AHL 1.15—Eigenvalues and eigenvectors

Most matrix-vector multiplications change both the direction and magnitude of a vector. But for certain special vectors , called eigenvectors , multiplication by a matrix only scales the vector, leaving its direction unchanged (or exactly reversed).

Eigenvector: A non-zero vector $\mathbf{v}$ is an eigenvector of a square matrix $A$ if multiplying $A$ by $\mathbf{v}$ produces a scalar multiple of $\mathbf{v}$:
$A\mathbf{v}=\lambda \mathbf{v}$
The scalar $\lambda$ is the corresponding eigenvalue.

Eigenvalue: The scalar $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$ for some non-zero vector $\mathbf{v}$. The eigenvalue tells you by how much the eigenvector is stretched, compressed, or reflected.

Key observations:

• If $\lambda >1$: the eigenvector is stretched
• If $0<\lambda <1$: the eigenvector is compressed
• If $\lambda <0$: the eigenvector is reversed in direction and scaled
• If $\lambda =1$: the eigenvector is unchanged by the transformation
• If $\lambda =0$: the eigenvector is mapped to the zero vector by $A$ (note: $A$ is singular in this case, and the eigenvector lies in the null space of $A$)

To find eigenvalues, we rearrange the equation $A\mathbf{v}=\lambda \mathbf{v}$:
$A\mathbf{v}-\lambda \mathbf{v}=0⟹(A-\lambda I)\mathbf{v}=0$

For this to have a non-zero solution $\mathbf{v}$, the matrix $(A-\lambda I)$ must be singular (non-invertible), which means its determinant must be zero:
$\det (A-\lambda I)=0$

This equation is called the characteristic equation, and the expression $\det (A-\lambda I)$ is the characteristic polynomial.

Characteristic Polynomial: For a $2\times 2$ matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the characteristic polynomial is:
$p(\lambda )=\det (A-\lambda I)={\lambda }^2-(a+d)\lambda +(ad-bc)$
Note that $(a+d)$ is the trace of $A$ and $(ad-bc)$ is the determinant of $A$, so:
$p(\lambda )={\lambda }^2-\text{tr}(A)\cdot \lambda +\det (A)$

The eigenvalues are the roots of this quadratic, which can be found by factoring or using:
$\lambda =\frac{\text{tr}(A)\pm \sqrt{[\text{tr}(A){]}^2-4\det (A)}}{2}$