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:

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


$V_{\lambda_i}$ = 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 $(T-\lambda_i) \leq m_i$?

  • $\begingroup$ 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 $\endgroup$
    – peek-a-boo
    Commented Apr 27, 2020 at 1:31

2 Answers 2


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.


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