0
$\begingroup$

In Hoffman and Kunze's linear algebra, right after the definition of an eigenvalue for an endomorphism $T: V \to V$ (i.e. $a\in F$ is an eigenvalue if there exists $\alpha\in V$, $\alpha \neq 0$, such that $T\alpha = a\cdot \alpha$) there is a theorem that states the equivalence of three conditions:

  1. $a$ is an eigenvalue of $T$
  2. $aI-T$ is singular
  3. $\det (aI-T) = 0$

I understand $1 \Rightarrow 2 \Rightarrow 3$, but I'm unable to understand the argument that underlies 3 $\Rightarrow$ 1, the book states (and I paraphrase):

If the space $V$ is of finite dimension, then $aI - T$ is non-injective just when its determinant is 0

I know that, if $aI - T$ is non-injective, the existence of an $\alpha$ such that $(aI - T)(\alpha) = 0$ (and, therefore, $T\alpha = a\cdot \alpha$) follows, but I don't understand the relation between $\det(aI-T)$ and the non-injectivity of $aI-T$. Any hints?

$\endgroup$
2
  • $\begingroup$ Well, one could prove it in another easier way. $2\iff 3$ isn't that bad, and $1\iff 2$ would do just fine, am I right? $\endgroup$
    – Miguelgondu
    Commented Feb 19, 2015 at 1:47
  • 1
    $\begingroup$ Denote by $a_1,...,a_n$ the columns of $aI-T$. If the determinant is zero then the columns of the matrix are linearly dependent. This means that there are $x_1,...,x_n$ not all zero such that $x_1a_1+...+x_na_n=0$. If $x$ is a vector with components $x_1,...,x_n$ then $x_1a_1+...+x_na_n=(aI-T)x$. Therefore $(aI-T)x=0$, with $x\neq0$. The equivalence with the non-injectivity (probably) passes through this argument, so it is not really needed. $\endgroup$
    – Tom
    Commented Feb 19, 2015 at 1:47

2 Answers 2

Reset to default
1
$\begingroup$

$\det(aI-T)\neq 0$ is equivalent to $aI-T$ is invertible, hence $(aI-T)\cdot\alpha=0$ implies $\alpha=0$, which means injectivity for linear mappings.

$\endgroup$
5
  • $\begingroup$ Very nice, thanks. So, the finite dimension of $V$ doesn't really matter?, or what am I losing? $\endgroup$
    – Miguelgondu
    Commented Feb 19, 2015 at 1:53
  • $\begingroup$ What is a determinant if the dimension is not finite? $\endgroup$
    – Bernard
    Commented Feb 19, 2015 at 1:55
  • $\begingroup$ V has to be of finite dimension so that we could talk about its matrix representation, is that correct? $\endgroup$
    – Miguelgondu
    Commented Feb 19, 2015 at 1:58
  • $\begingroup$ Infinite matrix also exist (in analysis) but I've never heard of determinants for endomorphisms of infinite dimensional vector spaces; $\endgroup$
    – Bernard
    Commented Feb 19, 2015 at 2:02
  • $\begingroup$ Might be relevant: mathoverflow.net/questions/126464/… $\endgroup$
    – Miguelgondu
    Commented Feb 19, 2015 at 2:03
-2
$\begingroup$

A matrix $M$ has $\det M = 0$ if and only if that matrix is not invertible. Thinking of $M$ as a linear transformation, that it's not invertible means it either fails to be injective or it fails to be surjective. But the rank-nullity theorem will tell you that because the dimensions of the domain and codomain are equal, failing to be surjective implies that it also fails to be injective. So either way it's not injective.

$\endgroup$
4
  • 2
    $\begingroup$ This is not proving anything, only re-stating the same claim as in the book the OP quoted. $\endgroup$
    – Tom
    Commented Feb 19, 2015 at 1:58
  • $\begingroup$ @Tom: This fills in intermediate steps, with considerably more detail than the accepted answer. $\endgroup$
    – Jim
    Commented Feb 19, 2015 at 5:51
  • 1
    $\begingroup$ This takes a huge detour to deduce non-injectivity using a completely unnecessary theorem and ends up not filling any more details than the accepted answer. The only missing detail in the accepted answer is saying that $u\neq v$ with $Mu=Mv$ (non-injectivity) is equivalent to $M(u-v)=0$, i.e. $Mx=0$ has non-zero solutions. $\endgroup$
    – Tom
    Commented Feb 19, 2015 at 12:18
  • $\begingroup$ Yes, indeed you're supposing that I know that "$[T]_B $ is singular $\iff$ $T$ is singular". And that is precisely the question. $\endgroup$
    – Miguelgondu
    Commented Feb 19, 2015 at 14:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .