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I have been reviewing some ideas about vector spaces and came upon a surprising fact. I am not quite sure how to begin the argument because the problem requires one to construct two bijective linear transformations whose difference is equal to a given linear transformation.

Let $V$ be a vector space over a filed $F$. Suppose $\phi:V \rightarrow V$ is a linear transformation that is not a bijection.

How do we show $\exists f,g :V \rightarrow V$ that are both bjiective linear transformations such that $\phi = f - g$.

I tried proving the fact using contradiction but have not been able to get to far so I am wondering if there is a standard constructive proof that applies directly.

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Take two invertible matrices $A,B$ such that $A-B$ is singular... – user13838 Sep 30 '11 at 17:00
@percusse- So this shows that the difference of bijections does not need to be one. I don't think that's what user7980 is asking for, or am I misunderstanding his question? – Chris Leary Sep 30 '11 at 17:27
@ChrisLeary I am sorry but in return, I did not understand your comment :) Are you commenting on the implication direction? I just gave a hint which is the compact version of Arturo's third hint. – user13838 Sep 30 '11 at 17:47
@user7980: Are you assuming anything about the dimension of $V$? Is it finite dimensional, or are you allowing infinite dimensional? – Arturo Magidin Sep 30 '11 at 18:36
There are no assumptions on the dimension of V. – user7980 Oct 1 '11 at 22:28
up vote 5 down vote accepted

Arturo's hints cover the case that $V$ is finite dimensional and $F$ is infinite. I add the case of $\dim V<\infty$ and a finite field $F$ with a solution that is very similar.

Let $|F|=q$ and $\dim V=n$. We fix once and for all an identification of $V$ with the field $GF(q^n)$ as vector spaces over $F$. Consider the set $$ S=\{\phi(x)/x \mid x\in V, x\neq0\}\subseteq GF(q^n). $$ Because $\ker\phi$ is non-trivial, $0\in S$, so the set $S$ contains at most $q^n-2$ non-zero elements. Let $\alpha\in GF(q^n)\setminus S$. Then for all non-zero $x\in V$ we have $\phi(x)\neq\alpha x$. In other words the mapping $g:V\rightarrow V$ defined by $$ g(x)=\alpha x-\phi(x) $$ has a trivial kernel, and is hence bijective. The mapping $f(x)=\alpha x$ is also bijective, because $\alpha\neq0$. For all $x\in V$ we have $$ f(x)-g(x)=\alpha x-\alpha x+\phi(x)=\phi(x) $$ as required.

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I think that there should be a simpler argument. And for the general case. I tried to come up with one using different bases in the domain and in the range, but the infinite dimensional case is trickier. – Jyrki Lahtonen Sep 30 '11 at 19:04
Thanks for the proof. It looks like the way this problem was written on a previous exam they are looking for a proof that is independent of dimension. Is there any hope of adapting something like this to the infinite dimensional case? – user7980 Oct 1 '11 at 23:23

Hint 1. Show that $a$ is an eigenvalue of $f$ if and only if $a+b$ is an eigenvalue of $f+bI$.

Hint 2. (Assuming $V$ is finite dimensional) Show that $f$ is bijective if and only if $0$ is not an eigenvalue of $f$.

Hint 3. (Assuming the field is infinite) Show that if $f$ is not bijective, then there is a nonzero scalar $b$ such that $f+bI$ is bijective.

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Thanks for the hints they are very clear and informative. Is there a way we can modify hint 2 to account for the infinite dimensional case? – user7980 Oct 1 '11 at 23:24
@user7980: The condition that $0$ is not an eigenvalue is necessary, but not sufficient (since in the infinite-dimensional case, you can have a one-to-one linear transformation that is not onto; e.g., the right-shift operator on the direct product (or sum) of countably many copies of the field. I've been thinking about it on and off, but can't yet think of a way of either ensuring it works, or coming up with a counterexample. – Arturo Magidin Oct 1 '11 at 23:27

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