Multiplicity of an Eigenvalue and the Minimal Polynomial

Question

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?

Thanks.

Answer 1

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.

Answer 2

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.

Answer 3

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?]