# Dimension and basis of kernel and image of linear transformation

From a bank of past prelim exams:

Let $V$ be the real vector space of real polynomials of degree at most $5$, and define $T: V \to \Bbb{R}^3$ by $T(f) = (f(0), f(1), f(2))$. Find the dimensions of the kernel and image of $T$, and find a basis for each.

I've attempted a solution, working with the standard basis of $V$ and concluded that the kernel is the set of polynomials divisible by $x(x-1)(x-2)$. Thus, I concluded that the dimension of the kernel is $3$, and the dimension of the image is $6-3=3$ by the rank-nullity theorem.

I'm just unsure of how to write the basis of the kernel and image. Any help would be greatly appreciated.

• Ummmm...what is $f$?
– user5137
Aug 25, 2012 at 21:51
• f is a polynomial of degree at most 5 Aug 25, 2012 at 22:00
• That's how I interpreted it. The problem is copied word for word from the exam Aug 25, 2012 at 22:01

The nullspace is all polynomials of the form $$x(x-1)(x-2)(a + bx + cx^2) \text{ for some } a, b, c \in \Bbb{R}.$$ Expand and rewrite as: $$ax(x-1)(x-2) + bx^2(x-1)(x-2) + cx^3(x-1)(x-2)$$ i.e. all the kernel elements are scalar linear combinations of $3$ polynomials. Hence the kernel basis elements are $$\{ x(x-1)(x-2), x^2(x-1)(x-2),x^3(x-1)(x-2)\}$$

Ops, I forgot the other part. The image is $\Bbb{R}^3$ whose basis is the canonical basis. Originally, I over-complicated things: consider every non-zero tuple $(r_0, r_1, r_2) \in \Bbb{R}^3$. We can show that we can construct $f(x) = ax^5 + bx^4 + cx^3 +dx^2 + ex + r_0$ such that $f(0) = r_0, f(1) = r_1, f(2) = r_2$. We can show that such $f$ always exists by an interpolation argument. Hence the image is all of $\Bbb{R}^3$.

Added: explicit construction for $f(x)$ such that $f(0) = r_0, f(1) = r_1, f(2) = r_2$: $$f(x) = \frac{r_2}{2}x(x-1) - r_1 x(x-2) + 2r_0(x-1)(x-2).$$

• Thanks Jennifer! So, to make sure I'm understanding this correctly, we could list the basis for the kernel as you have above, and then list the basis for the image as (1, x, x^2) since the construction you produced is a linear combination of these elements? Aug 26, 2012 at 0:01
• @NikKumar, no (the part about the image). Remember the image is range of the transformation. That is, subspace of $\Bbb{R}^3$. Which subspace it is? It's all of $\Bbb{R}^3 \setminus (0, 0, 0)$. The basis of this subspace is $3$ elements of $\Bbb{R}^3$. 3 because the dimension of the rank $= 3.$ Now, which 3 elements of $\Bbb{R}^3$ generate $\Bbb{R}^3 \setminus (0, 0, 0)$? There's an easy choice: $(1, 0, 0), (0, 1, 0)$ and $(0, 0, 1)$. These are the bases for the image. (important thing to remember: kernel is a subspace consisting of polynomials. image is a subspace of 3-tuples.)
– user2468
Aug 26, 2012 at 1:06
• Thanks so much Jennifer! That clears up all confusion! Aug 26, 2012 at 18:45
• $\mathbb{R}^3\setminus (0,0,0)$ is NOT a "subspace" of $\mathbb{R}^3$. May 30, 2014 at 4:18

Note: I'm going to assume that the underlying field here is $\mathbb{R}$.

Suppose $f\in ker(T)$. Then $T(f)=(f(0),f(1),f(2))=(0,0,0)$, hence $0$, $1$, and $2$ are all roots of $f$. So, $ker(T)\subseteq \{f\in V\,\vert\,f(0)=f(1)=f(2)=0\}$. The reverse containment is obvious.

So, any $f\in ker(f)$ is of the form $$f(x)=x(x-1)(x-2)g(x)$$ where $g(x)$ is a polynomial of degree at most 2. So, we have $$g(x)=ax^2 + bx + c$$ where $a,b,c\in \mathbb{R}$ (and, of course, one or more of $a$, $b$, or $c$ may be zero). In fact, $g\in P_2(\mathbb{R})$, the space of polynomials over $\mathbb{R}$ having degree at most 2.

So, consider the transformation $S:ker(T)\,\rightarrow\,P_2(\mathbb{R})$ defined by $$S(x(x-1)(x-2)g(x))=g(x).$$ I'll bet you can show that $S$ is an isomorphism.

Once you do that, you know $dim(ker(T))$ (and can find a basis easily enough). From there, it's a simple matter to find $dim(im(T))$.

First we consider the Kernel of $T$,

Suppose $f\in \operatorname{ker}(T)$ then we know that $f(0) = f(1) = f(2) = 0$,

so as you say $g = x(x-1)(x-2)$ must divide $f$.

so we can write $$\operatorname{ker}(T) = \{ g h : \operatorname{deg}(h) \leq 2 \}$$

we know that $\operatorname{deg}(h) \leq 2$ because we know that $\operatorname{deg}(f) \leq 5$.

So we can define a basis for $\operatorname{ker}(T)$ using the canonical basis for $\{h : \operatorname{deg}(h) \leq 2\}$

$$B = \{g, gx, gx^2\}$$

So now we have a basis for $\operatorname{ker}(T)$ and hence we know it has dimension 3, so by the rank nullity theorem $\operatorname{Im}(T)$ has dimension $6-3=3$ and thus the image is all of $\mathbb{R}^3$ and hence any basis for $\mathbb{R}^3$ will do the job for the image. (for instance $\{(1,0,0), (0,1,0), (0,0,1)\}$)