# Dimension of Image/Kernel of Linear Transformation

Consider the vector spaces $P_n$ (polynomials of degree no more than n). The differentiation D gives a linear transformation from $P_n$ to $P_{n-1}$. What is the dimension of the image and the kernel of D? Is D a valid linear transformation from $P_n$ to $P_n$? If it is, what is the dimension of the kernel and image in this case? It seems like this uses the rank-nullity theorem, but I'm not sure. Thanks

The transformation is a linear transformation. In order for a transformation $T$ to be a linear transformation, it has to implement 2 conditions: $$(1) \hspace{0.2cm} T(v + w) = T(v) + T(w) \\ (2) \hspace{0.2cm} T(\alpha v) = \alpha T(v)$$ for any $v, w \in V$.

The differentiation transformation $D$ satisfies those constraints because of the derivatives' properties.

As for dimensions:

$Im(D) = P_{n-1}$, because for any $p^{\prime}(x) \in P_{n-1}$ you can find some $p(x) \in P_n$ s.t. $D(p(x)) = p^{\prime}(x)$, namely $\int f^{\prime}(x) dx$. Therefore $dim(Im(D)) = dim(P_{n-1}) = n$.

By rank nullity theorem, $dim(Im(D)) + dim(Ker(D)) = dim(P_n)$, and therefore $n + dim(Ker(D)) = n+1$ and $dim(Ker(D)) = 1$

• So what would be the dimension of Im(D)? Would it just be n? – JanoyCresva Apr 15 '17 at 19:47
• Had a typo, edited. The dimension of $Im(D)$ is $n$ as explained in my answer – AsafHaas Apr 15 '17 at 19:49
• Ok I got it. I'm still slightly confused about the reasoning on why it is a valid linear transformation, though – JanoyCresva Apr 15 '17 at 19:53
• Edited my answer to contain an explanation of this as well. – AsafHaas Apr 15 '17 at 20:03

Yes, it's a valid linear transformation. If $f, g \in P_n$, $c \in \mathbb{R}$ or whatever field you're working over, $D(f + cg) = D(f) + cD(g)$.

$D$ is surjective. For any $f \in P_{n-1}$, you can integrate, $\int f \in P_n$, and then $D(\int f) = f$.

The kernel has dimension 1. You could do this by rank nullity, or just note that if $D(f) = 0$, $f$ is constant.

The mapping $D$ is a linear transformation:

Proof: Let $f,g \in P_n$, let $a,b \in \mathbb{R}$

$$(af+bg)'(c) = \lim\limits_{x\to c} \frac{(af+bg)(x) - (af + bg)(c)}{x-c}$$ $$= \lim\limits_{x\to c} \frac{af(x)+bg(x) - af(c) - bg(c)}{x-c}$$ $$= \lim\limits_{x\to c} \frac{af(x) - af(c) +bg(x) - bg(c)}{x-c}$$ $$= \lim\limits_{x\to c} \frac{af(x) - af(c) }{x-c} + \lim\limits_{x\to c} \frac{ bg(x) - bg(c)}{x-c}$$ $$= a\lim\limits_{x\to c} \frac{f(x) - af(c) }{x-c} + b\lim\limits_{x\to c} \frac{ g(x) - g(c)}{x-c}$$ $$=af'(c) + bf'(c)$$

Hence: $D(af + bg) = aD(f) + bD(g) \quad \triangle$

$Im(D) = P_{n-1}$

Proof:

Let $Q \in Im(D)$ Then, there is a $P \in P_n$ such that $D(P_n) = Q$. Because differentiation is an operation that reduces the exponent of a power function by $1$ (for example polynomials), $Q \in P_{n-1}$. Hence, $Im(D) \subset P_{n-1}$

Now, let $Q \in P_{n-1}$. Then, there is a $P$ in $P_n$ such that $D(P) = Q$. Simply take $$P = \int Q dx$$ Hence, $Q \in Im(D)$

We conclude that $P_{n-1} \subset Im(D)$

obtaining $P_{n-1} = Im(D) \quad \triangle$

From this, it follows that $dim(Im(D)) = n$, and by rank nullity theorem $dim(ker(D)) = (n+1)-n = 1$

One can also prove directly the kernel has dimension 1 (which is, in fact, easier), by showing $ker(D) = P_0$ ($P_0$ might be bad notation but I mean the constant polynomials)

Proof:

$P \in ker(D) \iff D(P) = 0\iff P \text{ constant}$ We deduce that $ker(D) = P_0 \quad \triangle$

Alternatively, by rank nullity theorem, it would follow that $dim(Im(D)) = n$