Finding dimensions of null and column space for different parameters of $t$

Question

Let $t$ be a parameter and consider $$Q_t = \begin{bmatrix} 1-t & t \\ 1 & 0 \end{bmatrix}$$ Find $\dim\big(\operatorname{Nul}(Q_t)\big) $ and $ \dim\big(\operatorname{col}(Q_t)\big)$.

So for $\dim \operatorname{col} Q_t$ I reduced the matrix to RRE and got that $\dim \operatorname{col} Q_t = 2$ when $t \ne 1$ and $t \ne 0$, and $1$ when $t = 0$. I wasn't really sure how to find $\dim \operatorname{Nul}Q_t$, but it seems logical if its $1$ when $t = 0$ and $0$ otherwise. Not sure if this is correct though. Also, is there a more methodical way to find $ \dim\operatorname{Nul}Q_t$?

Answer 1

The column space is spanned by the vectors $$\left\{\pmatrix{1-t \\ 1}, \pmatrix{t \\ 0} \right\}$$ which are linearly independent for all $t \in \mathbb{R}$ except $t = 0$ when the second vector is equal to $\pmatrix{0 \\ 0}$.

Therefore: $$\dim \operatorname{col} Q_t = \begin{cases} 2, & \text{if $t \ne 0$} \\ 1, & \text{if $t = 0$} \end{cases}$$

Rank-nullity theorem states that $$\dim \operatorname{null} Q_t + \dim \operatorname{col} Q_t = \dim \mathbb{R^2} = 2$$

hence $$\dim \operatorname{null} Q_t =2 -\operatorname{col} Q_t = \begin{cases} 0, & \text{if $t \ne 0$} \\ 1, & \text{if $t = 0$} \end{cases}$$

Answer 2

As long as the determinant of the matrix doesn't vanish, both dimensions are $2$. Now the determinant equals $-t$ and in case $t=0$ both spaces share dimension one. There's nothing special when $t=1$.