The proof that "u" is a subspace of "v" is given below.
Suppose that t:v->w is a linear transformation where "v" and "w" are vector spaces.
Let "z" be the subspace of "w".
Since "t" is a linear transformation, the zero vector of "w" is in "z".
Let x, y ∈ U.
Then t(x), t(y) ∈ z.
Since, "z" is a subspace of "w".
So, "z" is closed under vector addition.
Thus, t(x) + t(y) ∈ z.
Now, t(x) + t(y) = t(x + y), since, t is a linear transformation.
Therefore, t(x + y) ∈ z.
Thus, x + y ∈ u.
Therefore, "u" is closed under vector addition.
Hence proved.
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