From Stage Counts to a Projection Matrix
A population can change in size and composition at the same time. To model both, divide the population into stages, such as juveniles and adults, and record how many individuals occupy each stage at a census. A projection matrix describes how individuals in each current stage contribute to each stage at the next census. Repeated multiplication then models successive time steps.
We use column vectors. For a two-stage population, let \(\mathbf{x}_t=\begin{bmatrix}J_t\\A_t\end{bmatrix}\), where \(J_t\) and \(A_t\) are the stage counts at time \(t\). The entries may represent expected contributions per individual, so the model can describe expected counts even when a rate is not an integer.
In a simplified example, the first row might record new individuals entering the first stage, while the second row records contributions to the second stage. A positive entry means that the corresponding contribution is possible in the model. The matrix is called strictly positive when all four entries are greater than zero. This condition is stronger than merely requiring nonnegative entries, and it will be the hypothesis for the long-run result proved below.
Worked Example: Projecting One Census Forward
Suppose the model is \(\mathbf{P}=\begin{bmatrix}1&2\\1&1\end{bmatrix}\), with first stage \(J\) and second stage \(A\). Each column gives the next-census contributions from one individual in its current stage. Starting from \(\mathbf{x}_0=\begin{bmatrix}40\\10\end{bmatrix}\), compute the next population vector.
Multiply the matrix by the column vector, taking one row at a time:
The model projects 60 individuals in the first stage and 50 in the second, for a total of 110. The initial total was 50, but the stage counts did not both grow by the same factor: the first count changed by a factor of \(60/40=3/2\), while the second changed by a factor of \(50/10=5\).
Check: Multiplying the original matrix by the proposed answer’s input gives first component \(40+20=60\) and second component \(40+10=50\), so \(\mathbf{P}\mathbf{x}_0=\mathbf{x}_1\) as required.
The Dominant Eigenvalue and Its Population Profile
An eigenvector gives a stage composition that the projection matrix preserves up to a scalar factor. If \(\mathbf{v}\ne\mathbf{0}\) and \(\mathbf{P}\mathbf{v}=\lambda\mathbf{v}\), then a population with stage counts proportional to \(\mathbf{v}\) is projected to counts proportional to the same vector. The scalar \(\lambda\) is the change factor for that composition.
For a strictly positive two-by-two matrix, the eigenvalue of largest absolute value is positive and has a positive eigenvector. We can establish the relevant facts directly, rather than assume that every eigenvalue of every projection matrix has the same interpretation.
Proof. The characteristic equation is \(\det(\mathbf{P}-\lambda\mathbf{I}_2)=(a-\lambda)(d-\lambda)-bc=0\). Expanding gives \(\lambda^2-(a+d)\lambda+(ad-bc)=0\), and the quadratic formula gives the stated \(\lambda_+\) and \(\lambda_-\). Write \(\Delta=\sqrt{(a-d)^2+4bc}\). Since \(bc>0\), \(\Delta>|a-d|\). Therefore \(\lambda_+>0\), and \(\lambda_+-a=(d-a+\Delta)/2>0\). Both entries of \(\mathbf{v}_+\) are positive. Also, \(\lambda_+>\lambda_-\), and \(\lambda_++\lambda_-=a+d>0\). If \(\lambda_-\ge0\), these facts give \(|\lambda_-|<\lambda_+\); if \(\lambda_-<0\), the positive sum gives \(-\lambda_-<\lambda_+\), which again gives \(|\lambda_-|<\lambda_+\). Finally, the first component of \(\mathbf{P}\mathbf{v}_+\) is \(ab+b(\lambda_+-a)=b\lambda_+\). The second is \(bc+d(\lambda_+-a)=\lambda_+(\lambda_+-a)\), using \((\lambda_+-a)(\lambda_+-d)=bc\), which follows from the characteristic equation. Hence \(\mathbf{P}\mathbf{v}_+=\lambda_+\mathbf{v}_+\). \(\square\)
The positive eigenvector can be converted into proportions by dividing each entry by the sum of its entries. Those normalized values add to one and describe a stage composition, not a total population size. The eigenvalue remains the multiplier for counts when the population has exactly that composition.
Worked Example: Finding Stable Stage Proportions
For \(\mathbf{P}=\begin{bmatrix}1&2\\1&1\end{bmatrix}\), find the dominant eigenvalue and a corresponding positive eigenvector, then normalize the vector to proportions.
The characteristic equation is \((1-\lambda)^2-2=0\), so its roots are \(1+\sqrt{2}\) and \(1-\sqrt{2}\). The larger absolute value is \(\lambda_+=1+\sqrt{2}\), since \(|1-\sqrt{2}|=\sqrt{2}-1<1+\sqrt{2}\). The lemma gives the eigenvector \(\begin{bmatrix}2\\\sqrt{2}\end{bmatrix}\), which may be scaled to \(\mathbf{v}_+=\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}\).
Normalize by dividing by the sum \(\sqrt{2}+1\):
Thus the stable proportions are \(2-\sqrt{2}\) in the first stage and \(\sqrt{2}-1\) in the second. Their sum is \(1\). The long-run multiplier for counts is \(1+\sqrt{2}\) per time step.
Check: Multiplication verifies the eigenpair: \(\begin{bmatrix}1&2\\1&1\end{bmatrix}\begin{bmatrix}\sqrt{2}\\1\end{bmatrix} =\begin{bmatrix}\sqrt{2}+2\\\sqrt{2}+1\end{bmatrix} =(1+\sqrt{2})\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}\). The normalized entries also sum to \((2-\sqrt{2})+(\sqrt{2}-1)=1\).
Why the Long-Run Composition Emerges
A positive eigenvector is a composition preserved by one projection, but a population generally starts with a different composition. The next result explains why, for a strictly positive two-stage matrix, every nonzero nonnegative starting population approaches the dominant eigenvector’s proportions. Here “proportions” means each stage count divided by the total count at that time.
Proof. Use the notation in the lemma and set \(\mathbf{w}=\begin{bmatrix}c\\\lambda_+-a\end{bmatrix}\). Directly applying the lemma to \(\mathbf{P}^{\top}\) shows \(\mathbf{P}^{\top}\mathbf{w}=\lambda_+\mathbf{w}\), so \(\mathbf{w}\) has positive entries. The two eigenvalues are distinct because \(\Delta>0\). A nonzero eigenvector for \(\lambda_-\) is \(\mathbf{v}_-=\begin{bmatrix}b\\\lambda_--a\end{bmatrix}\); it is nonzero because \(b>0\), and the same characteristic-equation computation as in the lemma verifies its eigenvector equation. Eigenvectors for distinct eigenvalues are independent: if these two nonzero vectors were scalar multiples, applying \(\mathbf{P}\) would make the same nonzero vector have both eigenvalues. Thus they form a basis of \(\mathbb{R}^2\).
Write \(\mathbf{x}_0=\alpha\mathbf{v}_++\beta\mathbf{v}_-\). Since \(\mathbf{w}^{\top}\mathbf{P}=\lambda_+\mathbf{w}^{\top}\), the eigenvector equations imply \(\mathbf{w}^{\top}\mathbf{v}_-=0\): evaluating \(\mathbf{w}^{\top}\mathbf{P}\mathbf{v}_-\) in the two ways gives \(\lambda_+\mathbf{w}^{\top}\mathbf{v}_-=\lambda_-\mathbf{w}^{\top}\mathbf{v}_-\), and the eigenvalues differ. Consequently, \(\alpha=\frac{\mathbf{w}^{\top}\mathbf{x}_0}{\mathbf{w}^{\top}\mathbf{v}_+}>0\), because \(\mathbf{w}\) and \(\mathbf{v}_+\) are positive and \(\mathbf{x}_0\) is nonnegative and nonzero. Iterating now gives \(\mathbf{x}_t=\alpha\lambda_+^t\mathbf{v}_++\beta\lambda_-^t\mathbf{v}_-\). Divide by \(\lambda_+^t\). Since \(|\lambda_-/\lambda_+|<1\), the resulting vector converges to \(\alpha\mathbf{v}_+\). Its entry sum converges to the positive number \(\alpha\) times the entry sum of \(\mathbf{v}_+\). Dividing entries by their sum therefore gives the normalized proportions of \(\mathbf{v}_+\). Finally, the total count \(N_t\), the sum of the entries of \(\mathbf{x}_t\), satisfies \(N_t/\lambda_+^t\to\alpha\) times the positive entry sum of \(\mathbf{v}_+\). Taking the ratio of these limits for \(N_{t+1}/\lambda_+^{t+1}\) and \(N_t/\lambda_+^t\) gives \(N_{t+1}/N_t\to\lambda_+\). \(\square\)
The hypothesis that the starting vector is nonzero matters: a zero population remains zero, and its proportions are undefined. Strict positivity of the matrix also matters to this result as stated; it ensures the eigenvector used for the stable proportions is positive and the other eigenvalue has strictly smaller absolute value.
Worked Example: A Transient Difference in Composition
Use \(\mathbf{P}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\) with initial population \(\mathbf{x}_0=\begin{bmatrix}3\\1\end{bmatrix}\). Find two successive projections and compare their proportions with the stable composition.
The first projection is \(\mathbf{x}_1=\begin{bmatrix}2(3)+1(1)\\1(3)+2(1)\end{bmatrix}=\begin{bmatrix}7\\5\end{bmatrix}\). The second is \(\mathbf{x}_2=\begin{bmatrix}2(7)+1(5)\\1(7)+2(5)\end{bmatrix}=\begin{bmatrix}19\\17\end{bmatrix}\). The totals are \(4\), \(12\), and \(36\), so the ratios for these two steps are both \(3\).
To identify the long-run values, \(\mathbf{P}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\3\end{bmatrix}=3\begin{bmatrix}1\\1\end{bmatrix}\), while \(\mathbf{P}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}1\\-1\end{bmatrix}\). Thus the eigenvalues are \(3\) and \(1\), and the dominant eigenvector gives stable proportions \(\begin{bmatrix}1/2\\1/2\end{bmatrix}\). The initial vector decomposes as \(\begin{bmatrix}3\\1\end{bmatrix}=2\begin{bmatrix}1\\1\end{bmatrix}+\begin{bmatrix}1\\-1\end{bmatrix}\). Therefore \(\mathbf{x}_2=2(3^2)\begin{bmatrix}1\\1\end{bmatrix}+1^2\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}19\\17\end{bmatrix}\), agreeing with direct iteration. Its proportions are \(19/36\) and \(17/36\), not yet \(1/2\) each.
Check: Direct multiplication gives \(\mathbf{P}\mathbf{x}_1=\begin{bmatrix}14+5\\7+10\end{bmatrix}=\begin{bmatrix}19\\17\end{bmatrix}\), and the eigenvector checks above verify both eigenvalues used in the decomposition.
Interpreting the Model Carefully
The dominant eigenvalue is a multiplicative growth factor per projection step, not an amount to add to the total. If \(\lambda_+>1\), the long-run total grows by that factor per step; if \(0<\lambda_+<1\), the long-run total declines by that factor per step. The stable eigenvector describes the limiting distribution across stages after normalization. It does not say that every population begins in that distribution, or that the stage counts themselves stop changing.
A common mistake is to assume that the initial proportions remain fixed. For the first example, the initial proportions are \(40/50=4/5\) and \(10/50=1/5\). After one projection, they are \(60/110=6/11\) and \(50/110=5/11\), so the initial proportions were not preserved. The correction is to calculate the dominant eigenvector and normalize it: for this matrix the limiting proportions are \(2-\sqrt{2}\) and \(\sqrt{2}-1\), as verified above.
Another important convention is to multiply a column population vector on the left by the projection matrix. Reversing the order changes the model: for \(\mathbf{P}=\begin{bmatrix}1&2\\1&1\end{bmatrix}\), the correct product with \(\begin{bmatrix}40\\10\end{bmatrix}\) is \(\begin{bmatrix}60\\50\end{bmatrix}\), whereas \(\mathbf{P}^{\top}\begin{bmatrix}40\\10\end{bmatrix}=\begin{bmatrix}50\\90\end{bmatrix}\). Read the entries by columns: column \(j\) records what one individual currently in stage \(j\) contributes.
Check Your Understanding
Use column vectors throughout. When a question asks for a long-run conclusion, state the assumptions that make the conclusion applicable.
- For \(\mathbf{P}=\begin{bmatrix}1&1\\2&1\end{bmatrix}\), project \(\mathbf{x}_0=\begin{bmatrix}2\\3\end{bmatrix}\) forward one step and check the multiplication.
- For \(\mathbf{P}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\), verify that \(\begin{bmatrix}1\\1\end{bmatrix}\) is an eigenvector and state its normalized proportions.
- Explain why a dominant eigenvalue is interpreted as a long-run multiplicative factor rather than an additive increase.
- For a strictly positive two-by-two projection matrix, what condition on the initial population is needed for the stable-proportion conclusion in the theorem?
- In a projection matrix, what does column \(j\) represent?
