MHB Redundancy in Question about Linear Transformations

[Sudharaka]

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]

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]

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]

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 I am f and I am 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? :)