# Linear Transformations w/ Kernel & Range

Let $V=M_{2,3}(F)$ and $p_7=\{p(x)\in F[x]\mid \deg(p)\leq 7\}$

If $T: V \to p_7$is a linear transformation with

$$T \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\\end{bmatrix}=0$$

and $$T \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\\end{bmatrix} = 0$$

then the Range(T) is at most four-dimensional. True or False

Ok, so I understand you need to use the rank nullity thm for this $$\dim(\mathrm{Range}(T))+\dim(\ker(T))=\dim(V)$$ and $\dim(V)=6$, and the dim of range must be equal to or less than $4$. What I don't understand is how to find the kernel. Any help please?

• I have been told the statement is true, b/c the Kernel must be at LEAST 2 dimensional, but I still don't know how to find the kernel – Johnathon Svenkat Mar 6 '13 at 21:49
• What are $M_{23}$ and $p_7$? I'm guessing the first are $2 \times 3$ matrices? – Jim Mar 6 '13 at 21:52
• Yep, and P_7 is all polynomials of to the 7th degree (so 8th dimension I believe). and thanks for the edits, I'm still learning how to do all the formatting – Johnathon Svenkat Mar 6 '13 at 21:54

You are correct in that you need to use: $$\dim(\mathrm{Range}(T))+\dim(\ker(T))=\dim(V)$$ Specifically you will use: $$\dim(\mathrm{Range}(T))=\dim(V) - \dim(\ker(T))$$ As you've said $\dim(V) = 6$ so to prove that $\dim(\mathrm{Range}(T)) \leq 4$ we need to show that $\dim(\ker T) \geq 2$. We do this by showing that there are two linearly independent elements in the kernel. These are just the two elements that are given to be in the kernel by hypothesis.
It would be a good exercise for you to produce 5 examples of $T$ whose kernels have dimension 2, 3, 4, 5 and 6, respectively!