1. ## Non-Bijective Linear Maps

Do all non-bijective linear maps have an eigenvalue 0?

2. Originally Posted by Swlabr
Do all non-bijective linear maps have an eigenvalue 0?
Hi Swlabr.

For $\displaystyle A=\begin{pmatrix}a&b\\c&d\end{pmatrix},$

$\displaystyle \det(A-\lambda I)\ =\ 0$

$\displaystyle \implies\ \left|\begin{array}{cc}a-\lambda&b\\c&d-\lambda\end{array}\right|\ =\ 0$

$\displaystyle \implies\ (a-\lambda)(c-\lambda)-bc\ =\ 0$

$\displaystyle \implies\ \lambda^2-(a+d)\lambda+ad-bc\ =\ 0$

$\displaystyle \implies\ \lambda(\lambda-(a+d))\ =\ 0$ if $\displaystyle ad-bc=0$

$\displaystyle \implies\ \lambda\ =\ 0\ \mbox{or}\ \lambda\ =\ a+d$

So the answer is no, they can have a nonzero eigenvalue if $\displaystyle a+d\ne0.$

3. Originally Posted by TheAbstractionist
Hi Swlabr.

For $\displaystyle A=\begin{pmatrix}a&b\\c&d\end{pmatrix},$
$\displaystyle \det(A-\lambda I)\ =\ 0$
....
/snip
....
So the answer is no, they can have a nonzero eigenvalue if $\displaystyle a+d\ne0.$
I think the question was whether all non-bijective maps have a 0 eigenvalue, rather than whether non-bijective maps have all eigenvalues as zeros.But of course, then the question is trivial

4. Originally Posted by TheAbstractionist
Hi Swlabr.

For $\displaystyle A=\begin{pmatrix}a&b\\c&d\end{pmatrix},$

$\displaystyle \det(A-\lambda I)\ =\ 0$

$\displaystyle \implies\ \left|\begin{array}{cc}a-\lambda&b\\c&d-\lambda\end{array}\right|\ =\ 0$

$\displaystyle \implies\ (a-\lambda)(c-\lambda)-bc\ =\ 0$

$\displaystyle \implies\ \lambda^2-(a+d)\lambda+ad-bc\ =\ 0$

$\displaystyle \implies\ \lambda(\lambda-(a+d))\ =\ 0$ if $\displaystyle ad-bc=0$

$\displaystyle \implies\ \lambda\ =\ 0\ \mbox{or}\ \lambda\ =\ a+d$

So the answer is no, they can have a nonzero eigenvalue if $\displaystyle a+d\ne0.$

...but are they forced to have one that is 0? Alternatively, are non-bijective linear maps forced to have at least one eigenvalue?

I'm asking as apparently for $\displaystyle V$ a finite dimensional vector space, $\displaystyle T \in End(V)$ such that $\displaystyle T$ is nilpotent (there exists $\displaystyle n \in \mathbb{N}$ such that $\displaystyle T^n=0$) then zero is the only eigenvalue. It is clear that zero is the only possible eigenvalue, but I'm unsure if it is indeed an eigenvalue. (Note that if $\displaystyle T$ is nilpotent that it is not a bijection...which I hoped would be the route to the solution...)

5. Originally Posted by Isomorphism
I think the question was whether all non-bijective maps have a 0 eigenvalue, rather than whether non-bijective maps have all eigenvalues as zeros.But of course, then the question is trivial

Yes-that is what I was asking. What do you mean 'the answer is trivial'? Are you perhaps inferring that the zero-vector is an Eigenvector?...

6. Originally Posted by Swlabr
Yes-that is what I was asking. What do you mean 'the answer is trivial'? Are you perhaps inferring that the zero-vector is an Eigenvector?...
Eigenvalues are defined for square matrices. In such cases injectivity implies bijectivity, or rather non-bijectivity implies non-injectivity. But non-injective linear maps are non-invertible and hence 0 is an eigenvalue.

7. But why do non-invertible maps have a zero eigenvalue? Can you perhaps write up a short proof?

Thanks.

8. Originally Posted by Swlabr
But why do non-invertible maps have a zero eigenvalue? Can you perhaps write up a short proof?

Thanks.
$\displaystyle \det(T) = 0 \Leftrightarrow \det(T - 0I) = 0$