vector subspaces in $\mathbb{R}^{4}$

Find the dimension of the vector subspace in $\mathbb{R}^{4}$ spanned by the columns

$$\begin{pmatrix} 1 & 1 & 2 & 4 \\ t & 3 & 2 & 8 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

(discussion over real $t$).

Thanks :)

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Rows 1 and 3 are obviously linearly independent, and row 4 is all zeros, so the dimension of the span of the rows (which equals the dimension of the span of the columns) is either $2$ or $3$.
If $t\ne3$ then it's clear that the first three columns are linearly independent, making the required dimension $3$. So the only question is what happens when $t=3$: are the last three columns linearly independent then? But if you subtract the 3rd column from the 4th, you get twice the 2nd, so the answer is no, and the dimension is $2$ if $t=3$.
The dimension of the column space for a matrix $A$ (it's rank) is equivalent to the number of pivot columns that are in $A$. This is because each pivot column represents a distinct linearly independent vector which makes up the column space of $A$. If we manipulate your matrix to echelon form $R$ (which preserves the number of pivots), then you get: $$R=EA=E\begin{pmatrix} 1 & 1 & 2 & 4 \\ t & 3 & 2 & 8 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 2 & 4 \\ 0 & 3-t & 2-2t & 8-4t \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ Note: I multiplied by the elementary matrix $E=\begin{pmatrix} 1 & 0 & 0 & 0 \\ -t & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$. Note that if $3-t$ is nonzero, then there are 3 pivot columns, so $\text{dim } A = 3$. If $t=3$, then we have 2 pivot columns, so $\text{dim } A = 2$.