Let $A _{t} = \left[\begin{matrix}2 & t \\ 0 & 2\end{matrix}\right]$. Are the elements $I, A _{t}, A^2_{t}, A^3_{t}$ linearly independent in the set of matrices of type $2 \times 2$? What is the dimension of the subspace $V_{t}$ which they span? (The answer depends on $t$.)
I understand that I would have to solve equations to show whether the matrices are linearly independent. Such equations would be such that the first entry of the matricx would form an equation with certain coefficients, the second entry of the matrix would form an equation with the same coefficients as the first one and so on. But then what does it mean that the dimension depends on $t$?
If $t=0$, then $A_0=2I$, and the family of matrices that you study becomes $\{I,\,2I,\,4I,\,8I\}$, which is obviously linearly dependent. This family spans a vector space of dimension $1$.
Now, if $t\ne0$, then $$A_t^2 = \begin{pmatrix}4&4t\\0&4\end{pmatrix},\quad A_t^3 = \begin{pmatrix}8&12t\\0&8\end{pmatrix}.$$
Observe that $A_t^2 = 4A_t -4I$, hence your family of four matrices is linearly dependent; moreover, $$A_t^3 = A_t(4A_t -4I) = 4A_t^2-4A_t = 12A_t-16I.$$
Therefore, your family is nothing but $$\{I,A_t,4A_t -4I,12A_t-16I\}.$$ This implies that the dimension of $V_t$ is equal to $2$.