Eigenvalues & eigenvectors

Drag $v$ around. For most positions, $Av$ points somewhere completely different from $v$ — the rotation angle swings all over the place. But near two particular directions, that angle collapses toward $0°$: $A$ stops rotating and only scales.

A vector $v$ is an eigenvector of $A$ if $A$ doesn't rotate it at all — only scales it:

• $A$ — the matrix being analyzed.
• $v$ — the eigenvector: a direction $A$ only scales, never rotates.
• $\lambda$ — the eigenvalue: how much that direction gets stretched (or flipped, if negative).

1. Finding lambda first

   Rearranging gives $(A-\lambda I)v = 0$, which has a nonzero solution $v$ only when $\det(A-\lambda I)=0$ — a polynomial in $\lambda$ called the characteristic polynomial. Its roots are exactly the eigenvalues.

2. Then solving for each eigenvector

   Once $\lambda$ is known, $(A-\lambda I)v=0$ becomes a system with a whole line of solutions — any nonzero vector along that line is a valid eigenvector for that eigenvalue.

3. Eigenvectors need not be perpendicular

   Unlike a symmetric matrix, a general matrix's eigenvectors don't have to be at right angles to each other — this chapter's example has eigenvectors at $(1,1)$ and $(1,-2)$, which are not orthogonal.

Watch the stretch ratio $|Av|/|v|$ alongside the rotation angle. At each of the two special directions, the ratio locks onto a fixed number — that number is $\lambda$ for that direction, no matter how far out you drag $v$ along it.

For $A = \begin{pmatrix}4&1\\2&3\end{pmatrix}$:

1. Solve the characteristic equation

   $\det(A-\lambda I) = (4-\lambda)(3-\lambda) - (1)(2) = \lambda^2 - 7\lambda + 10 = (\lambda-5)(\lambda-2) = 0$, so $\lambda_1=5$ and $\lambda_2=2$.

2. Solve for the first eigenvector

   These equations come from the matrix-vector multiplication in $(A-\lambda I)v=0$, where $v=\begin{pmatrix}x\\y\end{pmatrix}$ and the right-hand side is the zero vector $\begin{pmatrix}0\\0\end{pmatrix}$. For $\lambda_1=5$:

   1. Set up the matrix $A-5I$:

      $$A - 5I = \begin{pmatrix}4&1\\2&3\end{pmatrix} - \begin{pmatrix}5&0\\0&5\end{pmatrix} = \begin{pmatrix}-1&1\\2&-2\end{pmatrix}$$

   2. Set up the system $(A-5I)v=0$:

      $$\begin{pmatrix}-1&1\\2&-2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$$

   3. Multiply row by row:

      • Row 1: $(-1)(x)+(1)(y)=0 \implies -x+y=0$
      • Row 2: $(2)(x)+(-2)(y)=0 \implies 2(-x+y)=0$ — the same relation, no new information (exactly what $\det(A-5I)=0$ guarantees)

   So $y=x$, giving direction $(1,1)$. Check: $A(1,1) = (4+1, 2+3) = (5,5) = 5(1,1)$. ✓

3. Solve for the second eigenvector

   For $\lambda_2=2$:

   1. Set up the matrix $A-2I$:

      $$A - 2I = \begin{pmatrix}4&1\\2&3\end{pmatrix} - \begin{pmatrix}2&0\\0&2\end{pmatrix} = \begin{pmatrix}2&1\\2&1\end{pmatrix}$$

   2. Set up the system $(A-2I)v=0$:

      $$\begin{pmatrix}2&1\\2&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$$

   3. Multiply row by row — both rows are identical here, so they give the same equation twice:

      • Row 1: $(2)(x)+(1)(y)=0 \implies 2x+y=0$
      • Row 2: $(2)(x)+(1)(y)=0 \implies 2x+y=0$ (identical to row 1)

   So $y=-2x$, giving direction $(1,-2)$. Check: $A(1,-2) = (4-2, 2-6) = (2,-4) = 2(1,-2)$. ✓