1. ## Rank and Nullity

Let T : P3 -> R

T(a0 + a1x + a2x^2 + a3x^3) = a0 - a1 + a2 - a3

Find the Rank and Nullity of T and find a basis for the kernel of T.

For my work :

I set a3 = a0 - a1 + a2
Plugged this into a3 for the left hand side and after some simplifications, I get

a0(1 + x^3) +a1(x - x^3) + a2(x^2 + x^3)

Basis = [ (1 + x^3), (x - x^3), (x^2 + x^3) ]

Now my professor says the nullity is 3 which makes the rank = 1 but i'm not understanding WHY it's 3. Also, my professor chose initially to set a3 = a0 - a1 + a2. If I had chosen to set say a1 = a0 + a2 - a3 would it still be correct even though the Basis would look different?

2. Originally Posted by JonathanEyoon
Let T : P3 -> R

T(a0 + a1x + a2x^2 + a3x^3) = a0 - a1 + a2 - a3

Find the Rank and Nullity of T and find a basis for the kernel of T.

For my work :

I set a3 = a0 - a1 + a2
Plugged this into a3 for the left hand side and after some simplifications, I get

a0(1 + x^3) +a1(x - x^3) + a2(x^2 + x^3)

Basis = [ (1 + x^3), (x - x^3), (x^2 + x^3) ]

Now my professor says the nullity is 3 which makes the rank = 1 but i'm not understanding WHY it's 3. Also, my professor chose initially to set a3 = a0 - a1 + a2. If I had chosen to set say a1 = a0 + a2 - a3 would it still be correct even though the Basis would look different?
you found a basis for $\ker T$ with 3 elements $1+x^3, x - x^3, x^2 + x^3,$ this exactly means that the nullity of $T$ is 3. (remember: $\text{nullity of} \ T = \dim \ker T$)

3. Thanks~!!