Eigenvalues

Why the determinant: $A\mathbf v = \lambda\mathbf v$ rearranges to $(A - \lambda I)\mathbf v = \mathbf 0$. A nonzero $\mathbf v$ can solve that only if $A - \lambda I$ has a nonzero null space, that is, only if it is singular — and singular means determinant $0$. So the search for special directions becomes a polynomial equation in one unknown, $\lambda$; each root then gives its direction by solving a singular linear system.

Along an eigenvector the matrix is just a number, so applying it $k$ times multiplies by that number $k$ times. A general vector is a combination of eigenvectors (when there are enough of them), and $A^k$ acts on each piece separately: $A^k(c_1\mathbf v_1 + c_2\mathbf v_2) = c_1\lambda_1^k\mathbf v_1 + c_2\lambda_2^k\mathbf v_2$. As $k$ grows, the term with the largest $|\lambda|$ takes over — which is why long-run behavior is read off the eigenvalues.

Check both directions by multiplying: $A(1, 1) = (3, 3) = 3(1, 1)$ and $A(1, -1) = (1, -1)$, so $(1, -1)$ is fixed outright ($\lambda = 1$). An ordinary vector is turned: $A(1, 0) = (2, 1)$ is not a multiple of $(1, 0)$. In the eigenvector coordinates the matrix is simply "triple along $(1, 1)$, leave $(1, -1)$ alone"; every other vector is a mix of the two and gets pulled toward the $(1, 1)$ line.

Step 2 always has a nonzero solution, because step 1 chose $\lambda$ to make $A - \lambda I$ singular; if row reduction ever shows only the zero solution, the eigenvalue was computed wrongly. The characteristic polynomial has degree $n$, so over the complex numbers it has exactly $n$ roots. A real matrix can have complex roots — then no real direction is merely stretched, and the matrix contains a rotation.

For $\lambda = 3$: $A - 3I = \begin{pmatrix}1&-2\\1&-2\end{pmatrix}$, whose rows both say $v_1 = 2v_2$, so $\mathbf v_1 = (2, 1)$. For $\lambda = 2$: $A - 2I = \begin{pmatrix}2&-2\\1&-1\end{pmatrix}$ gives $v_1 = v_2$, so $\mathbf v_2 = (1, 1)$. Checks: $A(2, 1) = (8 - 2, 2 + 1) = (6, 3) = 3(2, 1)$ ✓ and $A(1, 1) = (2, 2)$ ✓. The two eigenvectors are independent but not perpendicular ($(2, 1)\cdot(1, 1) = 3$), as expected for a non-symmetric matrix. An ordinary vector such as $(0, 1)$ is turned, to $(-2, 1)$.

For $\lambda = 3$: $A - 3I = \begin{pmatrix}-1&0&0\\1&0&0\\0&1&-2\end{pmatrix}$. Row 1 forces $v_1 = 0$ (row 2 then agrees), and row 3 gives $v_2 = 2v_3$. Take $v_3 = 1$: $\mathbf v = (0, 2, 1)$. Check: $A(0, 2, 1) = (0,\ 0 + 6 + 0,\ 0 + 2 + 1) = (0, 6, 3) = 3(0, 2, 1)$ ✓. The trace $2 + 3 + 1 = 6$ equals the sum of the eigenvalues, the same bookkeeping as in the $2\times2$ case.

The discriminant is negative, so there is no real eigenvalue and no real direction that $A$ merely stretches: every nonzero real vector is turned. The complex pair still carries geometry. Its modulus $|2 \pm i| = \sqrt5$ is the average stretch (and $\sqrt5\cdot\sqrt5 = 5 = \det A$), and its argument $\arctan\tfrac12 \approx 26.6^\circ$ is the rotation angle in a suitable basis. Iterating $A$ spirals vectors outward, growing by about $\sqrt5$ per step.

All of this follows from applying $A$ repeatedly to one vector. $A^2\mathbf v = A(\lambda\mathbf v) = \lambda^2\mathbf v$, sums of powers act by sums of powers of $\lambda$, and multiplying $A\mathbf v = \lambda\mathbf v$ by $\tfrac1\lambda A^{-1}$ gives $A^{-1}\mathbf v = \tfrac1\lambda\mathbf v$. Since $\det A$ is the product of the eigenvalues, a zero eigenvalue is the same thing as a zero determinant.

On each eigenvector, $A^2 + A$ acts as $\lambda^2 + \lambda$: that is $4 + 2 = 6$ and $9 - 3 = 6$. The same number on both independent directions means $A^2 + A = 6I$. Check directly: $A^2 = \begin{pmatrix}5&-2\\-2&8\end{pmatrix}$, and adding $A$ gives $\begin{pmatrix}6&0\\0&6\end{pmatrix}$ ✓. A free inverse falls out: $A(A + I) = 6I$, so $A^{-1} = \tfrac16(A + I) = \tfrac16\begin{pmatrix}2&2\\2&-1\end{pmatrix}$. Geometrically, the $-3$ says vectors along $(1, -2)$ are flipped and tripled.

For $\lambda = 4$: $A - 4I = \begin{pmatrix}-2&4\\1&-2\end{pmatrix}$ gives $v_1 = 2v_2$, so $(2, 1)$, and $A(2, 1) = (8, 4)$ ✓. Every input lands on the line through $(2, 1)$ — the columns $(2, 1)$ and $(4, 2)$ both lie on it — so the image is that line, and the kernel is the line through $(2, -1)$. For instance $(1, 1)$ goes to $(6, 3)$. $A$ is not invertible, as the zero eigenvalue announced.

The trace and determinant are the eigenvalues' fingerprints, readable without solving anything. $\blacksquare$

So along an eigenvector, iterating the matrix is just repeated scalar multiplication — the engine behind diagonalization, Fibonacci formulas, and Markov-chain limits. $\blacksquare$

Step two: for each root, solve $(A - \lambda I)\mathbf v = \mathbf 0$ — the matrix is now singular by design, so its rows are dependent and one row determines $\mathbf v$ up to scale. Step three: use them — along eigenvectors the matrix acts as a number, so $A^k$ acts as $\lambda^k$, systems decouple, and $A^k\mathbf x_0$ for large $k$ tilts toward the eigenvector of the largest $|\lambda|$ (the power method, and the reason Markov chains forget their starting point). Two structural gifts: symmetric matrices always have real eigenvalues with orthogonal eigenvectors, and distinct eigenvalues always give independent eigenvectors.

Step 1 — characteristic equation via trace and determinant: $\mathrm{tr} = 4$, $\det = 3$, so $\lambda^2 - 4\lambda + 3 = 0$, factoring as $(\lambda - 1)(\lambda - 3) = 0$: eigenvalues $1$ and $3$.

Step 2 — eigenvector for $\lambda = 3$: $(A - 3I)\mathbf v = \begin{pmatrix}-1&1\\1&-1\end{pmatrix}\mathbf v = \mathbf 0$ gives $v_1 = v_2$: take $(1, 1)$.

Step 3 — for $\lambda = 1$: $\begin{pmatrix}1&1\\1&1\end{pmatrix}\mathbf v = \mathbf 0$ gives $v_1 = -v_2$: take $(1, -1)$.

Step 4 — checks: $A(1,1) = (3,3) = 3(1,1)$ ✓; $A(1,-1) = (1,-1)$ ✓; and $(1,1)\cdot(1,-1) = 0$ — orthogonal, as symmetry promised.

Step 5 — payoff: $A^{10}(1, 0) = A^{10}\cdot\tfrac12\left[(1,1) + (1,-1)\right] = \tfrac12\left[3^{10}(1,1) + (1,-1)\right]$ — a closed form with no repeated multiplication anywhere.

Step 1 — eigenvalues: $\mathrm{tr} = 1.5$, $\det = 0.56 - 0.06 = 0.5$: $\lambda^2 - 1.5\lambda + 0.5 = 0$ factors as $(\lambda - 1)(\lambda - 0.5) = 0$.

Step 2 — the $\lambda = 1$ eigenvector: $(M - I)\mathbf v = \mathbf 0$ gives $-0.2v_1 + 0.3v_2 = 0$, so $\mathbf v \propto (3, 2)$.

Step 3 — conclusion: the $\lambda = 0.5$ component halves every year and dies; every starting split converges to the ratio $3 : 2$ — sixty percent at site one, forever, regardless of history.

Step 4 — check one iteration from $(1, 0)$: $M(1,0) = (0.8, 0.2)$, then $(0.7, 0.3)$, then $(0.65, 0.35)$ — marching monotonically toward $(0.6, 0.4)$ ✓. The eigenvalue $1$ holds the destination; the second eigenvalue sets the speed of forgetting.