Linear Algebra: abstract/general vector space (Determinant space) Dimension theorem

Question

I have this linear algebra question

Note the transformation $T:\mathbb{R}^2 \rightarrow \mathbb{R}$ \begin{equation} T(x_1,x_2) = \det\begin{bmatrix} x_1 & x_2 \\ -5 & 5 \end{bmatrix}\\ \end{equation}

So for this transformation we need to find the kernel and verify the Rank-Nullity Theorem.

So far, I can choose values of $x_1$ and $x_2$ to get the determinant to be zero. So from determinant we would get:

$x_1\cdot5 - (-5)\cdot x_2 = 0$

$5x_1+5x_2 = 0$

And cleaning this up, we get: $x_1 = -x_2$

So the vectors in the kernel form this vector: $$ (x_1,x_2) = (x_1,-x_1) = x_1(1,-1) $$ So we have this parameterized solution using $x_1$ as a parameter. So this I believe is 1-dimensional. So Nullity = 1.

But how to get the rank and with the dimension of the domain, not too sure.

Answer 1

Hint: Everything you have done is fine, now use the rank-nullity theorem