# Equivalent Linear Transformations

Given vector spaces $V$ and $W$ over a field $\mathbb{F}$. We'll call two transformations $T,S\in Hom_{\mathbb{F}}(V,W)$ equivalent if there exist invertible transformations $A,B$ in $Hom(W,W)$ and $Hom(V,V)$ respectively such that $S=ATB$ I want to show the following:

$S\sim T$ iff all three of the following hold:

$\bullet$ dim ker$S$=dim ker$T$

$\bullet$ dim im$S$=dim im$T$

$\bullet$ dim coker$S$=dim coker$T$

So I'm making use of the natural exact sequences $$0\longrightarrow Ker(T) \longrightarrow V\longrightarrow W\longrightarrow coker(T)\longrightarrow 0$$$$0\longrightarrow Ker(S) \longrightarrow V\longrightarrow W\longrightarrow coker(S)\longrightarrow 0$$

For the forward implication (I'm not quite sure how to drop the $A$ and $B$ arrows down in TeX...), I've managed to chase my way (I think) to the first two bullets. Starting with a basis for $ker(T)$ extending it, and so on. Then again for $Ker(S)$. I'm a little stuck on showing that the dimensions of the cokernels are the same, however. Naturally, in the finite case things are pretty straightforward once I have that the images have the same dimension, but I can't seem to see what further information I have at my disposal in the event that these vector spaces are infinite dimensional. Any help here would be much appreciated.

For the converse, I want to show that there exist such $A$ and $B$ given that all three bullets hold. So here I'm having trouble making the argument precise, and am also failing to see how the third bullet fits into the picture. Certainly there's easy isomorphisms from the kernels and cokernels. Is it as simple as extending the bases of the Kernels and chasing a little? If so, how can I show what I'm doing is well-defined? Again, apologies for the imprecision in the question...

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## 1 Answer

You're working too hard. Your notion of equivalence is equivalent to being able to freely choose bases in both $V$ and $W$. So:

Choose a basis of $\text{Ker}(T)$. Extend it to a basis of $V$. Use the image of this basis under $T$ to get a basis of $\text{Im}(T)$. Extend it to a basis of $W$. What does $T$ look like with respect to these bases?

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I don't know that anyone has ever accused me of that. You mean what it looks like as a matrix? I guess I can get as many 1s on the diagonal as there are elements of the basis for the image. The rest zeroes yes? – AsinglePANCAKE Oct 22 '12 at 0:12
@AsinglePANCAKE: yes. But the point is that as a matrix $T$ is completely determined by the three dimensions you specified. – Qiaochu Yuan Oct 22 '12 at 0:21
Ah, I see! So given two such transformations, how do I find $A$ and $B$? Just by choosing such bases for the first sequence and expressing them all in terms of the bases in the second sequence? – AsinglePANCAKE Oct 22 '12 at 0:32
@AsinglePANCAKE: yes. – Qiaochu Yuan Oct 22 '12 at 1:11
And I guess I'll press one question further...for the other implication, I've got that the dimensions of the respective kernels and images are the same, but I still can't quite see why the cokernels have to have the same dimension, in the infinite case anyway. – AsinglePANCAKE Oct 22 '12 at 7:00