Characteristic polynomial is divisible by $(x-\lambda)^r$

Question

Let $\lambda$ be an eigenvalue of a linear map $T: V \rightarrow V$, where $V$ is a vector space over $\Bbb{C}$.

Let $W$ be the associated eigenspace which has dimension $r$.

Then we need to prove that te characteristic polynomial $C_{A}(x)$ of a $n \times n$ matrix $A$ representing $t$ is divisible by $(\lambda - x)^r$ and $r \leq multiplicity(\lambda)$.

How do we prove this? also, it seems like $\lambda$ repeated for $r$ times, but I don't see why since the dimension of the eigenspace corresponding to $\lambda$ is $r$, so it must be having $r$ linearly independent directions?

Answer 1

We first prove that for any eigen-value $\alpha$ of a matrix $A\in M_n(\Bbb C)$ the algebraic multiplicity of $\alpha$ is not less than the geometric multiplicity of $\alpha$.

Let $\{x_1,x_2,...,x_k\}$ be a basis of $E=$eigen-space of $\alpha$ w.r.t. $A$. And $ \{x_1,....,x_k,x_{k+1},...,x_n\}$ an extension to a basis of $\Bbb C^n$ . Then $P:=[x_1:x_2:•••••:x_n]$ is non-singular and $P^{-1} AP=P^{-1}[Ax_1:Ax_2:•••••:Ax_n]=P^{-1}[\alpha x_1:•••••:\alpha x_k:Ax_{k+1}:•••••:Ax_n]$

Now $P^{-1}(\alpha x_j)=\alpha P^{-1}P_{*j}=\alpha e_j$ for $j=1,.....,k$.So $$P^{-1}AP=\begin{bmatrix}\alpha I_k&B\\0&C\end{bmatrix}$$ for some matrices $B,C$.

Hence characteristic poly. of $A$= characteristic poly. of $P^{-1}AP=(x-\alpha)^k \chi_C(x)$ where $\chi_C(x)$ is the characteristic polynomial of $C$. Therefore

$$geometric\ multiplicity\ of\ \alpha\ w.r.t.\ A:=dim(eigen\ space\ of\ \alpha\ w.r.t.\ A)≤ algebraic\ multiplicity\ of\ \alpha\ w.r.t.\ A$$

Now if we let algebraic multiplicity$=l$ then $l≥k$ so that $(x-\alpha)^k\ |\ (x-\alpha)^l$ and $(x-\alpha)^l\ |\ \chi_A(X)$ (definition of algebraic multiplicity). Hence $(x-\alpha)^k\ |\ \chi_A(X)$.

Now it is done for matrix but one can do it for a linear map $T:V\rightarrow V$ by considering it's matrix w.r.t. some fixed basis of $V$.

Answer 2

Let $A$ be a matrix representing $T$. Denote the characteristic polynomial of $A$ by $c(x)=c_A(x)$ (as in your question). The zeros of the characteristic polynomial are the eigenvalues of $A$. (I hope this is clear to you. Tell me if not and I will add a proof here.)
Now there are two multiplicities for an eigenvalue $\lambda$ that are of importance. The one is the algebraic multiplicity, i.e. the number of times that $(x-\lambda)$ appears when decomposing $c(x)$ into linear factors. (Since we are over $\mathbb C$, the fundamental theorem of algebra tells us that there is a unique such decomposition).
The other is the geometric multiplicity, defined as the dimension of the corresponding eigenspace.

It is a theorem that the algebraic multiplicity is always larger or equal to the geometric multiplicity.

Does this help you? Or are you rather looking for a proof of the above mentioned theorem that the algebraic multiplicity os always larger or equal to the geometric multiplicity?

Addendum: In case you are interested in a proof of the above mentioned fact, you will find good answers on MSE, for example here.