# 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.

-
Per chance you could try $a=T_1(1)/T_2(1)$?And use the fact that $Ker(T_1)=Ker(T_2)$ is either =$R^n$ or =$R^{n-1}$. – awllower Nov 21 '12 at 12:29
– wj32 Nov 21 '12 at 12:30
@awllower I can't rmemeber learning that Ker(T1)=Ker(T2) is either R^n or R^n-1. Can you show me a quick proof of that, is that is convenient? or maybe link me to a similar question. Thanks! – Yellow Skies Nov 21 '12 at 12:41
It was not supposed to be 1, but any generator of the complement of the kernel of the transformations. Sorry for that error. – awllower Nov 21 '12 at 12:45
I still don't quite understand. What is a generator? I am beginning my course in linear algebra haha. Sorry and thanks!@ – Yellow Skies Nov 21 '12 at 12:46

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, $T_1 = aT_2$.

-

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 $$T_2(x)=a(v_1\cdot x)=aT_1(x).$$

-
Is that inverted T a symbol for transpose? – Yellow Skies Nov 21 '12 at 12:50
No, it means the orthogonal complement of the subspace generated by that vector, at least so I suppose. – awllower Nov 21 '12 at 12:52
For a subset $U$ of $\mathbb{R}^n$ the notation $U^\perp$ stands for the orthogonal complement of $U$, i.e. $U^\perp=\{x \in \mathbb{R}^n:\ x\cdot u=0 \quad \forall\ u \in U\}$. – Mercy King Nov 21 '12 at 12:52
Nice geometric interpretation! – awllower Nov 21 '12 at 12:53
Ok this looks like what I am looking for, but just one last problem haha, either I forgot, or this was not on my textbook, but $$\ker(T_1)=v_1^\perp=\ker(T_2)=v_2^\perp,$$ I don't understand. – Yellow Skies Nov 21 '12 at 12:55

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 $T_2\left(v-\dfrac{T_1(v)}{T_1(u)}u\right)=0 \Rightarrow T_2(v)=\dfrac{T_1(v)}{T_1(u)}T_2(u)=\dfrac{1}{\alpha}T_1(v) \Rightarrow T_1(v)=\alpha T_2(v).$

-