# Eigenvalues of the unit matrix

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.

• Welcome to Mathematics Stack Exchange. Did you mean the eigenvalues of $A$ are $\lambda=0$ and $\lambda=n$? Commented Dec 16, 2020 at 3:16
• Yes, I did. I fixed it. Very sorry for that.
– user862302
Commented Dec 16, 2020 at 3:17
• Okay, and did you mean in the second case the components of the eigenvectors are all $1$? Commented Dec 16, 2020 at 3:18
• 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 $(1, \ldots, 1)$.
– user862302
Commented Dec 16, 2020 at 3:19

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 $$0$$.