# Dimension Theorem and nullity

I am currently reading Linear Algebra 4th edition Freidberg et al.

Dimension Theorem is defined as follows:

Let V and W be vector space, and let $$T:V \to W$$ be linear. If V is finite-dimensional, then $$nullity(T) + rank(T) = dim(V)$$

There is an example earlier in the book that states

Define the linear transformation $$T: P_2(R) \to M_{2x2}(R)$$ by $$T(f(x)) = \begin{bmatrix} f(1) - f(2) & 0 \\ 0 & f(0) \end{bmatrix}$$ where $$\beta = \{1,x,x^2\}$$ is a basis for $$P_2(R)$$ and we have $$R(T) = span(T(\beta)) = span(\{T(1), T(x), T(x^2)\})$$ $$R(T) = span(\{\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix}\})$$

as such rank(T) = 2, and dim(V) = 3 so nullity(T) = 1 . What is the role of rank nullity theorem which has stated above? Every element of the basis maps to a non null element of the transforms image. To me it seems that $$\{x,x^2\}$$ both map into the same basis element of the codomain, so the dimension is lost because the linear transform is not bijective and two domain basis elements map into a singular codomain basis element, not because of more things being mapped into the null element.

Can somebody please help with any misunderstandings I may have?

• You are more or less saying the correct ideas, what exactly are you confused about? – suchan Nov 10 '18 at 4:04
• Is your question about why nullity(T) = 1 and not nullity(T) = 0? You are right that x and x^2 both map into the same one-dimensional space. If you look at x^2 - 3x, that's an example of an element that is in the null space. – CJD Nov 10 '18 at 4:09

As you noticed, the basis elements $$x$$ and $$x^2$$ are mapped to the same subspace in the codomain: $$\text{Span}\left(\begin{bmatrix}-1 & 0 \\ 0 & 0\end{bmatrix}\right).$$ Another point of view is that the polynomial $$x^2 - 3x$$ is mapped to the zero matrix, and so it is a basis element of the null space of $$T$$.