IB Maths AI 1.12 Notes

Eigenvalues & eigenvectors

This section extends matrix work by introducing eigenvalues, eigenvectors, characteristic polynomials, diagonalization, and the use of matrices to model repeated processes. For this topic, calculations are restricted to $2\times2$ matrices when done by hand.

Let $A$ be a square matrix. A non-zero vector $\mathbf{v}$ is called an eigenvector of $A$ if multiplying by $A$ changes only its size, not its direction. This means:

$A\mathbf{v}=\lambda\mathbf{v}$

where $\lambda$ is a scalar called the eigenvalue corresponding to $\mathbf{v}$.

So, an eigenvector is a vector whose image under the matrix transformation lies on the same line, and the eigenvalue tells us the scale factor.

• If $\lambda\gt 0$, the direction is unchanged.
• If $\lambda\lt 0$, the direction is reversed.
• If $\lambda=0$, the vector is mapped to the zero vector.

To find eigenvalues, start from $A\mathbf{v}=\lambda\mathbf{v}$, which can be rearranged to $(A-\lambda I)\mathbf{v}=0$.

For a non-zero vector $\mathbf{v}$ to exist, the matrix $(A-\lambda I)$ must be singular, so $\det(A-\lambda I)=0$.

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

For a $2\times2$ matrix $A=\begin{pmatrix}a&&b\\c&&d\end{pmatrix}$, the characteristic equation is:

$\begin{vmatrix}a-\lambda&&b\\c&&d-\lambda\end{vmatrix}=0$

so $(a-\lambda)(d-\lambda)-bc=0$, which simplifies to:

$\lambda^2-(a+d)\lambda+(ad-bc)=0$

The sum of the eigenvalues is the trace, and the product is the determinant.

Let $A=\begin{pmatrix}4&&1\\2&&3\end{pmatrix}$. Find the eigenvalues.

Compute $\det(A-\lambda I)=\begin{vmatrix}4-\lambda&&1\\2&&3-\lambda\end{vmatrix}$.

$=(4-\lambda)(3-\lambda)-2$.

$=12-7\lambda+\lambda^2-2$.

$=\lambda^2-7\lambda+10$.

So the characteristic equation is $\lambda^2-7\lambda+10=0$.

Factorise: $(\lambda-5)(\lambda-2)=0$.

Hence the eigenvalues are $\lambda=5$ and $\lambda=2$.