# Limits of the dimension of an eigenspace

I am trying to prove that the dimension of an eigenspace is 1 $$\leq$$ dim $$V_{\lambda_i}$$ $$\leq m_i$$. Where $$m_i$$ is the multiplicity of the eigenvalue $$\lambda_i$$.

What I have done is the following:

Let $$V_{\lambda_i}$$ be defined as:

$$V_{\lambda_i} = \{v \in V | Tv=\lambda_iv\}$$

Therefore:

$$V_{\lambda_i}$$ = ker $$(T-\lambda_i)$$

Which would lead to:

dim $$V_{\lambda_i} =$$ nullity $$(T-\lambda_i)$$

But how can I prove that $$1\leq$$ nullity $$(T-\lambda_i) \leq m_i$$?

• By definition of $\lambda_i$ being an eigenvalue, the dimension is always $\geq 1$. To prove the other inequality, one approach is to consider a basis for the eigenspace, extend it to the a basis for the whole vector space $V$, and consider the matrix of $T$ relative to that basis. Calculate the characteristic polynomial; it will have $(t-\lambda_i)^{\dim V_{\lambda_i}}$ as a factor; hence the dimension of $V_i$ is $\leq$ the multiplicity $m_i$. But really, this should be proven in pretty much any decent linear algebra book Commented Apr 27, 2020 at 1:31

## 2 Answers

Here is one way of going about this problem:

1. If $$\vec{v}_1, \vec{v}_2, \cdots \vec{v}_k$$ are all linearly independent eigenvectors of $$T$$ with eigenvalue $$\lambda_i$$, show that $$T = S M S^{-1}$$, where $$M$$ takes form $$\begin{bmatrix} \lambda_i I_k & B \\ 0 & D \end{bmatrix}$$ where $$B$$ and $$D$$ are arbitrary, and $$I_k$$ is the $$k \times k$$ identity matrix. Hint: what is $$T$$ in a basis $$\beta$$ that contains all of $$\vec{v}_1$$ through $$\vec{v}_k$$?

2. Show that the characteristic polynomial of $$M$$ is of form $$p_M(\lambda) = (\lambda - \lambda_i)^k q(\lambda)$$, where $$q$$ is just some arbitrary polynomial in $$\lambda$$. Since $$M$$ is similar to $$T$$, it would follow they have the same characteristic polynomial (why?), and thus the algebraic multiplicity of $$\lambda_i$$ for $$T$$ would have to be at least $$k$$, by the definition of algebraic multiplicity using the characteristic polynomial.

Steps 1 and 2 would prove that the geometric multiplicity (dimension of the eigenspace) is always less than or equal to the algebraic multiplicity, as one can take all $$\dim V_{\lambda_i}$$ eigenvectors of $$T$$ with eigenvalue $$\lambda_i$$ in step 1.

In accordance to @paulinho's answer, given a linear operator $$T:V\rightarrow V$$ defined on a finite dimensional vector space $$V$$ over the field $$\textbf{F}$$, consider the eigenspace $$E_{\lambda}$$ associated to some eigenvalue $$\lambda$$, assuming that there is one.

Since $$E_{\lambda} = \ker(T-\lambda I)\subseteq V$$, we may consider a basis $$\mathcal{B}_{\lambda} = \{v_{1},v_{2},\ldots,v_{m}\}$$ for $$E_{\lambda}$$ and extend it to a basis $$\mathcal{B}_{V} = \{v_{1},v_{2},\ldots,v_{m},v_{m+1},\ldots,v_{n}\}$$ where $$n = \dim V$$.

Consequently, the matrix representation of $$T$$ associated to $$\mathcal{B}_{V}$$ is given by \begin{align*} [T]_{\mathcal{B}_{V}} & = [T(v_{1})^{T},T(v_{2})^{T},\ldots,T(v_{m})^{T},T(v_{m+1})^{T},\ldots,T(v_{n})^{T}]\\\\ & = [\lambda v^{T}_{1},\lambda v^{T}_{2},\ldots,\lambda v^{T}_{m},T(v_{m+1})^{T},\ldots,T(v_{n})^{T}]= \begin{bmatrix} \lambda I_{m} & A\\ O & B \end{bmatrix} \end{align*}

Therefore the characteristic polynomial of $$T$$ is given by \begin{align*} p(t) = \det\begin{bmatrix} (\lambda - t)I_{m} & A\\ O & B - tI_{n-m} \end{bmatrix} & = \det((\lambda - t)I_{m})\det(B - tI_{n-m})\\\\ & = (\lambda - t)^{m}g(t) \end{align*} where $$g(t)$$ is a polynomial of degree $$n - m$$, which may have $$\lambda$$ as a root.

Based on such considerations, we conclude that algebraic multiplicity of $$\lambda$$ is greater than or equal to $$m = \dim E_{\lambda}$$, and we are done.

Hopefully this helps.