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Consider the $n \times n$ unit matrix $A$ where $A_{ij} = 1$ for all $i$ and $j$ over a field $F$. Find the eigenvalues of $A$ and their geometric and algebraic multiplicities.

After having done some work, I found that the eigenvalues of $A$ are $\lambda = 0$ and $\lambda = n$. In the first case, the eigenvectors have the property that their components sum to $0$. In the second case, the components of the eigenvectors are all equal (and non-zero).

I cannot understand the geometric and algebraic multiplicities. Given eigenvalue $\lambda$, the geometric multiplicity is the dimension of $\text{null}(A - \lambda I)$ and the algebraic multiplicity is the dimension of $\text{null}(A - \lambda I)^n$. I am, in particular, struggling with algebraic multiplicity, without knowing the characteristic polynomial.

Any help would be appreciatied.

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. Did you mean the eigenvalues of $A$ are $\lambda=0$ and $\lambda=n$? $\endgroup$ Commented Dec 16, 2020 at 3:16
  • $\begingroup$ Yes, I did. I fixed it. Very sorry for that. $\endgroup$
    – user862302
    Commented Dec 16, 2020 at 3:17
  • $\begingroup$ Okay, and did you mean in the second case the components of the eigenvectors are all $1$? $\endgroup$ Commented Dec 16, 2020 at 3:18
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    $\begingroup$ This is another typo. I don't think they all have to be $1$, but only that if $v = (a_1, \ldots, a_n)$, then $a_1 = \ldots = a_n = 1$, but they are all non-zero scalar multiplies of $(1, \ldots, 1)$. $\endgroup$
    – user862302
    Commented Dec 16, 2020 at 3:19

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You don't need the characteristic polynomial, as the spectral theorem guarantees that $A$, being a symmetric matrix ($A = A^t$), is diagonalizable. In fact, it can be diagonalized with an orthogonal matrix (i.e. we can find a orthonormal basis of eigenvectors for $A$). That means that all the geometric and algebraic multiplicities are equal.

In this case, it seems pretty clear that a basis of eigenvectors could be given by an orthogonal basis for the space $\{ (x_1, ..., x_n) \in \Bbb{R}^n : x_1 + ... + x_n = 0 \} \cong \Bbb{R}^{n-1}$, together with a unit vector in the direction of $(1, ..., 1)$. That means the multiplicity (geometric or algebraic) of $\lambda = 0$ is $(n-1)$ and the multiplicity (geometric or algebraic) of $\lambda = n$ is $0$.

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