Latest update

# Kernal of a Transformation

2017-12-14 05:32:26

Let $T: P_2\to \mathbb R_3$ be given by $T(a,b,c)=(a+b)t^2+(a+b+c)t+c$

I have already shown that this is Linear. I need to find the Kernal. I know this requires me to show when $T(a,b,c)$ is equal to the zero vector.

I need help doing this as we have only ever done this with matrix transformation and not these polynomial transformations.

The zero of $P_2(\mathbb{R})$ (polynomials of degree at most $2$ with real coefficients) is the zero polynomial. So, we must find all elements of $\mathbb{R}^3$ that map to the zero polynomial.

To do this, suppose we have an arbitrary element of $\mathbb{R}^3$ that maps to $0$, say $(a, b, c)$. Then we have

$$0\cdot t^2 + 0 \cdot t + 0 = T(a,b,c) = (a+b)t^2 + (a+b+c)t + c.$$

This gives us a system of 3 equations and 3 unknowns (feel free to edit my system in MathJax as I wasn’t sure how to do so). Namely,

$$0 = a+b, 0 = a+b+c, 0 = c.$$

A solution to this is $c=0, b = -a, a \in \mathbb{R}$.

So, any vector that maps to th

• The zero of $P_2(\mathbb{R})$ (polynomials of degree at most $2$ with real coefficients) is the zero polynomial. So, we must find all elements of $\mathbb{R}^3$ that map to the zero polynomial.

To do this, suppose we have an arbitrary element of $\mathbb{R}^3$ that maps to $0$, say $(a, b, c)$. Then we have

$$0\cdot t^2 + 0 \cdot t + 0 = T(a,b,c) = (a+b)t^2 + (a+b+c)t + c.$$

This gives us a system of 3 equations and 3 unknowns (feel free to edit my system in MathJax as I wasn’t sure how to do so). Namely,

$$0 = a+b, 0 = a+b+c, 0 = c.$$

A solution to this is $c=0, b = -a, a \in \mathbb{R}$.

So, any vector that maps to the zero polynomial will have the form $(a, -a, 0)$.

This tells us $$Ker(T) = \{(a, -a, 0) | a \in \mathbb{R} \}.$$

2017-12-14 07:44:14