7.4Diagonalization

Example 7.4.4 (Diagonalizing a $2\times 2$ matrix). Let $𝐴=\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)\in {ℝ}^{2\times 2}$, the matrix of the introduction. By Theorem 7.2.9, $𝜆$ is an eigenvalue exactly when

vanishes, so the eigenvalues are $2$ and $3$. Row reducing,

so ${𝐸}_2=span((1,1))$ and ${𝐸}_3=span((2,1))$. The vectors $(1,1)$ and $(2,1)$ are not parallel, so they form an eigenbasis, and by Theorem 7.4.2

To check the work it is enough to verify $𝐴𝑆=𝑆𝐷$: both sides equal $\left(\begin{array}{cc}2& 6\\ 2& 3\end{array}\right)$. The order of the columns is our choice. Swapping them swaps the diagonal entries of $𝐷$, and rescaling a column of $𝑆$ by $𝑐\ne 0$ changes nothing in $𝐷$. So a diagonalization is far from unique, but $𝐷$ is unique up to the order of its entries (Theorem 7.4.16 will show why).

Example 7.4.6 (A repeated eigenvalue with enough eigenvectors). The matrix $𝐴=$$2{𝐼}_3+$$𝐽=$$\left(\begin{array}{ccc}3& 1& 1\\ 1& 3& 1\\ 1& 1& 3\end{array}\right)$ of Example 7.2.14 has only the eigenvalues $2$ and $5$, with

These three vectors form a basis of ${ℝ}^3$: the sum ${𝐸}_2+{𝐸}_5$ is direct (Corollary 7.2.23), so concatenating bases gives an independent list (Theorem 4.4.13), of length $3$. Hence

Example 7.4.8 (Shears and nilpotent matrices). Let $𝐽=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, a shear, over any field. Then $𝐽-$$𝜇{𝐼}_2=$$\left(\begin{array}{cc}1-𝜇& 1\\ 0& 1-𝜇\end{array}\right)$ has determinant ${(1-𝜇)}^2$, so $1$ is the only eigenvalue (Theorem 7.2.9). Since $𝐽\ne {𝐼}_2$, Proposition 7.4.7 shows that $𝐽$ is not diagonalizable. Directly: ${𝐸}_1=$$ker\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)=$$span({𝐞}_1)$ is a line, so there is no basis of eigenvectors.

More generally, let $𝑁\in {𝐹}^{𝑛\times 𝑛}$ be nilpotent and nonzero, say ${𝑁}^{𝑘}=0$. If $𝑁𝐯=𝜆𝐯$ with $𝐯\ne \mathrm{𝟎}$, then $\mathrm{𝟎}={𝑁}^{𝑘}𝐯={𝜆}^{𝑘}𝐯$, so $𝜆=0$; and $0$ is an eigenvalue, because $𝑁$ is not injective (otherwise ${𝑁}^{𝑘}$ would be injective too, while ${𝑁}^{𝑘}=0$ on ${𝐹}^{𝑛}\ne \{\mathrm{𝟎}\}$). So $0$ is the only eigenvalue, and $𝑁\ne 0=0\cdot {𝐼}_{𝑛}$ is not diagonalizable. The same argument works for an upper triangular matrix with all diagonal entries equal to $𝜆$: its only eigenvalue is $𝜆$, because $𝐴-𝜇{𝐼}_{𝑛}$ is triangular with diagonal entries $𝜆-𝜇$, hence invertible for $𝜇\ne 𝜆$ (Proposition 5.3.15, Theorem 5.4.7). So it is diagonalizable only if it equals $𝜆{𝐼}_{𝑛}$.

These are the minimal counterexamples of Technique 1.4.18 for diagonalizability. In particular, being invertible ($𝐽$) and being diagonalizable are unrelated: $𝐽$ is invertible but not diagonalizable, and $diag(1,0)$ is diagonal but not invertible.

Example 7.4.9 (Rotations: the field matters). Let $𝐴=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, the rotation by a right angle. By Example 7.2.18 its eigenvalues are the square roots of $-1$ in $𝐹$.

• Over $ℝ$ there are none, so there are no eigenvectors at all, and $𝐴$ is not diagonalizable over $ℝ$.

• Over $ℂ$ the eigenvalues $𝑖$ and $-𝑖$ are distinct, with eigenvectors $(1,-𝑖)$ and $(1,𝑖)$, so by Corollary 7.4.5

  $$𝐴=\left(\begin{array}{cc}1& 1\\ -𝑖& 𝑖\end{array}\right)\left(\begin{array}{cc}𝑖& 0\\ 0& -𝑖\end{array}\right){\left(\begin{array}{cc}1& 1\\ -𝑖& 𝑖\end{array}\right)}^{-1}.$$
• Over ${𝔽}_5$, where $2^2=3^2=-1$, it is diagonalizable with eigenvalues $2$ and $3$; over ${𝔽}_3$ there are no eigenvalues.
• Over ${𝔽}_2$, $𝐴$ is the swap $(𝑥,𝑦)\mapsto (𝑦,𝑥)$, whose only eigenvalue is $1$ (Example 7.2.17). Since $𝐴\ne {𝐼}_2$, it is not diagonalizable over ${𝔽}_2$, by Proposition 7.4.7. Over every field with $1+1\ne 0$ the swap is diagonalizable, with eigenbasis $(1,1),(1,-1)$.

So “diagonalizable” is a property of a matrix together with a field. The real matrix $𝐴$ fails for lack of eigenvalues, while over ${𝔽}_2$ the matrix fails for lack of eigenvectors.

Example 7.4.14 (The multiplicities of a diagonal matrix). Let $𝐷=$$diag({𝑑}_1,\dots ,{𝑑}_{𝑛})$ and $𝜆\in 𝐹$, and let $𝑚$ be the number of indices $𝑖$ with ${𝑑}_{𝑖}=𝜆$. Then $𝑥{𝐼}_{𝑛}-$$𝐷=$$diag(𝑥-{𝑑}_1,\dots ,𝑥-{𝑑}_{𝑛})$ is diagonal, so

and by Proposition 7.1.34, ${am}_{𝐷}(𝜆)=𝑚$. On the other hand $𝐷-$$𝜆{𝐼}_{𝑛}=$$diag({𝑑}_1-𝜆,\dots ,{𝑑}_{𝑛}-𝜆)$ has exactly $𝑛-𝑚$ nonzero columns, which are multiples of distinct standard basis vectors, so its rank is $𝑛-𝑚$, and ${gm}_{𝐷}(𝜆)=𝑚$ by Lemma 7.4.13. So for a diagonal matrix the two multiplicities agree: both count how often $𝜆$ occurs on the diagonal. Consequently, if $𝑇$ is diagonalizable, the number of times $𝜆$ occurs on the diagonal of a diagonal matrix of $𝑇$ does not depend on the eigenbasis: it is ${gm}_{𝑇}(𝜆)$. This is why $𝐷$ in a diagonalization $𝐴=𝑆𝐷{𝑆}^{-1}$ is unique up to order.

For the shear $𝐽=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, ${𝑝}_{𝐽}={(𝑥-1)}^2$, so ${am}_{𝐽}(1)=2$, while ${gm}_{𝐽}(1)=$$2-$$rank\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)=$$1$.

Example 7.4.19 (Two matrices that differ in one entry). Let

Both are upper triangular with diagonal $3,2,2$, so ${𝑝}_{𝐴}=$${𝑝}_{{𝐴}^{\prime }}=$$(𝑥-3){(𝑥-2)}^2$: the eigenvalue $3$ is simple, and $am(2)=2$ for both. For $𝐴$,

has rank $1$, so ${gm}_{𝐴}(2)=2$, with ${𝐸}_2=$$\{(𝑥,𝑦,𝑧)|𝑥+𝑦+𝑧=0\}=$$span((1,-1,0),(1,0,-1))$. Since $3$ is simple, ${gm}_{𝐴}(3)=1$, by Theorem 7.4.15, and ${𝐸}_3=span({𝐞}_1)$. The dimensions add up to $3$, and

For ${𝐴}^{\prime }$,

has rank $2$, so ${gm}_{{𝐴}^{\prime }}(2)=$$1<$$2=$${am}_{{𝐴}^{\prime }}(2)$, and ${𝐴}^{\prime }$ is not diagonalizable. The single entry ${𝑎}_{23}^{\prime }=1$ destroys one eigenvector for $2$. In particular $𝐴$ and ${𝐴}^{\prime }$ are not similar, although they have the same characteristic polynomial (Lemma 7.4.13).

Example 7.4.20 (The Fibonacci matrix over several fields). Let $𝑄=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$, so that ${𝑝}_{𝑄}=$$det\left(\begin{array}{cc}𝑥-1& -1\\ -1& 𝑥\end{array}\right)=$${𝑥}^2-$$𝑥-$$1$. Each failure mode of Theorem 7.4.16 occurs over some field.

• Over ${𝔽}_2$, ${𝑝}_{𝑄}={𝑥}^2+𝑥+1$ has no roots (${𝑝}_{𝑄}(0)={𝑝}_{𝑄}(1)=1$). So $𝑄$ has no eigenvalues and is not diagonalizable: ${𝑝}_{𝑄}$ does not split.
• Over ${𝔽}_5$, ${(𝑥+2)}^2=$${𝑥}^2+$$4𝑥+$$4=$${𝑥}^2-$$𝑥-$$1$, so ${𝑝}_{𝑄}$ splits with the single root $-2=3$, of algebraic multiplicity $2$. But $𝑄-$$3{𝐼}_2=$$\left(\begin{array}{cc}-2& 1\\ 1& -3\end{array}\right)=$$\left(\begin{array}{cc}3& 1\\ 1& 2\end{array}\right)$ has rank $1$ (its second row is $2$ times its first in ${𝔽}_5$), so $gm(3)=1<2$. Not diagonalizable: the multiplicities differ.
• Over ${𝔽}_{11}$, $(𝑥-4)(𝑥+3)=$${𝑥}^2-$$𝑥-$$12=$${𝑥}^2-$$𝑥-$$1$, so $𝑄$ has the distinct eigenvalues $4$ and $-3=8$, and is diagonalizable by Corollary 7.4.5.
• Over $ℝ$, the roots $𝜑=(1+\sqrt{5})/2$ and $𝜓=(1-\sqrt{5})/2$ are distinct, and $𝑄$ is diagonalizable; Section 7.4.4 uses this to find the Fibonacci numbers.

The discriminant $5$ of ${𝑥}^2-𝑥-1$ explains the pattern: the roots are $(1\pm \sqrt{5})/2$ whenever $1+1\ne 0$, they coincide when $5=0$, and they lie in ${𝔽}_{𝑝}$ when $5$ is a square there, as $4^2=16=5$ is in ${𝔽}_{11}$.

Example 7.4.24 (A closed form for ${𝐴}^{𝑘}$). For $𝐴=$$\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)=$$𝑆diag(2,3){𝑆}^{-1}$ from Example 7.4.4,

for every $𝑘\in \mathbb{Z}$. At $𝑘=1$ this returns $𝐴$, and at $𝑘=0$ it returns ${𝐼}_2$. Every entry of ${𝐴}^{𝑘}$ is a combination $𝛼2^{𝑘}+𝛽3^{𝑘}$, and for large $𝑘$ the term $3^{𝑘}$ dominates.

Often only ${𝐴}^{𝑘}𝐯$ is needed, and then ${𝑆}^{-1}$ is unnecessary. Expand $𝐯$ in the eigenbasis, $𝐯={𝑐}_1{𝐬}_1+{𝑐}_2{𝐬}_2$; then ${𝐴}^{𝑘}𝐯=$${𝑐}_1{2}^{𝑘}{𝐬}_1+$${𝑐}_2{3}^{𝑘}{𝐬}_2$, because ${𝐴}^{𝑘}{𝐬}_{𝑗}=$${𝜆}_{𝑗}^{𝑘}{𝐬}_{𝑗}$ (Proposition 7.2.28). For instance $(0,1)=$$2(1,1)-$$(2,1)$, so ${𝐴}^{𝑘}(0,1)=$$(2^{𝑘+1}-2\cdot 3^{𝑘},2^{𝑘+1}-3^{𝑘})$, the second column above.