Here's a problem I have. Consider having 4 vectors in $R^5$ such that $3\vec{v_{1}}+\vec{v_{2}}=4\vec{v_{4}}$. Let $T$ be a linear transformation from $R^4$ to $R^5$ such that $T(e_{j})=v_{j}$ where $e$ is a normal vector.
(a) Find a nonzero vector in kernel of $T$ -- this I solved as $\begin{bmatrix} 3\\ 1\\ 0\\ -4 \end{bmatrix}$
(b) What is the largest possible dimension for $Im(T)$? -- This is the question I am having trouble with. So the matrix for t is clearly 5 by 4, and it has four vectors, relationship between three of which is given to me. By rank-nullity theorem, the dimension of the image is 4 minus the number of dependent columns/free variables. However, I am struggling to understand how many dependent columns/free variables I have from $3\vec{v_{1}}+\vec{v_{2}}-4\vec{v_{4}}=0$ -- clearly, I am given a relationship between three vectors, but does it mean that one of them is dependent on the other two, hence two free variables, or 1 dependent column?! I guess I am just confused by the connection -- I know the number of free variables and dependent columns should be the same, so I would appreciate if someone could clarify how to get both from the equation I am given (I realize that the third vector is not defined, so it could or could not be dependent).