Eigenvalues of a Rank-1 Matrix
Let $A = \mathbf{u}\mathbf{v}^T$ be the outer product of two vectors $\mathbf{u}, \mathbf{v} \in \mathbb{R}^n$.
What are all the eigenvalues of $A$, and what are their corresponding eigenvectors? Justify your answer.
Hints
- What is the rank of $A = \mathbf{u}\mathbf{v}^T$, and what does rank tell you about the number of zero eigenvalues?
- Compute $A\mathbf{u}$ directly -- factor out $\mathbf{v}^T \mathbf{u}$ and notice what remains.
- Use the trace identity $\text{tr}(A) = \sum_i \lambda_i$ to confirm: $\text{tr}(\mathbf{u}\mathbf{v}^T) = \mathbf{v}^T \mathbf{u}$.
Worked Solution
How to Think About It: $A = \mathbf{u}\mathbf{v}^T$ is a rank-1 matrix -- it collapses all of $\mathbb{R}^n$ onto a one-dimensional subspace spanned by $\mathbf{u}$. That geometric fact immediately tells you almost everything: rank 1 means the null space has dimension $n-1$, so $\lambda = 0$ must be an eigenvalue with multiplicity $n-1$. The only remaining eigenvalue is pinned down by the trace. Start there and the problem solves itself.
Quick Estimate: For a concrete check, try $n = 2$, $\mathbf{u} = (1, 2)^T$, $\mathbf{v} = (3, 1)^T$. Then $A = \begin{pmatrix} 3 & 1 \\ 6 & 2 \end{pmatrix}$. The trace is $5 = \mathbf{v}^T \mathbf{u} = 3 + 2 = 5$. The determinant is $0$ (rank 1). So eigenvalues are $5$ and $0$. That matches $\lambda_1 = \mathbf{v}^T \mathbf{u}$ and $\lambda_2 = 0$.
Approach: Use rank to count zero eigenvalues, then find the nonzero one via direct computation of $A\mathbf{u}$.
Formal Solution:
Step 1 -- The zero eigenvalues. Since $A = \mathbf{u}\mathbf{v}^T$ has rank at most 1, its null space has dimension at least $n-1$. Any vector $\mathbf{x}$ orthogonal to $\mathbf{v}$ satisfies $A\mathbf{x} = \mathbf{u}(\mathbf{v}^T \mathbf{x}) = \mathbf{u} \cdot 0 = \mathbf{0}$. So $\lambda = 0$ is an eigenvalue with multiplicity $n-1$, and the corresponding eigenspace is $\{\mathbf{x} : \mathbf{v}^T \mathbf{x} = 0\}$.
Step 2 -- The nonzero eigenvalue. Try $\mathbf{u}$ as an eigenvector: $$A\mathbf{u} = (\mathbf{u}\mathbf{v}^T)\mathbf{u} = \mathbf{u}(\mathbf{v}^T \mathbf{u})$$ Since $\mathbf{v}^T \mathbf{u}$ is a scalar, this gives $A\mathbf{u} = (\mathbf{v}^T \mathbf{u})\,\mathbf{u}$. So $\mathbf{u}$ is an eigenvector with eigenvalue $\lambda_1 = \mathbf{v}^T \mathbf{u}$.
Step 3 -- Sanity check via trace. The trace equals the sum of eigenvalues: $$\text{tr}(A) = \sum_i (\mathbf{u}\mathbf{v}^T)_{ii} = \sum_i u_i v_i = \mathbf{v}^T \mathbf{u} = \lambda_1 + 0 + \cdots + 0 \checkmark$$
Answer: - One nonzero eigenvalue $\lambda_1 = \mathbf{v}^T \mathbf{u}$ with eigenvector $\mathbf{u}$ (assuming $\mathbf{u} \neq \mathbf{0}$ and $\mathbf{v}^T \mathbf{u} \neq 0$) - $n-1$ zero eigenvalues with eigenspace $\ker(\mathbf{v}^T) = \{\mathbf{x} : \mathbf{v}^T \mathbf{x} = 0\}$
If $\mathbf{v}^T \mathbf{u} = 0$, then $A$ is nilpotent and all eigenvalues are zero.
Intuition
The key insight here is that rank and eigenvalues are intimately connected. A rank-1 matrix can only stretch vectors along one direction -- everything else gets annihilated. So once you know the rank, you know $n-1$ eigenvalues immediately (they are all zero), and you only need to find one more.
This shows up constantly in quant work. Principal component analysis produces low-rank covariance approximations, and understanding how eigenvalues behave under rank-1 updates is central to Kalman filtering, online regression, and covariance estimation. The Sherman-Morrison formula -- which tells you how the inverse of a matrix changes after a rank-1 update -- is built on exactly this structure. Any time you see an outer product $\mathbf{u}\mathbf{v}^T$ in a financial model, your first instinct should be: rank 1, one interesting eigenvalue equal to the inner product $\mathbf{v}^T \mathbf{u}$, everything else is noise.