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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).

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The Rank-Nullity Theorem tells you that if you have a linear transformation $T\colon \mathbf{V}\to\mathbf{W}$, then the dimension of the kernel of $T$ plus the dimension of the image of $T$ equals the dimension of $\mathbf{V}$. In this case, since your $T$ maps $\mathbb{R}^4$ to $\mathbb{R}^5$, you know that the dimension of the kernel of $T$ plus the dimension of $\mathrm{Im}(T)$ equals $4$ (the dimension of $\mathbb{R}^4$). Since you know that $\dim(\mathrm{ker}(T))\geq 1$ (you already produced at least one nonzero element in the kernel), that tells you that $\dim(\mathrm{Im}(T))\leq \cdots$.

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Thanks Arturo, I see how having at least one non-trivial element in Kernel makes my Image's dimension $\leq$ 3. I was just wondering if there was a way to deduce from the equation that I was given (or any other equation of this type) if there is more than one dependent column. –  LinAlgStudent Jan 31 '11 at 1:12
    
@user5157: No, because while you are told that the relation $3\mathbf{v}_1 + \mathbf{v}_2 = 4\mathbf{v}_4$ holds, you are not told whether any other relations hold. So you aren't really given enough information to figure out the dimension of $\mathrm{ker}(T)$. –  Arturo Magidin Jan 31 '11 at 1:15
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