Question

Recall that Pn is the vector space of polynomials with degree at most n.

The elements in Pn are: p(t)=a0+a1t+⋯+antⁿ,
where a0,a1,…,an are real numbers. Here are four maps
(1) T1:Pn→Pn,p(t)→p(t+1)
(2) T2:P2→P3,p(t)→∫1tp(x)dx
(3) T3:P3→P3,p(t)→p(t)−t2
(4) T4:P2→P4,p(t)→t2p(t)

Please select all linear transformations among these four maps.
a. (1)
b. (2)
c. (3)
d. (4)

Answer 1

Maps T1 (shifting polynomial), T2 (integration), and T3 (subtracting a fixed polynomial) are linear transformations as they maintain both additivity and homogeneity.

However, T4 (multiplication by t²) is not linear since it may change the polynomial's degree, thus failing the additivity property.

Therefore, the correct options are: a. (1), b. (2) and c. (3).

Step-by-step explanation:

To determine which of the maps (T1, T2, T3, and T4) are linear transformations, we must check whether they satisfy both linearity properties: the ability to preserve addition (additivity) and scalar multiplication (homogeneity).

Let's examine each map in turn.

• T1 shifts all the coefficients in the polynomial without altering the scalar factors or the sum of polynomials, thus maintaining both linearity properties. Therefore, T1 is a linear transformation.

• T2 involves the integration of a polynomial, which is a linear operation since it respects both addition and scalar multiplication. As such, T2 is also a linear transformation.

• T3 subtracts a fixed polynomial, t², from the input polynomial. Subtraction of a fixed polynomial does not affect the additivity and homogeneity of the map, implying that T3 is linear too.

• T4 multiplies the polynomial by t², which changes the degrees of the terms. While it does preserve scalar multiplication within the terms of the polynomial, it is not guaranteed to preserve the degree of the resulting polynomial, thus it does not always satisfy the additivity property for polynomials specifically in the vector space P2. Therefore, T4 is not a linear transformation.