# Finding the multiplicity of eigenvalues

Let $$T: V \to V$$ for a finite-dimensional vector space $$V$$ be a linear operator whose matrix relative to the standard basis consists of all $$1's$$. find all eigenvalues and eigenvectors of $$T$$ and their geometric and algebraic multiplicities.

I didn't have much trouble finding the eigenvectors and eigenvalues, but I am not totally certain on the multiplicities. Here is what I have.

Let $$A$$ denote the matrix of $$T$$ relative to the standard basis, so $$A_{ij} = 1$$ for all $$i,j$$. If $$v = (v_1, \ldots, v_n)$$ is an eigenvector of $$A$$ with eigenvalue $$\lambda$$, then $$Av = \begin{pmatrix} \sum\limits_{i} v_i \\ \vdots \\ \sum\limits_{i} v_i \end{pmatrix} = \lambda v = \begin{pmatrix} \lambda v_1 \\ \vdots \\ \lambda v_n \end{pmatrix}.$$ So $$\sum\limits_{i} v_i = \lambda v_k$$ for all $$k$$, so $$\lambda v_1 = \lambda v_2 = \ldots = \ldots,$$ which is true if and only if $$\lambda = 0$$ or $$v_1 = v_2 = \ldots = v_n$$. If $$\lambda = 0$$, then eigenvectors must satisfy $$\sum\limits_{i} v_i = 0$$. If $$\lambda \neq 0$$, then eigenvectors must satisfy $$v_1 = \ldots = v_n$$. Let $$t$$ equal this common term, then we have $$\sum_i t = tn = \lambda t$$, so $$\lambda = n$$.

So the eigenvalues are $$\lambda_1 = 0$$, in which case the sum of the eigenvectors must be $$0$$, and $$\lambda_2 = n$$, in which case all components of the eigenvector must sum to $$1$$.

By definition, the algebraic multiplicity is the dimension of the nullspace of $$(T - \lambda I)^{\dim V}$$ and the geometric multiplicity is the dimension of the nullspace of $$T - \lambda I$$. I cannot figure out how to compute these from my above work, though.

• If you know that the only possible eigenvalues are $0$ and $n$, then use the fact that the trace of the matrix is the sum of the eigenvalues counted with their multiplicity. Dec 15, 2020 at 21:16
• So this is algebraic multiplicity, right? If so, the trace of $A$ is $1 + \ldots + 1 = n$, so that would require that $\lambda_2 = n$ has multiplicity $1$, but it wouldn't tell me anything about the multiplicity of $0$, right? Since the geometric multiplicity is less than or equal to the algebraic multiplicity, that means that the geometric multiplicity of $\lambda_2$ can't exceed one. It still wouldn't tell me much about $\lambda_1$, though.
– user862302
Dec 15, 2020 at 21:20
• For the geometric multiplicity of 0, you can use direct computations: you have to solve the equation $v_1+v_2+\ldots+v_n=0$. You have $n$ variables for one equation. Can you conclude from there? Dec 15, 2020 at 21:25
• That means I have one leading variable and $n-1$ free variable, so the geometric multiplicity is $n-1$?
– user862302
Dec 15, 2020 at 21:29
• Indeed: $\ker(A)=\mathrm{span}\Big\{(1,-1,0,0\ldots,0,0), (1,0,-1,0,\ldots,0,0),\ldots,(1,0,0,0,\ldots,0,-1)\Big\}$. This was Robert's hint: $\dim(\ker A) = n-\mathrm{rank} A = n-1$. Dec 15, 2020 at 21:32

Hint: the matrix has rank $$1$$.
• So if the matrix has rank $1$, then its kernel has dimension $n-1$ by the rank-nullity theorem, but how do I relate this back to multiplicities?
• The (geometric) multiplicity of eigenvalue $0$ is the dimension of ... Dec 16, 2020 at 6:29