Rank-Nullity Possible Pairs

Question

Let $T : \mathbb{R}^3 → \mathbb{R}^3$ be a linear map of vector spaces over the reals. Assume that $T(1,2,0) = (2,3,0)$ and $T(−1,1,0) =(1, 1,0)$. What are the possible pairs of Rank(T) and Nullity(T)?

I know that Rank(T)+Nullity(T)=3. How do I find specific values of Rank(T) and Nullity(T)? Any guidance is much appreciated

Answer 1

Since $T(\Bbb R^3)$ contains $(2,3,0)$ and $(1,1,0)$, which are linearly independent, $\operatorname{rank}(T)$ is at least $2$. If $T(0,0,1)=(0,0,0)$, then $T(\Bbb R^3)=\operatorname{span}\bigl(\{(2,3,0),(1,1,0)\}\bigr)$ and therefore $\operatorname{rank}(T)=2$ and $\operatorname{null}(T)=1$. And if $T(0,0,1)=(0,0,1)$, then $T(\Bbb R^3)=\Bbb R^3$. So, $\operatorname{rank}(T)=3$ and $\operatorname{null}(T)=0$.

Answer 2

It's correct that by the dimension theorem we have that $$\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim(\mathbb{R}^{3})=3.$$ Now, you need to use the first hypothesis of the problem: we need that $T$ satisfy:

1. $T(1,2,0)=(2,3,0)$.
2. $T(-1,1,0)=(1,1,0)$.

Now, note that $$B=\left\{\color{red}{\begin{pmatrix} 1\\ 2 \\ 0 \end{pmatrix}}, \color{red}{\begin{pmatrix} -1\\ 1 \\ 0 \end{pmatrix}}, \begin{pmatrix} 0\\ 0 \\ 1 \end{pmatrix} \right\}\subset \mathbb{R}^{3}$$ is a linearly independent set and generates $\mathbb{R}^{3}$; that is, $B$ is a basis for $\mathbb{R}^{3}$ (the starting vector space). We know the transformation of $(1,2,0)$ and $(-1,1,0)$, but we have added a vector but do we know the transformation of that vector? Sure not, but we know we only have two options:

1. $T(0,0,1)=(0,0,0) \implies (0,0,1)\in \operatorname{ker}(T)\implies \operatorname{nullity}(T)=1.$
2. $T(0,0,1)\not=(0,0,0) \implies (0,0,1) \in \operatorname{im}(T) \implies \operatorname{rank}(T)=3. $

Therefore, the answer is:

1. $(\operatorname{rank}, \operatorname{nullity}(T))=(2,1).$ (First choice).
2. $(\operatorname{rank}, \operatorname{nullity}(T))=(3,0).$ (Second choice).

Also, note that always we have that $$2+1=3, \quad \text{and} \quad 3+0=3.$$ no matter what choice we make.

Answer 3

$$Rank(T)+Nullity(T)=3$$

The span of the transformation already contains $\left\{\begin{pmatrix}2\\3\\0\end{pmatrix},\begin{pmatrix}1\\1\\0\end{pmatrix}\right\}$

Thus if there exists any vector $v=\begin{pmatrix}a\\b\\c\end{pmatrix}$ such that $T(v)=\begin{pmatrix}a'\\b'\\c'\end{pmatrix},c'\ne0$ then Rank(T)=3 and Nullity(T)=0. Otherwise, if $T(v)=\begin{pmatrix}a'\\b'\\0\end{pmatrix}$ for all v where a' and b' change, then Rank(T)=2 and Nullity(T)=1.