Nullity of a matrix transformation

Question

I'm having some trouble with one part of this question.

Which of the following statements are true?

$(1)$ For $\theta\epsilon\mathbb{R}$ fixed, $R:\mathbb{R^2} \rightarrow \mathbb{R^2}, R(x,y)=(x\cos(\theta)-y\sin(\theta),x\sin(\theta)+y\cos(\theta))$ is linear.

$(2)$ The transformation $T:\mathbb{R^2} \rightarrow \mathbb{R^2},T(x,y)=(1+x^2,1+y^2)$ is linear.

$(3)$ For $n\geq1$, every linear transformation $L:R:\mathbb{R}^n \rightarrow \mathbb{R}^n$ satisties $L(\mathbf{0})=\mathbf{0}$.

$(4)$ If $L:\mathbb{R^3} \rightarrow \mathbb{R^2}$ is linear, then the matrix which represents $L$ with respect to the standard coordinates must have positive nullity.

$(5)$ If $L:\mathbb{R^2} \rightarrow \mathbb{R^3}$ is linear, then the matrix which represents $L$ with respect to the standard coordinates must have positive nullity.

$(A)$ Only $(1)$,$(3)$, and $(5)$ are true.

$(B)$ Only $(1)$,$(3)$, and $(4)$ are true.

$(C)$ Only $(2)$,$(3)$, and $(4)$ are true.

$(D)$ Only $(2)$,$(3)$, and $(5)$ are true.

$(E)$ Only $(1)$,$(2)$, and $(5)$ are true.

I know that $(1)$ is true, $(2)$ is false and $(3)$ is true. What is the difference between $(4)$ and $(5)$ though? How do I know which one is correct? Apparently statement $(4)$ is the correct one but why is that?

Answer 1

Recall that the rank-nullity theorem states that the rank plus the nullity of a linear transformation equals the dimension of its domain.

In terms of the matrix of the linear transformation, the dimension of its columnspace plus the dimension of its nullspace equals the number of columns.

In (4), the matrix is $2 \times 3$, so the rank plus nullity equals $3$. The rank must be $\le 2$ (do you see why?), so the nullity must be $\ge 1$. Thus (4) is true.

In (5), the matrix is $3 \times 2$, so the rank plus nullity equals $2$. It is possible for the rank to equal $2$ while the nullity equals $0$; for example consider $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}.$$ So (5) is false.

Answer 2

As an easy counterexample to $(5)$, take the embedding of $\mathbb{R}^{2}$ into $\mathbb{R}^{3}$ given by $T(x_{1},x_{2})=(x_{1},x_{2},0)$. This is obviously linear and the null space consists of only the zero vector $(0,0$).

However, if $T:\mathbb{R}^{3}\to\mathbb{R}^{2}$ is linear, the null space must have positive dimension. This follows immediately from the rank-nullity theorem, since the rank of the matrix representing $T$ can be at most $2$, but the rank plus the nullity must equal $3$, the number of columns of this matrix.

Intuitively, if you are going from a bigger space to a smaller one in a linear fashion, a lot of elements are going to have to get killed (mapped to $0$). In particular, the dimension of the elements that are killed (i.e. the nullity of the mapping) has to be at least as big as the difference in dimension between the domain and codomain, which in this case is $3-2=1$.