Question

let v be the vector space of polynomial with real coefficients and of degree at most 2. If D=d/dx is the differential operator on v and B={1+2x^2, x+x^2, x^2} is an ordered basis of V, find [D]b. find the rank and nullity of D. Is D invertible? justify your answer.

Answer 1

Answer on Question #40911, Math, Linear Algebra

Let $v$ be the vector space of polynomial with real coefficients and of degree at most 2. If $D = d / dx$ is the differential operator on $v$ and $B = \{1 + 2x^2, x + x^2, x^2\}$ is an ordered basis of $V$ , find [D]b. find the rank and nullity of $D$ . Is $D$ invertible? Justify your answer.

Solution

Let $B_{1} = 1 + 2x^{2}, B_{2} = x + x^{2}, B_{3} = x^{2}$ . Then

$$D B _ {1} = \frac {d}{d x} (1 + 2 x ^ {2}) = 4 x, D B _ {2} = \frac {d}{d x} (x + x ^ {2}) = 1 + 2 x, D B _ {3} = \frac {d}{d x} (x ^ {2}) = 2 x.$$

We can see that

$$D B _ {1} = 4 B _ {2} - 4 B _ {3},$$ $$D B _ {2} = \left(B _ {1} - 2 B _ {3}\right) + 2 B _ {2} - 2 B _ {3} = B _ {1} + 2 B _ {2} - 4 B _ {3},$$ $$D B _ {3} = 2 B _ {2} - 2 B _ {3}.$$

So we get

$$D \left( \begin{array}{c} B _ {1} \\ B _ {2} \\ B _ {3} \end{array} \right) = \left( \begin{array}{c c c} 0 & 4 & - 4 \\ 1 & 2 & - 4 \\ 0 & 2 & - 2 \end{array} \right) \left( \begin{array}{c} B _ {1} \\ B _ {2} \\ B _ {3} \end{array} \right).$$

The differential operator on $v$ with basis $B = \{1 + 2x^2, x + x^2, x^2\}$ expressed by matrix

$$D = \left( \begin{array}{c c c} 0 & 4 & - 4 \\ 1 & 2 & - 4 \\ 0 & 2 & - 2 \end{array} \right).$$

Observe that rows 1 and 3 are linearly dependent, with

$$\operatorname {r o w} _ {1} (D) = 2 \cdot \operatorname {r o w} _ {3} (D).$$

The rank of $D$ is number of linearly independent rows of $D$ . So

$$R (D) = 2.$$

The nullity of $D$ is

$$N (D) = 3 - R (D) = 1.$$

The determinant of $D$ is zero, because rows 1 and 3 are linearly dependent. That's why differential operator $D$ is not invertible.