Let $V$ be a finite dimensional vector space over $\mathbf C$ and $T:V\to V$ be a linear transformation. Let $p(x)=(x-\lambda_1)^{k_1}\cdots(x-\lambda_m)^{k_m}$ be the minimal polynomial of $T$.

Then is it true that the algebraic multiplicity of $\lambda_i$ is $\dim \ker(T-\lambda_i I)^{k_i}$?

I suspect that this is true since on the one hand, by the Primary Decomposition Theorem, we have

$$V=\bigoplus_{i=1}^m \ker(T-\lambda_iI)^{k_i}$$

On the other hand we also have

$$V=\bigoplus_{i=1}^m(T-\lambda_iI)^{\dim V}$$

Now since $\ker (T-\lambda_iI)^{k_i}\subseteq \ker(T-\lambda_iI)^{\dim V}$, we must have from the above two decompositions of $V$, that $\ker(T-\lambda_iI)^{k_i}=\ker(T-\lambda_iI)^{\dim V}$.

From where it follows that $\dim\ker(T-\lambda_iI)^{\dim V}=\dim\ker(T-\lambda_iI)^{k_i}$, validating the assertion.

Am i making a mistake somewhere or is this alright?


  • $\begingroup$ You should be more clear about what facts you are using. The algebraic multiplicity of $\lambda$ is by definition its multiplicity as root of the characteristic polynomial, yet your argument never mentions that. It is also the dimension of the generalised eigenspace for$~\lambda$, which is what you may be using; however that generalised eigenspace is $\ker(T-\lambda_iI)^{k_i}$, so in this case there remains nothing left to prove. $\endgroup$ – Marc van Leeuwen Nov 11 '14 at 18:50
  • $\begingroup$ The definition of the space of generalized eigenvectors corresponding to an eigenvalue $\lambda$ of $T$ which I have read is $\ker(T-\lambda)^{\dim V}$. (I have read linear algebra only from Axler's Linear Algebra Done Right). I am sorry if my post poses any ambiguity. $\endgroup$ – caffeinemachine Nov 11 '14 at 18:56
  • $\begingroup$ Well, that explains your question. What is not clear is why one should choose that exponent in the definition; what one needs is a sufficiently large exponent (so that increasing changes nothing) and $\dim V$ is a very crude upper bound. The algebraic multiplicity of $\lambda$ (from the char. pol.) itself is a much better bound, and $k_i$ is a tight bound. It's not the first time I'm surprised by Axler's Book With A Pretentions Title. $\endgroup$ – Marc van Leeuwen Nov 12 '14 at 5:09
  • $\begingroup$ See my answer for my point of view on this issue. $\endgroup$ – Marc van Leeuwen Nov 12 '14 at 6:12

Your reasoning seems fine. My one complaint is that your statement

Now since $\ker (T-\lambda_iI)^{k_i}\subseteq \ker(T-\lambda_iI)^{\dim V}$, we must have from the above two decompositions of $V$, that $\ker(T-\lambda_iI)^{k_i}=\ker(T-\lambda_iI)^{\dim V}$

is true, but should be justified.

  • $\begingroup$ You are right. I am using the fact that the degree of the minimal polynomial is no more than $\dim V$. Anyway. So the result is true right? It came to me as a bit of a shock since I hadn't seen it in the three years of my acquaintance with linear algebra. $\endgroup$ – caffeinemachine Nov 11 '14 at 18:46
  • 1
    $\begingroup$ Yes, the result is true. $\endgroup$ – Ben Grossmann Nov 11 '14 at 18:51

The main issue here is that the subspace $\ker(T-\lambda_iI)^k$ is the same for all $k\geq k_i$. If this were not true, there would exist a vector $v$ with $(T-\lambda_iI)^{k_i}(v)\neq0$ but $(T-\lambda_i I)^{k_i+1}(v)=0$. Then $w=(T-\lambda_iI)^{k_i}(v)$ is an eigenvector for$~\lambda$. Writing $p=(x-\lambda_i)^{k_i}q$, the quotient $q$ is a polynomial with $q[\lambda_i]\neq0$. But then $p[T](v)=q[T](w)=q[\lambda_i]w\neq0$ contradicts the fact that $p[T]=0$.

It is clear that $\dim V\geq k_i$, but I cannot imagine that proving $\ker(T-\lambda_iI)^k$ stabilises for $k\geq\dim V$ would be easier or more natural than above proof. In other words I cannot see why one would want to prefer using the exponent $\dim V$ rather than $k_i$ in the definition of a generalised eigenvector or eigenspace.

  • $\begingroup$ Your argument for the fact that $\ker (T-\lambda_i I)^k$ is same for all $k\geq k_i$ is very nice. To address "but I cannot imagine...". Actually, for $T\in\mathcal L(V)$, $V$ being a finite dimensional vector space, $\ker T^k$ is same for all $k\geq \dim V$. The proof is easy. Write $n=\dim V$ and suppose that $\dim \ker T^{n+1}>\dim \ker T^n$. Then it easily follows that $\ker T\subsetneq \ker T^{2}\subsetneq \cdots \subsetneq T^{n+1}$. Hence $\dim \ker T^{n+1}>n$, which is a contradiction. So I don't think it is very bad to use $\dim V$ there. $\endgroup$ – caffeinemachine Nov 13 '14 at 13:38
  • $\begingroup$ I used to like Axler's book a lot but seeing that he hasn't discussed the "Primary Decomposition Theorem", which I discovered very recently, is a very bad thing. A LOT of Axler's working can be summarized in that one single theorem. Especially Chapter 9, which is very messy. $\endgroup$ – caffeinemachine Nov 13 '14 at 13:42

Since you are using the Primary Decomposition Theorem, you can also simply reason as follows.

The restriction of $T$ to $V_i=\ker(T-\lambda_i)^{k_i}$ clearly has only $\lambda_i$ as eigenvalue. Therefore the characteristic polynomial of this restriction is $(X-\lambda_i)^{\dim(V_i)}$. Since $V=\bigoplus_{i=1}^mV_i$, the characteristic polynomial of$~T$ then is the product of these those of these restrictions:$$\chi_T=\prod_{i=1}^m(X-\lambda_i)^{\dim(V_i)},$$ so clearly the algebraic multiplicity of $\lambda_i$ is $\dim(V_i)$.

[This uses that algebraic multiplicity is by definition the multiplicity as root of the characteristic polynomial; from what is in the question it might seem you are using a different (geometric) definition?]

  • $\begingroup$ I was using the algebraic multiplicity. $\endgroup$ – caffeinemachine Nov 14 '14 at 10:57
  • $\begingroup$ I was asking what definition of the algebraic multiplicity you were using. The ordinary definition uses multiplicity as root of the characteristic polynomial, but Axler (who doesn't like determinants) may be using another one? My point is that with the usual definition you get multiplicity $\dim(V_i)$ immediately from the direct sum and properties of the characteristic polynomial, so your question becomes rather trivial then. $\endgroup$ – Marc van Leeuwen Nov 14 '14 at 11:20
  • $\begingroup$ I am using the following: The algebraic multiplicity of an eigenvalue $\lambda$ of a linear operator $T$ on a finite dimensional vector space $V$ is defined as $\dim \ker (T-\lambda I)^{\dim V}$. My first comment to your first answer suggests that this is not a bad definition. $\endgroup$ – caffeinemachine Nov 14 '14 at 14:22
  • $\begingroup$ It is not a bad definition, but it is geometric (dimension of a subspace). Of course this does not mean it is the geometric multiplicity, that is the dimension of another subspace (the eigenspace). Much can be said against the algebraic/geometric terminology, but the fact that it is used suggests that the more algebraic concept of root multiplicity is normally used to define the algebraic multiplicity. Anyway, knowing sufficiently many facts it is obvious that the definitions of algebraic multiplicity coincide, and my confusion is cleared up. $\endgroup$ – Marc van Leeuwen Nov 14 '14 at 15:54

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