T = (x, y, z) : -1 ≤ x + y + z ≤ 1

is not a vector subspace of R3

Thanks!

- Thread starter markovchain
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- #1

- 1

- 0

Show that T = (x, y, z) : -1 ≤ x + y + z ≤ 1

is not a vector subspace of R3

Thanks!

- #2

- 606

- 1

Show that T = (x, y, z) : -1 ≤ x + y + z ≤ 1

is not a vector subspace of R3

Thanks!

$$(1,0,0)\in T\,\,\,but\,\,\,2(1,0,0)=(2,0,0)\notin T$$

Thus T cannot be v. subspace as it isn't closed under scalar multiplication

DonAntonio

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