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(b) Assume that {v1, v2, · · · , vk} is a basis of Y . Show that the linear transform T : X → Y from (a) is invertible. (Hint: Use (a) to define a suitable linear transform U : Y → X and show that it is the inverse of T .)

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(b) Assume that {v1, v2, · · · , vk} is a basis of Y . Show that the linear transform T : X → Y from (a) is invertible. (Hint: Use (a) to define a suitable linear transform U : Y → X and show that it is the inverse of T .)

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Solution

To show that the linear transformation T: X → Y is invertible, we need to define a suitable linear transformation U: Y → X and show that it is the inverse of T.

Step 1: Define the linear transformation U: Y → X

We know that {v1, v2, ..., vk} is a basis of Y. This means that any vector y in Y can be expressed as a linear combination of these basis vectors. We can define a linear transformation U: Y → X such that for any y in Y, U(y) is the vector in X that was mapped to y by T. In other words, if T(x) = y, then U(y) = x.

Step 2: Show that U is the inverse of T

To show that U is the inverse of T, we need to show that applying U after T (or T after U) results in the original vector. That is, we need to show that for any x in X, U(T(x)) = x, and for any y in Y, T(U(y)) = y.

For any x in X, since T is a linear transformation, T(x) is a vector in Y. By the definition of U, U(T(x)) is the original vector x. So, U(T(x)) = x.

For any y in Y, by the definition of U, U(y) is the vector in X that was mapped to y by T. So, T(U(y)) is the same as T applied to the vector that was mapped to y, which is y. So, T(U(y)) = y.

Therefore, U is the inverse of T, and so T is invertible.

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