# Calculating the dimension of a vector space in 2 different ways

Let $T$ be the linear transformation represented by the matrix

$$\left( \begin{array}{ccc} 1 & 1 & 0 & 3 \\ 1 & 1 & 1 & 5 \\ 2 & 2 & 1 & 8 \end{array} \right)$$

One can easily calculate that the image of this as a map from $\mathbb{R}^4\to\mathbb{R}^3$ is 3.

Call the above matrix $A$. Now consider the space $V$ of linear maps $B$ from $\mathbb{R}^2\to\mathbb{R}^4$ satisfying $AB=0$. Plainly $B$ is in $V$ iff the image of $B$ is in the kernel of $T$.

What is the dimension of $V$? Directly calculating, I get 4. However, it seems that one could argue that the dimension is the dimension of the kernel of $T$, which is $2$. What is the flaw in reasoning?

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To rephrase you slightly, $V$ is simply the space of linear maps from $\mathbb{R}^2 \rightarrow ker(T)$. Note that $ker(T)$ is 2 dimensional, as you've said, and then linear maps from a 2 dimensional space to a 2 dimensional space form a vector space of $2 \cdot 2 = 4$ dimensions.
It is not simply the dimension of $ker(T)$ because you've got to account for the dimension of the domain as well.
If the rank of $A$ is $3$, as you say, then it's kernel has dimension $4-3 = 1$, by the rank-nullity-theorem. But the rank is two (the third row is the sum of the first 2 and hence the kernel has dimension 2). Now let's $V$ is, as you say, the space of all linear maps $\mathbb R^2\to \mathbb R^4$, whose image is contained in $\ker T$. So we are asked for the dimension of $L(\mathbb R^2, \ker T)$, that is the space of linear maps from $\mathbb R^2$ to another two dimensional space. The dimension of $L(X,Y)$ is known to be $\dim X \cdot \dim Y$ (if you represent them as matrices, a basis is easily given as those matrices which have exactly one 1 and otherwise zeros), so $\dim V = 2 \cdot 2 = 4$.