Eigenvalues and Eigenvectors of the Matrix [[2, 1], [1, 2]]
Let $$A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}.$$ Find all eigenvalues of $A$ and a corresponding (normalized) eigenvector for each. Check your answer using the trace and the determinant.
Hints
- Eigenvalues solve the characteristic equation $\det(A - \lambda I) = 0$. For a $2 \times 2$ matrix that is $\lambda^2 - (\operatorname{tr}A)\lambda + \det A = 0$.
- Here $\operatorname{tr}A = 4$ and $\det A = 3$, so the eigenvalues satisfy $\lambda_1 + \lambda_2 = 4$ and $\lambda_1\lambda_2 = 3$.
- For each $\lambda$, solve $(A - \lambda I)v = 0$. For a symmetric matrix the two eigenvectors must be orthogonal, which is a quick check.
Worked Solution
How to Think About It: For a $2 \times 2$ matrix the characteristic polynomial is determined by the trace and determinant, so the eigenvalues can be read off almost without computation. Because $A$ is symmetric, its eigenvectors are orthogonal, which both guides the search and checks the result.
Quick Estimate: $A = I + \mathbf{1}\mathbf{1}^{T}$ with $\mathbf{1} = (1, 1)^{T}$. The rank-one part $\mathbf{1}\mathbf{1}^{T}$ has eigenvalues $\mathbf{1}^{T}\mathbf{1} = 2$ (along $\mathbf{1}$) and $0$ (orthogonal to $\mathbf{1}$); adding $I$ shifts both by $1$, giving $3$ and $1$.
Formal Solution:
*Step 1 -- Characteristic equation.* $$\det(A - \lambda I) = \det\begin{pmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{pmatrix} = (2 - \lambda)^{2} - 1 = \lambda^{2} - 4\lambda + 3 = (\lambda - 1)(\lambda - 3).$$ So $\lambda_1 = 3$ and $\lambda_2 = 1$.
*Step 2 -- Trace and determinant check.* $\lambda_1 + \lambda_2 = 4 = \operatorname{tr}A$ and $\lambda_1\lambda_2 = 3 = \det A = 4 - 1$. $\checkmark$
*Step 3 -- Eigenvector for $\lambda_1 = 3$.* Solve $(A - 3I)v = 0$: $$\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \;\Longrightarrow\; v_1 = v_2, \qquad v^{(1)} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}.$$
*Step 4 -- Eigenvector for $\lambda_2 = 1$.* Solve $(A - I)v = 0$: $$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \;\Longrightarrow\; v_1 = -v_2, \qquad v^{(2)} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}.$$
*Step 5 -- Verify.* $A v^{(1)} = \frac{1}{\sqrt2}(3, 3)^{T} = 3v^{(1)}$ and $A v^{(2)} = \frac{1}{\sqrt2}(1, -1)^{T} = 1 \cdot v^{(2)}$. The eigenvectors are orthogonal, as they must be for a symmetric matrix, and $A = 3\,v^{(1)}v^{(1)T} + 1\,v^{(2)}v^{(2)T}$ is the spectral decomposition.
Answer: Eigenvalues $\lambda = 3$ and $\lambda = 1$, with normalized eigenvectors $\frac{1}{\sqrt2}(1, 1)^{T}$ and $\frac{1}{\sqrt2}(1, -1)^{T}$ respectively (any nonzero scalar multiples also work).
Intuition
$A = I + \mathbf{1}\mathbf{1}^{T}$ is a one-factor structure: the all-ones direction is stretched by $1 + 2 = 3$ and every direction orthogonal to it is left at $1$. This is exactly the shape of an equicorrelation covariance matrix, where the top eigenvector $(1,1)/\sqrt2$ is the market mode and the orthogonal eigenvector $(1,-1)/\sqrt2$ is the spread or pairs-trade direction, with the eigenvalues giving the variances of those two portfolios.