Let \(A\) be the \(n\times n\) matrix with zeros on the main diagonal and ones everywhere else. Find the eigenvalues and eigenspaces of \(A\), and compute \(\det A\).
Solution
Let \[ \mathbf{1}\coloneqq(1,\ldots,1)^T \] and \[ H\coloneqq \left\{ x=(x_1,\ldots,x_n)^T\in\RR^n: \sum_{i=1}^n x_i=0 \right\}. \]
1. The vector \(\mathbf{1}\) is an eigenvector of \(A\) with eigenvalue\(n-1\).
Every row of \(A\) contains exactly \(n-1\) entries equal to \(1\). Hence \[ A\mathbf{1}=(n-1)\mathbf{1}. \]
2. Every vector in \(H\) is an eigenvector with eigenvalue \(-1\), unless itis zero.
If \(x\in H\), then for each \(i\), \[ (Ax)_i = \sum_{j\neq i}x_j = \sum_{j=1}^n x_j-x_i = -x_i. \] Thus \[ Ax=-x. \]
3. For \(n\geq2\), the complete eigenspace decomposition is\[ \boxed{ E_{n-1} = \{c\mathbf{1}:c\in\RR\}, \qquad E_{-1} = H }. \]
The subspace \(H\) has dimension \(n-1\), while \[ \{c\mathbf{1}:c\in\RR\} \] has dimension \(1\). Their intersection is zero because a scalar multiple of \(\mathbf{1}\) lies in \(H\) only when \[ nc=0, \] hence \(c=0\). Therefore \[ \RR^n = \{c\mathbf{1}:c\in\RR\} \oplus H. \] Steps 1 and 2 show that \(A\) acts on these summands by \(n-1\) and \(-1\), respectively, so there are no other eigenvalues or eigenspaces.
If \(n=1\), then \(A=(0)\), so the only eigenvalue is \(0\) and its eigenspace is all of \(\RR\).
4. The determinant is\[ \boxed{ \det A = (n-1)(-1)^{n-1} }. \]
For \(n\geq2\), step 3 gives the eigenvalue \(n-1\) with multiplicity \(1\) and the eigenvalue \(-1\) with multiplicity \(n-1\). Their product is the displayed determinant. If \(n=1\), then \(A=(0)\) and the same formula gives \(0\).
5. Q.E.D.
Steps 3 and 4 give the requested eigenspaces, eigenvalues, and determinant.