Linear transformation with special properties

Question

how should I do that please (I had this in my test yesterday)?

Linear transformation $f:\mathbf{R}^{10} \to \mathbf{R}^7$ has an attribute that every vector $\mathbf{v}$ for which is true that $f(\mathbf{v})=\mathbf{o}$ is in linear span $<(1,2,…,10)^T,(1,1,…,1)^T>$. Create such transformation or prove that it doesn't exists.

Thanks for helping.

Answer 1

The linear transformation should satisfy that $$7\geq\underbrace{\dim(\text{Im}(f))=\dim \mathbf{R}^{10}-\dim(\text{Ker}(f))}_{\text{Rank-Nullity theorem}}=10-\dim(\text{Ker}(f))$$

Therefore $$\dim(\text{Ker}(f))\geq3.$$

Answer 2

there is no such transformation $T$ given that $\ker(T) \subset span\{(1,2,\cdots, 10)^T, (1,1, \cdots, 1)^T\}$ because the dimension row space of $T$ is the same as the dimension of the column space which is less or equal to seven. by the nullity theorem, dimension of $ker(T)$ is greater or equal to three. and that contradicts that the dimension of $ker(T)$ is two.