Characteristic polynomials and diagonalization tests

Proof of the finite real matrix product laws

Let $m,n,p,q$ be positive integers, let $A\in\mathbb R^{m\times n}$, $B\in\mathbb R^{n\times p}$, and $C\in\mathbb R^{p\times q}$. All the following arguments use the row-by-column definition and finite real sums. For any output position $(i,\ell)$,

$$\begin{aligned} ((AB)C)_{i\ell} &=\sum_{k=1}^{p}\left(\sum_{j=1}^{n}a_{ij}b_{jk}\right)c_{k\ell}\\ &=\sum_{j=1}^{n}a_{ij}\left(\sum_{k=1}^{p}b_{jk}c_{k\ell}\right) =(A(BC))_{i\ell}. \end{aligned}$$

The middle equality only reorders two finite sums and uses real distributivity. Both products are $m\times q$, so equality at every position proves associativity. No factor has changed its order.

For $D\in\mathbb R^{n\times p}$, the same definition gives

$$[A(B+D)]_{ik}=\sum_{j=1}^{n}a_{ij}(b_{jk}+d_{jk}) =(AB)_{ik}+(AD)_{ik}.$$

For $E\in\mathbb R^{m\times n}$, it gives

$$[(A+E)B]_{ik}=\sum_{j=1}^{n}(a_{ij}+e_{ij})b_{jk} =(AB)_{ik}+(EB)_{ik}.$$

Thus both distributive laws hold with the displayed matching sizes. For every real scalar $c$,

$$[(cA)B]_{ik}=\sum_j(ca_{ij})b_{jk} =c\sum_j a_{ij}b_{jk} =\sum_j a_{ij}(cb_{jk})=[A(cB)]_{ik},$$

proving $(cA)B=c(AB)=A(cB)$ by entrywise equality. In particular, for $x,y\in\mathbb R^n$ and real $s,t$,

$$[A(sx+ty)]_i=\sum_j a_{ij}(sx_j+ty_j) =s(Ax)_i+t(Ay)_i.$$

This proves the matrix-vector linearity used when comparing solutions of the same linear system. All identities concern compatible finite real matrices; they do not assert commutativity or cancellation of arbitrary matrix factors.

Throughout, $V$ is a nonzero finite-dimensional vector space over a field $F$, and $T\in\operatorname{End}_F(V)$. No algebraic closure is assumed unless stated.

Polynomial division and the root factor identity over a field

Let $F$ be any field, and use formal polynomials in $F[t]$. The zero polynomial is a separate case whenever degree is used. For nonzero $f,g$, the leading coefficient of $fg$ is the product of their nonzero leading coefficients, so $\deg(fg)=\deg f+\deg g$; in particular there are no zero divisors.

For every $f$ and every nonzero $g$ there are unique polynomials $q,r$ with $f=qg+r$, where $r=0$ or $\deg r<\deg g$. Start with residual $f$ and quotient zero. If a nonzero residual has degree $d\ge e=\deg g$, subtract $(a/b)t^{d-e}g$, where $a,b$ are its and $g$'s leading coefficients, and add $(a/b)t^{d-e}$ to the quotient. The new residual is zero or has strictly smaller degree. Nonnegative integer degrees cannot strictly decrease indefinitely, so the process stops. If two answers exist, $g(q-q')=r'-r$; a nonzero left side has degree at least $e$, whereas a nonzero right side has degree below $e$. Thus both differences are zero.

For $\lambda\in F$, the finite identity

$$t^k-\lambda^k=(t-\lambda)\sum_{j=0}^{k-1}t^{k-1-j}\lambda^j\quad(k\ge1)$$

gives $f(t)-f(\lambda)=(t-\lambda)h(t)$. Consequently $f(\lambda)=0$ exactly when $t-\lambda$ divides $f$. This does not assert that any root exists in $F$.

Leibniz determinant and alternation over a commutative ring

Let $R$ be a commutative ring with identity, $n\ge1$, and $A\in M_n(R)$. Define

$$\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{i,\sigma(i)},\qquad \det[\ ]=1\quad(n=0).$$

The sign is $(-1)^{\operatorname{inv}(\sigma)}$, interpreted in $R$. Exchanging adjacent entries in a permutation changes its inversion count by one in parity. Sorting by adjacent inversions therefore gives its sign as the product of the signs of those exchanges. Applying an adjacent-exchange decomposition of $\tau$ to the positions of $\sigma$ proves $\operatorname{sgn}(\sigma\tau)=\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)$; a transposition has sign $-1$. These are integer parity facts before mapping the signs into $R$.

The finite formula is multilinear in each row and column. If two rows are equal, pair each permutation term with the term obtained by exchanging the two assigned columns. The pair is distinct, has the same product and opposite signs; its sum is zero in every characteristic. Thus the determinant vanishes on repeated rows; the column statement is identical. This is a paired cancellation, not an inference from $2\det A=0$. Exchanging rows or columns changes sign by the same permutation substitution. Also $\det I=1$. Substituting $\sigma^{-1}$ in the sum gives $\det A^{\mathsf T}=\det A$.

Here is the needed uniqueness principle, valid even with zero divisors. If $D$ is an alternating multilinear function of $n$ column vectors in $R^n$, expand each input in the standard basis. Terms with a repeated basis vector vanish. The other index lists are permutations; swapping two slots negates $D$, since expansion of $D(\ldots,u+v,\ldots,u+v,\ldots)=0$ gives $D(\ldots,u,\ldots,v,\ldots)+D(\ldots,v,\ldots,u,\ldots)=0$. Hence the expansion equals $D(e_1,\ldots,e_n)$ times the Leibniz determinant of the input columns. No division or field assumption was used.

Multiplicativity, triangular blocks and the adjugate

For fixed $A$, the function $D(v_1,\ldots,v_n)=\det(Av_1,\ldots,Av_n)$ is alternating multilinear, with $D(e_1,\ldots,e_n)=\det A$. Applying the preceding uniqueness principle to the columns of $B$ proves

$$\det(AB)=\det A\det B.$$

For $\begin{pmatrix}B&C\\0&D\end{pmatrix}$ with square block sizes $p,q$, a nonzero Leibniz term must assign all last $q$ rows to the last $q$ columns; otherwise one lower-left entry is zero. Those rows use all those columns, so the first $p$ rows use the first $p$ columns. The remaining permutations are exactly pairs of permutations inside the two blocks, with product sign and product monomial. Summing gives $\det B\det D$, including an empty block using determinant one. This argument also covers terms that vanish for other reasons; it does not require absence of zero divisors.

Let $A_{ij}^{\widehat{\ } }$ delete row $i$ and column $j$, and put $C_{ij}=(-1)^{i+j}\det A_{ij}^{\widehat{\ } }$. Grouping the Leibniz sum by the value $\sigma(i)=j$ gives row cofactor expansion $\det A=\sum_j a_{ij}C_{ij}$. The sign factor follows by moving row $i$ and column $j$ to the first positions, making $i+j-2$ adjacent exchanges; the remaining permutation is on the undeleted indices. Transpose gives the column expansion. Define $\operatorname{adj}(A)_{ji}=C_{ij}$. The $(i,j)$ entry of $A\operatorname{adj}(A)$ is $\sum_k a_{ik}C_{jk}$, the cofactor expansion of the matrix obtained by replacing row $j$ by row $i$. It is $\det A$ if $i=j$, and zero otherwise by repeated-row cancellation. The column argument gives the other product. Therefore

$$A\operatorname{adj}(A)=\operatorname{adj}(A)A=(\det A)I.$$

For $n=1$, the empty minor has determinant one, so the adjugate is $[1]$ and the identity still holds. For $n=0$, define the adjugate to be the empty matrix; the identity is the equality of the unique empty matrices, while the determinant remains one. All results in this section apply to $R=F[t]$; no division by a determinant has occurred.

A vector space is a set with addition and scalar multiplication satisfying the usual abelian-group and distributive axioms over a field $F$. Columns are one model. Polynomials of degree at most $d$, denoted $P_d(F)$, and linear maps between spaces are others. The zero polynomial is included in $P_d(F)$ although its degree is not ordinarily defined.

A subspace contains zero and is closed under sums and scalar multiples. A list spans if every vector is a linear combination of its entries; it is independent if only the zero coefficients give zero. A basis is both. A finite spanning list can be reduced to a basis by deleting dependent entries; an independent list can be extended to a basis by adding vectors outside its span until the given finite spanning list is spanned.

Exchange lemma. An independent list of $r$ vectors in the span of $s$ vectors has $r\le s$. To see this, express the first independent vector in the spanning list, use a nonzero coefficient to replace one spanning vector by it, and repeat. At each stage independence ensures the next independent vector has a nonzero coefficient on an as-yet unreplaced spanning vector. There must be at least $r$ such slots. Comparing two bases in both directions proves that their lengths agree; this common length is $\dim V$.

Consequently, in dimension $n$, any $n$ independent vectors form a basis, as do any $n$ spanning vectors. Counting a list is useful only after its independence or spanning has been justified.

A map $T:V\to W$ is linear if $T(\alpha v+\beta u)=\alpha Tv+\beta Tu$. Its kernel and image are subspaces. Choose a basis $k_1,\ldots,k_r$ of $\ker T$ and extend to a basis with $v_1,\ldots,v_s$. Then $Tv_1,\ldots,Tv_s$ form a basis of $\operatorname{im}T$: spanning follows by expanding any input; a dependence among these images places the corresponding input combination in the kernel, and independence of the extended basis makes all coefficients zero. Thus

$$\dim V=\dim\ker T+\dim\operatorname{im}T.$$

This is rank–nullity. In equal finite dimensions, injectivity and surjectivity are equivalent, but the equivalence can fail in infinite-dimensional spaces.

The field bridge: singularity and characteristic polynomials

Now let $R=F$ be a field. If $\det A\ne0$, the adjugate identity gives inverse $\operatorname{adj}(A)/\det A$. If $A$ is invertible, multiplicativity gives $\det A\det A^{-1}=1$, so $\det A\ne0$. Rank–nullity, proved in C01, shows a finite square linear map has zero kernel exactly when it is invertible. Thus $\det A=0$ exactly when its kernel is nonzero.

For a fixed $A\in M_n(F)$, apply the ring construction to $tI-A\in M_n(F[t])$. Its determinant is monic of degree $n$: the identity-permutation product contributes $t^n$, whereas every other permutation uses an off-diagonal constant entry, so its product is either zero or has degree less than $n$. Multiplicativity proves invariance under $A\mapsto P^{-1}AP$ with invertible constant $P$; the block-triangular formula proves restriction–quotient factorization. Finally every adjugate entry is zero or has degree at most $n-1$, so their coefficient comparison is legitimate in the C02 Cayley–Hamilton proof. In dimension zero, the characteristic polynomial is the empty determinant one; evaluating it gives $I$ on the zero space, which is its zero operator, so Cayley–Hamilton is separate and immediate. This supplies the actual commutative-ring determinant prerequisites without extending any real-only determinant authority.

For $p(t)=\sum_{j=0}^dc_jt^j$, define $p(T)=\sum c_jT^j$, with $T^0=I$. Because powers of the same operator commute, $(pq)(T)=p(T)q(T)$ and evaluation is a ring homomorphism $F[t]\to\operatorname{End}(V)$. This does not permit replacing variables by arbitrary noncommuting matrices in every matrix-coefficient identity.

Theorem. $\chi_T(T)=0$.

Proof. Let $\chi_A(t)=\sum_{k=0}^nc_kt^k$ with $c_n=1$. The adjugate identity over the commutative polynomial ring $F[t]$ gives

$$(tI-A)B(t)=\chi_A(t)I,\qquad B(t)=\operatorname{adj}(tI-A)=\sum_{k=0}^{n-1}B_kt^k.$$

Set $B_{-1}=B_n=0$. Comparing coefficients yields $B_{k-1}-AB_k=c_kI$ for $0\le k\le n$. Multiply the $k$th identity on the left by $A^k$ and sum. The terms telescope:

$$\sum_{k=0}^nc_kA^k=\sum_{k=0}^n\left(A^kB_{k-1}-A^{k+1}B_k\right)=0.$$

This is $\chi_A(A)=0$. It uses coefficient identities and associativity only, not an unexplained substitution into a matrix polynomial with potentially noncommuting coefficients. ∎

The null-space criterion

$$\det(A-\lambda I)=0$$

tells you how to test a proposed eigenvalue. The next step is to package that test into one polynomial whose roots are exactly the eigenvalues of the matrix.

That polynomial is strong enough to limit how many eigenvalues a matrix can have, describe repeated eigenvalues, and give practical tests for diagonalization.

Throughout this note, the base field is $F=\mathbb{R}$. Unless a complex extension is explicitly mentioned, matrices, vectors, scalars, and eigenvalues are over $F$; every determinant is taken for a square matrix with $n\ge1$. We use $t$ for the polynomial indeterminate, so the convention is $p_A(t)=\det(A-tI_n)$.

Characteristic polynomial

Definition

Characteristic polynomial

Let $A\in M_n(F)$ with $F=\mathbb{R}$ and $n\ge1$. The polynomial

$$p_A(t)=\det(A-tI_n)$$

is called the characteristic polynomial of $A$.

The variable $t$ is only an indeterminate. Once you substitute a scalar $\lambda$, the value $p_A(\lambda)$ becomes the determinant of $A-\lambda I$.

Theorem

Eigenvalues are exactly the roots of the characteristic polynomial

Let $A$ be an $n\times n$ matrix and $\lambda$ a scalar. Then the following are equivalent:

1. $\lambda$ is an eigenvalue of $A$;
2. $\lambda$ is a root of $p_A(t)$.

This theorem is just the determinant test from the previous note written in a more organized way:

$$\lambda\text{ is an eigenvalue} \iff \det(A-\lambda I)=0 \iff p_A(\lambda)=0.$$

A root finds a value; an eigenspace counts directions

The roots of $p_A(t)=\det(A-tI)$ identify the eigenvalues. The corresponding eigenspace dimensions determine whether they supply enough independent vectors for diagonalization.

Theorem

Basic shape of the characteristic polynomial

If $A$ is an $n\times n$ matrix, then $p_A(t)$ is a degree-$n$ polynomial whose leading coefficient is $(-1)^n$ and whose constant term is $\det(A)$.

So the characteristic polynomial is never an arbitrary expression. Its degree is fixed by the matrix size, and its constant term already remembers the determinant.

Proof

Degree and leading sign from column multilinearity

Write the columns of $A$ as $a_1,\ldots,a_n$. The columns of $A-tI_n$ are $a_j-t e_j$. Column multilinearity expands the determinant by choosing either $a_j$ or $-t e_j$ in each position. Choosing $-t e_j$ in every column gives

$$(-t)^n\det(I_n)=(-1)^n t^n.$$

Every other choice contains at least one $a_j$, so it has at most $n-1$ factors of $t$. Hence the degree is exactly $n$ and the leading coefficient is $(-1)^n$. The term with no $t$ is $\det(A)$, so $p_A(0)=\det(A)$. This proof keeps the convention $\det(A-tI_n)$; the alternative $\det(tI_n-A)$ is $(-1)^n p_A(t)$, with the same roots and multiplicities but different odd-degree coefficient signs.

First examples

Worked example

A 2×2 characteristic polynomial

Let

$$A= \begin{bmatrix} 3&2\\ 3&-2 \end{bmatrix}.$$

Then

$$p_A(t)= \det \begin{bmatrix} 3-t&2\\ 3&-2-t \end{bmatrix} =(3-t)(-2-t)-6.$$

Simplifying gives

$$p_A(t)=t^2-t-12=(t-4)(t+3).$$

The coefficient calculation is worth checking rather than treating the factorization as a pattern to memorize:

$$(3-t)(-2-t)-6=t^2-t-6-6=t^2-t-12.$$

The constant term is $-12$, which agrees with $\det(A)=-12$, and the leading coefficient is $+1$ because this is a 2 by 2 instance of the sign $(-1)^n$. The factors therefore give the roots $4$ and $-3$, and the root theorem turns those scalar roots into the two eigenvalues. Notice that the sign convention does not change this root set: replacing $\det(A-tI)$ by $\det(tI-A)$ multiplies the whole polynomial by $(-1)^n$, rather than changing its zeros. It does change the leading coefficient when $n$ is odd, which is why the convention must be stated before reading coefficient identities such as Cayley-Hamilton. Under our convention $\det(A-tI)$, the constant term remains $\det(A)$, as the shape proof predicted.

Worked example

A repeated root

Let

$$B= \begin{bmatrix} 2&3\\ 0&2 \end{bmatrix}.$$

Then

$$p_B(t)= \det \begin{bmatrix} 2-t&3\\ 0&2-t \end{bmatrix} =(2-t)^2=(t-2)^2.$$

So $2$ is the only eigenvalue, but it appears with multiplicity $2$. The factorization alone does not say that there are two independent eigenvectors: that question belongs to the null space of $B-2I$, which is why algebraic and geometric multiplicity must be kept separate.

Algebraic and geometric multiplicity

Repeated roots need extra language.

Definition

Algebraic multiplicity and geometric multiplicity

Let $\lambda\in F$ be a root of $p_A(t)$. Its algebraic multiplicity $m_a(\lambda)$ is the greatest positive integer $m$ for which $(t-\lambda)^m$ divides $p_A(t)$. Equivalently, $p_A(t)=(t-\lambda)^m q(t)$ for a polynomial $q$ with $q(\lambda)\ne0$. Its geometric multiplicity $m_g(\lambda)$ is $\dim_F E_A(\lambda)=\dim_F N(A-\lambda I_n)$.

If $p_A$ splits completely over $F$, its distinct roots can be listed as $\lambda_1,\dots,\lambda_s$ and

$$p_A(t)=(-1)^n(t-\lambda_1)^{m_1}\cdots(t-\lambda_s)^{m_s},\qquad\sum_i m_i=n.$$

The factorization and sum require complete splitting over the chosen field; the local multiplicity definition itself does not.

Algebraic multiplicity comes from the polynomial. Geometric multiplicity comes from the null space $N(A-\lambda I)$. They measure different things, and they do not always agree.

Theorem

Multiplicity inequality

If $\lambda$ is an eigenvalue of an $n\times n$ matrix $A$, then

$$1\le m_g(\lambda)\le m_a(\lambda)\le n.$$

So every eigenspace has dimension at least $1$, but it can never be larger than the algebraic multiplicity of its eigenvalue.

Proof

Why geometric multiplicity cannot exceed algebraic multiplicity

Fix an eigenvalue $\lambda$ and let $g=\dim E_A(\lambda)$. Choose a basis of that eigenspace and extend it to a basis of $F^n$. Let $P$ have this extended basis as columns. The first $g$ columns of $B=P^{-1}AP$ are $\lambda e_1,\dots,\lambda e_g$, so

$$B=\begin{bmatrix}\lambda I_g&C\\0&D\end{bmatrix}.$$

Similarity preserves characteristic polynomials. If $g<n$, the first $g$ columns of $B-tI$ have at most one potentially nonzero entry, namely $\lambda-t$; at $t=\lambda$ that entry is zero, and the repeated column expansion is a polynomial identity giving

$$p_A(t)=\det(B-tI)=(\lambda-t)^g\det(D-tI).$$

Thus $(t-\lambda)^g$ divides $p_A(t)$, so $g\le m_a(\lambda)$. If $g=n$, then $B=\lambda I_n$ and $p_A(t)=(\lambda-t)^n$. This composition uses basis extension and similarity already established earlier and does not assume splitting.

Worked example

A repeated eigenvalue with too small an eigenspace

For

$$B= \begin{bmatrix} 2&3\\ 0&2 \end{bmatrix},$$

the characteristic polynomial is $(t-2)^2$, so the eigenvalue $2$ has algebraic multiplicity $2$.

Now solve $(B-2I)x=0$:

$$B-2I= \begin{bmatrix} 0&3\\ 0&0 \end{bmatrix}.$$

So $x_2=0$ and $x_1$ is free. The eigenspace is

$$\operatorname{span}\left\{ \begin{bmatrix} 1\\0 \end{bmatrix} \right\},$$

which has dimension $1$. Therefore

$$m_a(2)=2,\qquad m_g(2)=1.$$

This mismatch is exactly why the matrix is not diagonalizable.

Distinct eigenvalues force independence

Theorem

Eigenvectors of distinct eigenvalues are linearly independent

If $v_1,\dots,v_k$ are eigenvectors of $A$ corresponding to pairwise distinct eigenvalues $\lambda_1,\dots,\lambda_k$, then

$$v_1,\dots,v_k$$

are linearly independent.

Proof

Induction without assuming independence

Suppose $\lambda_1,\dots,\lambda_s$ are pairwise distinct and $w_j\in E_A(\lambda_j)$. We prove by induction on $s$ that $\sum_{j=1}^s w_j=0$ forces every $w_j=0$, allowing some $w_j$ to be zero. The case $s=1$ is immediate. For the induction step, apply $A-\lambda_s I$ to the relation. The last term vanishes and we obtain

$$\sum_{j=1}^{s-1}(\lambda_j-\lambda_s)w_j=0.$$

Each scaled vector remains in its own eigenspace. The induction hypothesis forces every $(\lambda_j-\lambda_s)w_j$ to be zero; distinctness makes each scalar nonzero, so $w_j=0$ for $j<s$, and the original relation gives $w_s=0$. Grouping a relation by eigenvalue then proves that the union of bases chosen inside the eigenspaces is independent: the grouped vector in each eigenspace is zero, and independence inside that eigenspace kills every coefficient.

This theorem has two major consequences.

Theorem

Upper bound on the number of distinct eigenvalues

An $n\times n$ matrix can have at most $n$ distinct eigenvalues.

Theorem

Distinct eigenvalue test for diagonalizability

If an $n\times n$ matrix has exactly $n$ distinct eigenvalues, then it is diagonalizable.

The preceding independence result gives both consequences: there can be at most $n$ distinct eigenvalues, and when there are $n$, one nonzero eigenvector from each one is an independent list of length $n$, hence a basis of $F^n$.

The converse is false. A diagonalizable matrix may still have repeated eigenvalues. The identity matrix is the simplest example: every nonzero vector is an eigenvector with eigenvalue $1$, yet the matrix is already diagonal.

A sharper diagonalization criterion

Distinct eigenvalues are a sufficient test, but not the only one.

Theorem

Dimension-sum test for diagonalizability

List all distinct eigenvalues of $A$ that lie in the chosen field $F$ as $\lambda_1,\dots,\lambda_s$; the list may be empty. Then $A$ is diagonalizable over $F$ if and only if

$$\dim_F E_A(\lambda_1)+\cdots+\dim_F E_A(\lambda_s)=n.$$

This criterion is a basis count and does not require the characteristic polynomial to split. Choose a basis in each nonzero eigenspace and concatenate these finite lists. The distinct-eigenspace proof shows that the union of eigenspace bases is independent. If the sum is $n$, that union is a basis and diagonalization follows. Conversely, a diagonalizing basis is partitioned by its diagonal eigenvalues; its vectors lie in the corresponding eigenspaces. The independent union of eigenspace bases has dimension at most $n$; its span contains every vector of the diagonalizing basis, so it spans $F^n$ and has dimension at least $n$. Thus the sum is exactly $n$. If there are no eigenvalues in $F$, the sum is $0$, and an $n\ge1$ space cannot be diagonalizable over $F$.

A sharper multiplicity version does require splitting:

Theorem

Algebraic-geometric multiplicity test

If $p_A$ splits completely over $F$, then $A$ is diagonalizable over $F$ if and only if $m_g(\lambda)=m_a(\lambda)$ for every eigenvalue $\lambda$ in $F$.

Indeed, complete splitting gives $\sum_i m_a(\lambda_i)=n$, while the multiplicity inequality gives $m_g(\lambda_i)\le m_a(\lambda_i)$ term by term. Thus equality of every pair is equivalent to the dimension sum being $n$, and the preceding criterion applies. Without splitting, the condition is not a safe replacement. Over the reals,

$$R=\begin{bmatrix}1&-1\\1&1\end{bmatrix},\qquad p_R(t)=(t-1)^2+1$$

has no real eigenvalue. The statement “$m_g=m_a$ for every real eigenvalue” would be vacuous, but $R$ is not diagonalizable over $\mathbb R$. Over $\mathbb C$, its roots $1+i$ and $1-i$ are distinct, so the field-qualified criterion applies there.

Worked example

Repeated eigenvalue but still diagonalizable

Let

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

Put $d=2-t$. Expanding along the first row, the first minor is $\det\begin{bmatrix}d&1\\1&d\end{bmatrix}=d^2-1$ with positive sign. The second minor is $\det\begin{bmatrix}1&1\\1&d\end{bmatrix}=d-1$ and receives the cofactor sign $-$; the third is $\det\begin{bmatrix}1&d\\1&1\end{bmatrix}=1-d$ with sign $+$. Thus the three signed terms are $d(d^2-1)$, $-(d-1)$, and $+(1-d)$, so

$$d(d^2-1)-(d-1)+(1-d)=d^3-3d+2=(d-1)^2(d+2).$$

Substituting back gives

$$p_A(t)=(1-t)^2(4-t)=-(t-1)^2(t-4).$$

The factorization records root $1$ twice and root $4$ once, while the odd degree explains the negative leading coefficient. Since the polynomial has degree three, these are all of its roots over $\mathbb R$, so the later eigenspace count accounts for every real eigenvalue of this matrix, before we inspect eigenspace dimensions.

For $\lambda=1$, solving $(A-I)x=0$ gives $x+y+z=0$, so

$$E_A(1)=\operatorname{span}\left\{\begin{bmatrix}1\\-1\\0\end{bmatrix},\begin{bmatrix}1\\0\\-1\end{bmatrix}\right\}.$$

For $\lambda=4$, the equations give $x=y=z$, hence $E_A(4)=\operatorname{span}\{(1,1,1)^T\}$. The two vectors in $E_A(1)$ are independent because their second and third coordinates force both coefficients to vanish. The distinct-eigenspace lemma then makes their union with the $E_A(4)$ vector independent, and

$$\dim E_A(1)+\dim E_A(4)=2+1=3.$$

For an audit of the eigenspace calculations,

$$A-I=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}, \qquad A-4I=\begin{bmatrix}-2&1&1\\1&-2&1\\1&1&-2\end{bmatrix}.$$

The first matrix gives exactly the single equation $x+y+z=0$, so its null space has dimension $2$; the second has row equations equivalent to $x=y=z$, so its null space has dimension $1$. The polynomial and these dimensions therefore agree with the factorization: the repeated root $1$ contributes two independent directions, while root $4$ contributes one. Their union is a basis of $F^3$, which is the concrete diagonalization test in this example.

Thus $A$ is diagonalizable even though $1$ is a repeated root.

Cayley-Hamilton gives a polynomial identity

One further theorem relates the characteristic polynomial back to the matrix itself.

Theorem

Cayley-Hamilton theorem

If

$$p_A(t)=c_0+c_1t+c_2t^2+\cdots+c_nt^n,$$

then

$$c_0I+c_1A+c_2A^2+\cdots+c_nA^n=0.$$

Equivalently, $p_A(A)=0$. The theorem is general; the proof supplied here handles the diagonalizable case, while the general proof is outside this section.

For diagonalizable matrices this is easy to believe: once $A$ is written as $SDS^{-1}$, the identity reduces to applying the scalar polynomial $p_A$ to each diagonal entry of $D$, and every diagonal entry is an eigenvalue, hence a root of $p_A$.

Proof

Cayley-Hamilton in the diagonalizable case

Assume $A=SDS^{-1}$ with $D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$. For a scalar polynomial $q(t)=c_0+\cdots+c_rt^r$, define $q(A)=c_0I+c_1A+\cdots+c_rA^r$; this is scalar substitution into matrix powers, not substitution inside a determinant. Since powers respect similarity, $q(A)=S q(D)S^{-1}$, and $q(D)$ is diagonal with entries $q(\lambda_1),\dots,q(\lambda_n)$. Taking $q=p_A$, every diagonal entry is zero by the eigenvalue-root theorem. The constant term uses $I=SIS^{-1}$, so $p_A(A)=0$.

The practical consequence is that high powers of $A$ can be rewritten in terms of lower powers $I,A,\dots,A^{n-1}$. If

$$p_A(t)=c_0+c_1t+\cdots+c_{n-1}t^{n-1}+c_nt^n,$$

then $c_n=(-1)^n$ is nonzero, so the Cayley-Hamilton identity solves for $A^n$. Multiplying that expression by $A$ and reducing again gives an algorithm for all later powers. When $c_0=\det(A)\ne0$, multiplying the identity by $A^{-1}$ and solving gives the explicit formula

$$A^{-1}=-\frac{c_1I+c_2A+\cdots+c_nA^{n-1}}{c_0}.$$

This use is valid only after invertibility has been established.

Worked example

A small Cayley-Hamilton identity

Let

$$J= \begin{bmatrix} 1&4\\ 0&1 \end{bmatrix}$$

Its characteristic polynomial is

$$p_J(t)=(1-t)^2=t^2-2t+1$$

Therefore Cayley-Hamilton says

$$J^2-2J+I=0$$

Directly,

$$J^2=\begin{bmatrix}1&8\\0&1\end{bmatrix}, \qquad J^2-2J+I=0$$

Multiplying this verified identity by $J$ gives

$$J^3=2J^2-J=2(2J-I)-J=3J-2I$$

More explicitly, multiplying $J^2-2J+I=0$ by $J^k$ gives the recurrence $J^{k+2}=2J^{k+1}-J^k$ for every $k\ge0$. Thus the next terms are determined without multiplying two large matrices from scratch; for example, $J^4=2J^3-J^2=4J-3I$. This is the concrete reduction mechanism behind the polynomial identity. In fact, for every positive integer $m$,

$$J^m=mJ-(m-1)I=\begin{bmatrix}1&4m\\0&1\end{bmatrix}.$$

The case $m=1$ is the definition of $J$. If the formula holds for $m$, then $J^{m+1}=mJ^2-(m-1)J=m(2J-I)-(m-1)J=(m+1)J-mI$, proving the next case. This makes the reduction concrete even though $J$ is not diagonalizable.

Earlier, the coefficient formula can now be applied concretely: here $c_0=1$, $c_1=-2$, and $c_2=1$, so

$$J^{-1}=-\frac{-2I+J}{1}=2I-J=\begin{bmatrix}1&-4\\0&1\end{bmatrix}.$$

Both cancellation orders can be checked directly:

$$J(2I-J)=2J-J^2=I, \qquad (2I-J)J=2J-J^2=I.$$

Common mistake

Counterexample mode

A repeated eigenvalue can still be diagonalizable

False claim: every repeated eigenvalue prevents diagonalization. Take $A=2I_2$. Then $p_A(t)=(2-t)^2$, so $2$ has algebraic multiplicity $2$, but $E_A(2)=\mathbb R^2$ has basis $e_1,e_2$; the matrix is already diagonal. The earlier matrix $B$ has the same repeated-root pattern but only a one-dimensional eigenspace, so it is not diagonalizable. The repair is to compare $m_g(\lambda)$ with $m_a(\lambda)$ under the stated splitting condition, or to use the unconditioned dimension-sum criterion over the chosen field.

Quick check

Checkpoint

What is the characteristic polynomial of $\begin{bmatrix}2&0\\0&5\end{bmatrix}$?

Use $p_A(t)=\det(A-tI)$.

Solution · Answer

$$p_A(t)= \begin{vmatrix} 2-t&0\\ 0&5-t \end{vmatrix} =(2-t)(5-t).$$

Checkpoint

If an $n\times n$ matrix has n distinct eigenvalues, what can you conclude immediately?

Use the distinct-eigenvalue theorem.

Solution · Answer

It is diagonalizable, because eigenvectors corresponding to distinct eigenvalues are linearly independent.

Checkpoint

Can geometric multiplicity ever exceed algebraic multiplicity?

Use the multiplicity inequality.

Solution · Answer

No. For every eigenvalue $\lambda$,

$$m_g(\lambda)\le m_a(\lambda).$$

Exercises

Checkpoint

Find the characteristic polynomial of $\begin{bmatrix}0&1\\-2&3\end{bmatrix}$ and list its eigenvalues.

Compute $\det(A-tI)$ and factor the result.

Solution · Guided solution

$$p_A(t)= \begin{vmatrix} -t&1\\ -2&3-t \end{vmatrix} =t^2-3t+2=(t-1)(t-2).$$

So the eigenvalues are $1$ and $2$.

Checkpoint

A real 4×4 matrix has four distinct real eigenvalues. What is the dimension of the span of one chosen eigenvector from each eigenvalue?

Use linear independence of eigenvectors with distinct eigenvalues.

Solution · Guided solution

The four chosen eigenvectors are linearly independent, so their span has dimension $4$. Therefore they already form a basis of $\mathbb{R}^4$.

Checkpoint

Suppose $p_A(t)=-t^3+6t^2-11t+6$. What matrix identity does Cayley-Hamilton give?

Replace the variable $t$ by the matrix $A$.

Solution · Guided solution

$$-A^3+6A^2-11A+6I=0,$$

or, after multiplying by $-1$,

$$A^3-6A^2+11A-6I=0.$$

That is the Cayley-Hamilton identity for this characteristic polynomial.

Related notes

Keep 8.1 Eigenvalues, eigenvectors, and eigenspaces open for the null-space interpretation of eigenvalues.

Keep 8.2 Diagonalization and similarity nearby, because this note supplies the polynomial tests that feed that chapter.

The determinant machinery inside $p_A(t)=\det(A-tI)$ depends on 7.1 Determinants and cofactor expansion.