5.3 Diagonalization

Big idea

Diagonal matrices are easy. They scale each coordinate on its own, their powers are the powers of the diagonal entries, and their determinant is a product. Almost nothing else about a matrix is that simple.

Diagonalization asks whether a given matrix is a diagonal matrix in disguise. The disguise is a change of coordinates: if $A$ has enough eigenvectors to form a basis, then in the coordinates those eigenvectors set up, $A$ does nothing but scale. Written out, $A = PDP^{-1}$, with the eigenvectors as the columns of $P$ and the eigenvalues down the diagonal of $D$, matched in order.

The payoff is powers. $A^k = PD^kP^{-1}$, because the inner $P^{-1}P$ pairs cancel, and raising a diagonal matrix to the hundredth power costs $n$ multiplications. Everything that follows in this chapter, from population models to systems of differential equations, is that identity used on a different problem.

Decoder

$A$ is diagonalizable when it is similar to a diagonal matrix, which happens exactly when $A$ has $n$ linearly independent eigenvectors.

Read “similar to a diagonal matrix” as “equal to $PDP^{-1}$ for some invertible $P$ and some diagonal $D$”. The second half of the sentence says how to find out: collect a basis for each eigenspace, pool them, and count. If you end up with $n$ vectors, $A$ is diagonalizable and those vectors are the columns of $P$. If you end up short, it is not, and no cleverness will repair it.

Definitions and results

Diagonalizable. A square matrix $A$ is diagonalizable when $A = PDP^{-1}$ for some invertible $P$ and diagonal $D$.

The factorization theorem. An $n \times n$ matrix is diagonalizable exactly when it has $n$ linearly independent eigenvectors. In that case the columns of $P$ are those eigenvectors and the diagonal entries of $D$ are the matching eigenvalues, in the same order. Change the order of the columns of $P$ and you must change the order of the entries of $D$ to match.

Why it works. The equation $A = PDP^{-1}$ is the same as $AP = PD$. Column by column, the left side is $A$ applied to each column of $P$, and the right side is each column of $P$ scaled by the matching diagonal entry. So $AP = PD$ says precisely that every column of $P$ is an eigenvector. Requiring $P$ to be invertible is requiring those columns to be independent.

Checking without inverting. To verify a claimed factorization, check $AP = PD$. That is two matrix products and no inverse, and it catches ordering mistakes immediately.

Distinct eigenvalues suffice. If an $n \times n$ matrix has $n$ different eigenvalues, it is diagonalizable. Eigenvectors for distinct eigenvalues are independent, so one from each eigenvalue already gives a basis. The converse fails: a diagonalizable matrix can have repeated eigenvalues, and the identity matrix is the extreme case.

The general test with repeats. Suppose the characteristic polynomial factors completely over the reals. Then $A$ is diagonalizable exactly when every eigenvalue has geometric multiplicity equal to its algebraic multiplicity. Equivalently, the eigenspace dimensions add up to $n$. Bases drawn from different eigenspaces never overlap and always pool into an independent set, so counting dimensions is enough.

Not diagonalizable is not defective arithmetic. A matrix can fail the test and still be perfectly well behaved. A shear matrix has one eigenvalue and one eigenvector line, and that is a true fact about the geometry, not a computational failure. Two escape routes appear later: allowing complex eigenvalues rescues rotations, and the singular value decomposition handles every matrix.

Powers. If $A = PDP^{-1}$, then

$$ A^k = PD^kP^{-1} $$

where $D^k$ raises each diagonal entry to the $k$th power. The same identity works for negative $k$ when every eigenvalue is nonzero, and $A^{-1} = PD^{-1}P^{-1}$.

Reading a power qualitatively. Written out, $A^k$ is a sum of terms, one per eigenvalue, each carrying a factor $\lambda^k$. Eigenvalues with $|\lambda| < 1$ contribute terms that shrink to nothing, eigenvalues with $|\lambda| > 1$ contribute terms that blow up, and an eigenvalue of exactly $1$ contributes a term that never changes. That reading is the whole content of the applications later in the chapter.

Worked examples

Diagonalizing a two by two

Let

$$ A = \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix} $$

The trace is $5$ and the determinant is $4 + 2 = 6$, so the characteristic polynomial is $\lambda^2 - 5\lambda + 6 = (\lambda - 2)(\lambda - 3)$. Two distinct eigenvalues in a $2 \times 2$ matrix, so it is diagonalizable before any eigenvector work.

The eigenspaces were found earlier in the chapter: $(1,1)$ for $\lambda = 2$ and $(2,1)$ for $\lambda = 3$. Set

$$ P = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}, \qquad D = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} $$

Check with $AP = PD$:

$$ AP = \begin{bmatrix} 4 - 2 & 8 - 2 \\ 1 + 1 & 2 + 1 \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 2 & 3 \end{bmatrix}, \qquad PD = \begin{bmatrix} 2 & 6 \\ 2 & 3 \end{bmatrix} $$

They agree. Since $\det P = 1 - 2 = -1$, the inverse is

$$ P^{-1} = \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix} $$

A fifth power without multiplying five times

Keep $A$, $P$, $D$ from above. Then $D^5 = \operatorname{diag}(32, 243)$ and

$$ A^5 = P D^5 P^{-1} = \begin{bmatrix} 32 & 486 \\ 32 & 243 \end{bmatrix} \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 454 & -422 \\ 211 & -179 \end{bmatrix} $$

Check the trace and determinant against the eigenvalues: $A^5$ should have eigenvalues $32$ and $243$, so its trace should be $275$ and its determinant $7776$. Indeed $454 - 179 = 275$, and $454(-179) - (-422)(211) = -81266 + 89042 = 7776$. Both match, so the arithmetic is right.

Repeated eigenvalue, still diagonalizable

Let

$$ B = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} $$

Its characteristic polynomial factors as $-(\lambda - 5)(\lambda + 1)^2$, so the eigenvalues are $5$ once and $-1$ twice. The eigenspace for $5$ is spanned by $(1,1,1)$, since $B(1,1,1) = (5,5,5)$. The eigenspace for $-1$ is the null space of $B + I$, a matrix with every entry $2$, so it is the plane $x_1 + x_2 + x_3 = 0$, spanned by $(1,-1,0)$ and $(1,0,-1)$.

Dimensions $1$ and $2$ add to $3$, so $B$ is diagonalizable. Take

$$ P = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 0 & -1 \end{bmatrix}, \qquad D = \begin{bmatrix} 5 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} $$

Check $AP = PD$ one column at a time. $B(1,1,1) = (5,5,5)$, and the first column of $PD$ is $5(1,1,1)$. $B(1,-1,0) = (1-2, \, 2-1, \, 2-2) = (-1,1,0)$, and the second column of $PD$ is $-1(1,-1,0)$. $B(1,0,-1) = (1-2, \, 2-2, \, 2-1) = (-1,0,1)$, matching the third column of $PD$. Also $\det P = 3$, so $P$ is invertible and the factorization stands.

Compare this with the earlier $2 \times 2$ shear, whose repeated eigenvalue came with only one eigenvector line. A repeated eigenvalue is a warning, not a verdict: you have to compute the eigenspace dimension to know.

Practice

Find the eigenvalues, build a basis for each eigenspace, and assemble $P$ and $D$, or state why no such pair exists.

Then use the factorization for what it is best at: large powers, and the long-run shape of $A^k$.

Videos

The first video shows why an eigenvector basis makes a transformation look diagonal. The second carries the same idea straight into powers; watch how the middle terms collapse when you multiply $PDP^{-1}$ by itself.

Quiz

Five items on building a factorization, deciding when one exists, and computing powers with it.