# Redundancy in Question about Linear Transformations

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone, Take a look at this question.

Show that if two linear transformations $$f,\,g$$ of rank 1 have equal $$\mbox{Ker f}=\mbox{Ker g},\,\mbox{Im f}=\mbox{Im g},$$ then $$fg=gf$$.
Now the problem is that I feel this question is not properly worded. If the linear transformations have rank = 1 then it is obvious that $$\mbox{Im f}=\mbox{Im g}=\{0\}$$. So restating that is not needed. Don't you think so? Correct me if I am wrong. If I am correct the answer is also obvious. Since the image space of the linear transformations contain only the identity element, $$fg=gf$$.

#### Deveno

##### Well-known member
MHB Math Scholar
A linear transformation of rank 1 maps to a 1-dimensional subspace of the co-domain (which is isomorphic to the underlying field, at least as a vector space). The trivial subspace $\{0\}$ has null basis (and thus 0 dimension), as it is ALWAYS a linearly dependent set.

Two such functions need not have the same image, consider $f,g:\Bbb R^2 \to \Bbb R^2$:

$f(x,y) = (x,0)$

$g(x,y) = (0,y)$.

#### Sudharaka

##### Well-known member
MHB Math Helper
A linear transformation of rank 1 maps to a 1-dimensional subspace of the co-domain (which is isomorphic to the underlying field, at least as a vector space). The trivial subspace $\{0\}$ has null basis (and thus 0 dimension), as it is ALWAYS a linearly dependent set.

Two such functions need not have the same image, consider $f,g:\Bbb R^2 \to \Bbb R^2$:

$f(x,y) = (x,0)$

$g(x,y) = (0,y)$.
Thanks very much. I don't know, but somehow I have forgotten that the trivial subspace has null basis and zero dimension. Back to the square one.

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone, Do you have any ideas how to use the fact that $$\mbox{Ker f}=\mbox{Ker g}$$ in this question? I am finding it hard to see how this fact could be connected to the question. My initial thought was to use the Homomophism theorem for vector spaces, however I don't see how it could be linked. Let me write down what I did for the moment,

It is given that, $$\mbox{rank f} = \mbox{rank g} = 1$$

Then, $$\mbox{Im f} = \mbox{Im g} = <u>$$ where each element of Im f and Im g can be written as a scaler multiple of $$u$$. Now consider $$fg(v)$$ and $$gf(v)$$ for any $$v\in\mbox{Im f}=\mbox{Im g}$$.

$fg(v)=f[g(v)]=f(ku)=kf(u)=(kl)u$

where $$k\mbox{ and }l$$ are scalers. Similarly,

$gf(v)=(mn)u$

Therefore, $$fg(v)=\lambda \,gf(v)$$ where $$\lambda = \lambda (v)$$

What I feel is that if I could use the fact that $$\mbox{Ker f}=\mbox{Ker g}$$ then I might be able to show that $$\lambda=1$$. What do you think? 