Diagonalization

A = PDP^-1 — eigenvalues in D, eigenvectors in P, and A^k = PD^kP^-1.

The idea

Diagonalization factors a matrix into a diagonal matrix seen through a change of coordinates. A diagonal matrix is the easiest kind to work with — it scales each axis by its own factor, its eigenvalues are its diagonal entries, and its $k$-th power just raises each entry to the $k$-th power — so the factorization transfers those conveniences to the matrix being factored.

Suppose the $n \times n$ matrix $A$ has $n$ linearly independent eigenvectors. Make them the columns of a matrix $P$, and put the matching eigenvalues down the diagonal of $D$. Then $AP = PD$ column by column, since $A$ times the $j$-th eigenvector is $\lambda_{j}$ times it. Because the columns of $P$ are independent, $P$ is invertible, and the equation rearranges to $A = P D P^{-1}.$ A matrix with such a factorization is diagonalizable. Read from right to left, the product changes to eigenvector coordinates ($P^{-1}$), scales along each eigenvector direction ($D$), and changes back ($P$).

A matrix is diagonalizable exactly when it has $n$ independent eigenvectors. Distinct eigenvalues guarantee this; a repeated eigenvalue whose eigenspace has too small a dimension defeats it.

The factorization pays off in powers: the inner $P^{-1}P$ pairs cancel, so $A^{k} = P D^{k} P^{-1}$, with $D^{k}$ computed entry by entry.

Ways to work on it

• Walkthrough. Read off D from eigenvalues and learn when A is diagonalizable.
• Practice. Compute a power of a diagonal matrix, or decide whether a matrix is diagonalizable.
• Hardest. Compute A^k via D^k, or analyze a defective shear.

Not sure where to start? Take the ten-question placement test.