The eigenvalues are the numbers that make a certain determinant zero, and that determinant, written out, is a polynomial. Finding the eigenvalues of a matrix is finding the roots of that polynomial, and then everything the course knows about polynomials applies.

A lock opens only at certain settings of its dial. Writing down the condition for opening as one equation in the setting turns the search for the settings into solving that equation.

The characteristic polynomial of an $n\times n$ matrix $A$ is $p_A(\lambda)=\det(A-\lambda I)$. Expanding the determinant shows it is a polynomial of degree $n$ in $\lambda$, and by the last lesson its real roots are exactly the eigenvalues of $A$. So an $n\times n$ matrix has at most $n$ eigenvalues.

• For a $2\times2$ matrix: $p_A(\lambda)=\lambda^2-(\operatorname{tr}A)\lambda+\det A$.
• For a triangular matrix, $A-\lambda I$ is triangular, so $p_A(\lambda)=(a_{11}-\lambda)\cdots(a_{nn}-\lambda)$: the eigenvalues are the diagonal entries.
• When a $2\times2$ matrix has real eigenvalues $\lambda_1,\lambda_2$, they add up to $\operatorname{tr}A$ and multiply to , since .
• A polynomial with real coefficients may have no real roots: the quarter turn has $p(\lambda)=\lambda^2+1$ and no real eigenvalue.
• $A$ and $A^{T}$ have the same characteristic polynomial, because $\det\big((A-\lambda I)^{T}\big)=\det(A-\lambda I)$.

$\lambda I$ commutes with every matrix, so it can be written as $P^{-1}(\lambda I)P$, and then $P^{-1}AP-\lambda I=P^{-1}(A-\lambda I)P$. The product rule turns the determinant into $\det P^{-1}\det(A-\lambda I)\det P$, and the two outer factors cancel. So similar matrices share the characteristic polynomial and the eigenvalues: they belong to the map, not to the basis used to write it.

$\lambda I=P^{-1}(\lambda I)P$, because $\lambda I$ commutes with every matrix; then factor.

The product rule for determinants.

The outer factors cancel.

So similar matrices have the same characteristic polynomial, and the same eigenvalues.

For a $2\times2$ matrix the determinant of $A-\lambda I$ expands to a quadratic, whose coefficients are the trace and the determinant.

$A-\lambda I$ is triangular when $A$ is, and the determinant of a triangular matrix is the product of its diagonal. So the eigenvalues of a triangular matrix are its diagonal entries.

• $p_A(\lambda)=\det(A-\lambda I)$ has degree $n$; its real roots are the eigenvalues.
• For $2\times2$: $\lambda^2-(\operatorname{tr}A)\lambda+\det A$.
• A triangular matrix has its eigenvalues on the diagonal.
• Similar matrices, and a matrix and its transpose, have the same eigenvalues.

Great work! You now know how to:

• write the characteristic polynomial of a matrix;
• find eigenvalues as its roots, from trace and determinant for 2 by 2;
• read the eigenvalues of a triangular matrix off its diagonal;
• explain why similar matrices have the same eigenvalues.