Let T: Vy is defined by T (1,5. ) = (x + y + 3) then the kernel of T
has dimension
a.0
b.1
c.2
d.3
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Answer:
Since Im(φ) ⊂ R2, its dimension is at most 2, so that dim(Ker(φ)) ≥ 3. The ... So it cannot possibly be the kernel of a linear map φ : ... Again T is onto because if B ∈ U then B = − BT and 1.
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