Define T:P2​→P2​by T(p)=p(0)−p(1)t+p(2)t2

Define T:P
2
​
→P
2
​
by T(p)=p(0)−p(1)t+p(2)t
2
.
a. Show that T is a linear transformation. b. Find T(p) when p(t)=−2+t . Is p an eigenvector of T ? c. Find the matrix for T relative to the basis {1,t,t
2
} for P
2
​
.

The Correct Answer and Explanation is :

To solve the problem, we need to analyze the transformation ( T: P_2 \to P_2 ) defined by ( T(p) = p(0) – p(1)t + p(2)t^2 ). Here, ( P_2 ) is the vector space of polynomials of degree at most 2. We will break down the solution into parts:

a. Show that T is a linear transformation.

To show that ( T ) is a linear transformation, we need to verify two properties:

1. Additivity: ( T(p + q) = T(p) + T(q) ) for all ( p, q \in P_2 ).
2. Homogeneity: ( T(cp) = cT(p) ) for all ( p \in P_2 ) and all scalars ( c ).

Let ( p(t) ) and ( q(t) ) be polynomials in ( P_2 ):

• Additivity:
  [
  T(p + q) = (p + q)(0) – (p + q)(1)t + (p + q)(2)t^2
  ]
  Using the linearity of evaluation:
  [
  = (p(0) + q(0)) – (p(1) + q(1))t + (p(2) + q(2))t^2
  ]
  [
  = (p(0) – p(1)t + p(2)t^2) + (q(0) – q(1)t + q(2)t^2) = T(p) + T(q)
  ]
• Homogeneity:
  [
  T(cp) = (cp)(0) – (cp)(1)t + (cp)(2)t^2 = c \cdot p(0) – c \cdot p(1)t + c \cdot p(2)t^2 = cT(p)
  ]

Since both properties are satisfied, ( T ) is a linear transformation.

b. Find ( T(p) ) when ( p(t) = -2 + t ) and determine if ( p ) is an eigenvector of ( T ).

First, we evaluate ( T(p) ):

• Calculate ( p(0) = -2 ), ( p(1) = -1 ), and ( p(2) = 0 ).
  [
  T(p) = -2 – (-1)t + 0 \cdot t^2 = -2 + t
  ]

Now, check if ( p(t) = -2 + t ) is an eigenvector. A polynomial ( p ) is an eigenvector of ( T ) if ( T(p) = \lambda p ) for some scalar ( \lambda ). We have:
[
T(p) = -2 + t \quad \text{and} \quad p = -2 + t
]
This implies ( T(p) = p ), meaning ( \lambda = 1 ). Hence, ( p ) is indeed an eigenvector of ( T ) with eigenvalue ( \lambda = 1 ).

c. Find the matrix for T relative to the basis ( {1, t, t^2} ) for ( P_2 ).

To find the matrix representation, we compute ( T ) for each basis element:

1. ( T(1) ):
   [
   T(1) = 1(0) – 1(1)t + 1(2)t^2 = 0 – t + 0 = -t \rightarrow \begin{pmatrix} 0 \ -1 \ 0 \end{pmatrix}
   ]
2. ( T(t) ):
   [
   T(t) = t(0) – t(1)t + t(2)t^2 = 0 – t + 0 = -t \rightarrow \begin{pmatrix} 0 \ -1 \ 0 \end{pmatrix}
   ]
3. ( T(t^2) ):
   [
   T(t^2) = t^2(0) – t^2(1)t + t^2(2)t^2 = 0 – t^3 + 2t^2 \rightarrow \text{(Not applicable since this exceeds } P_2)
   ]
   Here, we compute it correctly:
   [
   T(t^2) = 0 – 0 + 2t^2 = 0 + 0 + 2t^2 \rightarrow \begin{pmatrix} 0 \ 0 \ 2 \end{pmatrix}
   ]

Combining these results, the matrix representation of ( T ) with respect to the basis ( {1, t, t^2} ) is:
[
[T] = \begin{pmatrix}
0 & 0 & 0 \
-1 & -1 & 0 \
0 & 0 & 2
\end{pmatrix}
]

Summary

In conclusion, we demonstrated that ( T ) is a linear transformation by confirming it satisfies the properties of additivity and homogeneity. Evaluating ( T ) at a specific polynomial showed that ( p(t) = -2 + t ) is an eigenvector with eigenvalue ( 1 ). Finally, we derived the matrix representation of ( T ) in the given basis, providing a complete overview of the transformation’s structure in ( P_2 ).