Show that there is a non-zero constant $a$ such that $T_1(v)=aT_2(v)$ for all $v\in R^n$ Let $T_1$ and $T_2$ be two linear transformations from $R^n$ to $R^1$. If $ker(T_1)=ker(T_2)$, show that there is a non-zero constant $a$ such that $T_1(v)=aT_2(v)$ for all $v\in R^n$

My Attempt-

Analysis: I need to achieve $T_1(v)=aT_2(v)$, that is $Av=aBv$
Since $T_1$ and $T_2$ are mappings from $R^n$ to $R^1$, they must both be $1\ X\  n$ row  matrices. (Not sure how this will help.)
Let the standard matrix for $T_1$ and $T_2$ be A and B respectively. Therefore, $ker(T_1)$ is the solution space of $Ax=0$ and similarily, $ker(T_2)$ is the solution space of $Bx=0$


I am not making any progress. Can anyone guide me in the right direction? Thanks Stack!@ Owe you one again and again. 
 A: We consider two cases: If $\ker T_1 = \mathbb R^n$, we have $T_1 = 0$, and by $\ker T_2 = \mathbb R^n$ also $T_2 = 0$. Then $a = 1$ will do.
Otherwise, $\ker T_1$ is a $n-1$ dimensional subspace of $\mathbb R^n$, let $x \in \mathbb R^n \setminus \ker T_1$. As $\ker T_1$ is $(n-1)$-dimensional, we have $\ker T_1 \oplus \langle x \rangle = \mathbb R^n$. We have $T_2(x) \ne 0$, as $x \not\in \ker T_2 = \ker T_1$, now let $a := \frac{T_1(x)}{T_2(x)}$. Let $y \in \mathbb R^n$, then $y$ can be written as $y = u + \lambda x$ with $u \in \ker T_1$, $\lambda \in \mathbb R$. We have 
\begin{align*}
  T_1(y) &= T_1(u) + \lambda T_1(x)\\
         &= \lambda T_1(x)\\
         &= \lambda a T_2(x)\\
         &= a T_2(u) + a\lambda T_2(x)\\
         &= aT_2(u+ \lambda x)\\
         &= a T_2(y).
\end{align*}
As $y$ was arbitrary, $T_1 = aT_2$.
A: We have
$$
T_1(x)=v_1\cdot x,\  T_2(x)=v_2\cdot x \quad \forall x \in \mathbb{R}^n
$$
for some $v_1,v_2 \in \mathbb{R}^n$. Since
$$
\ker(T_1)=v_1^\perp=\ker(T_2)=v_2^\perp,
$$
it follows that $v_2=a v_1$ for some nonzero real constant $a$. Hence for every $x \in \mathbb{R}^n$ we have
$$
T_2(x)=a(v_1\cdot x)=aT_1(x).
$$
A: If $\dim( \ker( T_1))=n$ is easy. So assume $\dim( \ker( T_1))\neq n \Rightarrow T_1 \neq0$. The exercise suggest to define $\alpha$ as $\alpha=\dfrac{T_1(u)}{T_2(u)}$ for some $u$ with $T_1(u)\neq 0$. If $T_1(v)\neq 0$ then $v-\dfrac{T_1(v)}{T_1(u)}u \in \ker( T_1)=\ker( T_2)$. Thus $T_2\left(v-\dfrac{T_1(v)}{T_1(u)}u\right)=0 \Rightarrow T_2(v)=\dfrac{T_1(v)}{T_1(u)}T_2(u)=\dfrac{1}{\alpha}T_1(v) \Rightarrow T_1(v)=\alpha T_2(v).$
