Mathematical Methods for Economics · Linear Algebra
Eigenvalues & diagonalisation
Eigenvalues capture the directions in which a matrix acts by pure stretching. They make matrix powers tractable, they decide the stability of dynamic economic systems, and — through symmetric matrices — they are the gateway to the definiteness tests used in optimisation. This page defines eigenvalues and eigenvectors, shows how to find them from the characteristic equation, and builds up to diagonalisation.
These notes assume the matrices and linear systems material, particularly determinants and the inverse. They belong to the linear algebra part of a mathematical economics course (module code EC3304).
Why powers motivate eigenvalues
For \(k\in\mathbb{N}\), the power \(A^k\) is \(A\) multiplied by itself \(k\) times. Powers appear naturally whenever a system is iterated forward — a transition matrix applied period after period, for instance. If \(A\) is diagonal the computation is instant:
$$A=\operatorname{diag}(a_{11},\ldots,a_{nn}) \implies A^k=\operatorname{diag}(a_{11}^{k},\ldots,a_{nn}^{k}).$$
But most matrices are not diagonal. The strategy of this page is to find a change of coordinates in which \(A\) looks diagonal — and the coordinates that achieve this are the eigenvectors.
A square matrix \(A\) is diagonalisable if there exist an invertible matrix \(P\) and a diagonal matrix \(D\) with
$$P^{-1}AP = D.$$
Then \(P^{-1}A^kP = D^k\), so \(A^k = PD^kP^{-1}\) — powers become easy again.
Eigenvalues and eigenvectors
A scalar \(\lambda\) is an eigenvalue of a square matrix \(A\) if there is a nonzero vector \(\mathbf{x}\) with
$$A\mathbf{x} = \lambda\mathbf{x}.$$
Then \(\mathbf{x}\) is an eigenvector of \(A\) associated with \(\lambda\).
Geometrically, \(A\) leaves the direction of an eigenvector unchanged and merely scales it by \(\lambda\). Eigenvectors are only defined up to scale: if \(A\mathbf{x}=\lambda\mathbf{x}\) then for any scalar \(\alpha\),
$$A(\alpha\mathbf{x}) = \lambda(\alpha\mathbf{x}),$$
so any nonzero multiple of an eigenvector is again an eigenvector for the same \(\lambda\).
The characteristic equation
Rewriting \(A\mathbf{x}=\lambda\mathbf{x}\) as \((A-\lambda I)\mathbf{x}=\mathbf{0}\), a nonzero solution \(\mathbf{x}\) exists only when \(A-\lambda I\) is singular. Hence:
The eigenvalues of a square matrix \(A\) are the roots of its characteristic equation
$$\lvert A-\lambda I\rvert = \begin{vmatrix} a_{11}-\lambda & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22}-\lambda & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn}-\lambda\end{vmatrix} = 0.$$
Eigenvalues may be repeated, and they may be complex numbers.
Find the eigenvalues of
$$A=\begin{pmatrix}5&-6&-6\\-1&4&2\\3&-6&-4\end{pmatrix}.$$
Expanding the determinant \(\lvert A-\lambda I\rvert\) and factorising gives the characteristic equation
$$\begin{vmatrix}5-\lambda&-6&-6\\-1&4-\lambda&2\\3&-6&-4-\lambda\end{vmatrix} = -(\lambda-2)^2(\lambda-1) = 0.$$
Hence the eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = \lambda_3 = 2\) (a repeated root of multiplicity two).
Trace and determinant shortcuts
Two properties let you check eigenvalue calculations quickly, without recomputing determinants.
If \(A\) is an \(n\times n\) matrix with eigenvalues \(\lambda_1,\lambda_2,\ldots,\lambda_n\), then
$$\prod_{i=1}^{n}\lambda_i = \lvert A\rvert, \qquad \sum_{i=1}^{n}\lambda_i = \operatorname{tr}(A) = \sum_{i=1}^{n} a_{ii}.$$
The product of the eigenvalues equals the determinant; their sum equals the trace (the sum of the diagonal entries).
For the \(3\times3\) example above, \(\operatorname{tr}(A) = 5 + 4 + (-4) = 5\) matches \(1 + 2 + 2 = 5\), and \(\lvert A\rvert = 1\cdot2\cdot2 = 4\). A zero determinant is therefore equivalent to having \(0\) as an eigenvalue — the same singularity condition seen for the inverse.
Finding eigenvectors
For each eigenvalue \(\lambda\), solve the homogeneous system \((A-\lambda I)\mathbf{x}=\mathbf{0}\). There may be several linearly independent eigenvectors for a repeated eigenvalue.
For \(\lambda=1\), \(A\mathbf{x}=\mathbf{x}\) becomes, after moving everything to one side,
$$\begin{cases}4x_1 - 6x_2 - 6x_3 = 0\\ -x_1 + 3x_2 + 2x_3 = 0\\ 3x_1 - 6x_2 - 5x_3 = 0\end{cases} \implies x_1 = x_3 = -3x_2.$$
The eigenvectors for \(\lambda=1\) are therefore
$$\mathbf{x} = \alpha\begin{pmatrix}3\\-1\\3\end{pmatrix}, \quad \alpha\neq 0.$$
For the repeated eigenvalue \(\lambda=2\), the single condition \(\tilde x_1 = 2(\tilde x_2 + \tilde x_3)\) leaves two free parameters, giving a two-dimensional family of eigenvectors:
$$\mathbf{x} = \alpha\begin{pmatrix}2\\1\\0\end{pmatrix} + \beta\begin{pmatrix}2\\0\\1\end{pmatrix}, \quad \alpha,\beta \text{ not both zero}.$$
Diagonalisation
An \(n\times n\) matrix \(A\) is diagonalisable if and only if it has a set of \(n\) linearly independent eigenvectors \(\mathbf{x}_1,\ldots,\mathbf{x}_n\). Then
$$P^{-1}AP = \operatorname{diag}(\lambda_1,\lambda_2,\ldots,\lambda_n),$$
where \(P\) has the eigenvectors \(\mathbf{x}_1,\ldots,\mathbf{x}_n\) as its columns and \(\lambda_1,\ldots,\lambda_n\) are the corresponding eigenvalues.
The \(3\times3\) matrix above is diagonalisable because, even though \(\lambda=2\) is repeated, it supplies two independent eigenvectors — three in total. Placing them as columns of \(P\) yields
$$\begin{pmatrix}-1&2&2\\-1&3&2\\3&-6&-5\end{pmatrix}\begin{pmatrix}5&-6&-6\\-1&4&2\\3&-6&-4\end{pmatrix}P = \operatorname{diag}(1,2,2),$$
where the left factor is \(P^{-1}\). Once \(A=PDP^{-1}\) is in hand, \(A^{k}=PD^{k}P^{-1}\) computes any power cheaply.
Symmetric matrices
Symmetric matrices — those with \(A=A^{\top}\), such as covariance matrices and the matrices of quadratic forms — are especially well behaved.
If an \(n\times n\) matrix \(A\) is symmetric, then:
- all eigenvalues \(\lambda_1,\ldots,\lambda_n\) are real numbers;
- eigenvectors associated with different eigenvalues are orthogonal;
- \(A\) can be diagonalised by an orthogonal matrix \(P\) whose columns are eigenvectors of unit length (so \(P^{-1}=P^{\top}\)).
For \(A=\begin{pmatrix}1&2\\2&1\end{pmatrix}\), the eigenvalues are \(\lambda_1=3\) and \(\lambda_2=-1\), with unit eigenvectors
$$\lambda_1=3:\ \begin{pmatrix}\tfrac{\sqrt2}{2}\\[2pt]\tfrac{\sqrt2}{2}\end{pmatrix}, \qquad \lambda_2=-1:\ \begin{pmatrix}\tfrac{\sqrt2}{2}\\[2pt]-\tfrac{\sqrt2}{2}\end{pmatrix}.$$
These are orthogonal, so \(P=\begin{pmatrix}\sqrt2/2&\sqrt2/2\\ \sqrt2/2&-\sqrt2/2\end{pmatrix}\) is orthogonal and
$$P^{\top}AP = \begin{pmatrix}3&0\\0&-1\end{pmatrix}.$$
Because a symmetric matrix diagonalises as \(A=PDP^{\top}\) with real eigenvalues, the sign pattern of those eigenvalues completely determines whether the associated quadratic form is positive or negative (semi)definite. That is the bridge to second-order conditions in optimisation and to the concavity or convexity of objective functions — developed on the quadratic forms and definiteness page. Eigenvalues of a system's coefficient matrix also govern the stability of dynamic models: whether an economy converges to a steady state or diverges depends on their signs.
Self-check questions
1. Find the eigenvalues of \(A=\begin{pmatrix}2&1\\1&2\end{pmatrix}\), and verify them against the trace and determinant.
The characteristic equation is \(\begin{vmatrix}2-\lambda&1\\1&2-\lambda\end{vmatrix} = (2-\lambda)^2 - 1 = 0\), i.e. \((2-\lambda)^2 = 1\), so \(2-\lambda = \pm 1\) and \(\lambda = 1\) or \(\lambda = 3\).
Check: \(\operatorname{tr}(A) = 2+2 = 4 = 1+3\) ✓ and \(\lvert A\rvert = 2(2)-1(1) = 3 = 1\times 3\) ✓.
2. An eigenvalue of a matrix \(A\) equals \(0\). What does this imply about \(A\)?
If \(0\) is an eigenvalue then \(\prod_i \lambda_i = \lvert A\rvert = 0\), so \(A\) is singular (not invertible). Equivalently, there is a nonzero vector \(\mathbf{x}\) with \(A\mathbf{x} = 0\cdot\mathbf{x} = \mathbf{0}\), so the columns of \(A\) are linearly dependent and \(A\) is not full rank.
3. Find an eigenvector of \(A=\begin{pmatrix}2&1\\1&2\end{pmatrix}\) for the eigenvalue \(\lambda=3\).
Solve \((A-3I)\mathbf{x}=\mathbf{0}\): \(\begin{pmatrix}-1&1\\1&-1\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\mathbf{0}\) gives \(-x_1+x_2=0\), i.e. \(x_1=x_2\).
So any nonzero multiple of \(\begin{pmatrix}1\\1\end{pmatrix}\) is an eigenvector. Check: \(A\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}3\\3\end{pmatrix} = 3\begin{pmatrix}1\\1\end{pmatrix}\). ✓ Notice it is orthogonal to the \(\lambda=1\) eigenvector \((1,-1)\), as expected for a symmetric matrix.
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