Characteristic polynomial

Given a square matrix A, we want to find a polynomial whose zeros are the eigenvalues of A. For a diagonal matrix A, the characteristic polynomial is easy to define: if the diagonal entries are a₁, a₂, a₃, etc. then the characteristic polynomial will be:

${\displaystyle (t-a_{1})(t-a_{2})(t-a_{3})\cdots .}$

This works because the diagonal entries are also the eigenvalues of this matrix.

For a general matrix A, one can proceed as follows. A scalar λ is an eigenvalue of A if and only if there is a nonzero vector v, called an eigenvector, such that

${\displaystyle A\mathbf {v} =\lambda \mathbf {v} ,}$

or, equivalently,

${\displaystyle (\lambda I-A)\mathbf {v} =0}$

(where I is the identity matrix). Since v must be non-zero, this means that the matrix λI – A has a nonzero kernel. Thus this matrix is not invertible, and the same is true for its determinant, which must therefore be zero. Thus the eigenvalues of A are the roots of det(λI – A), which is a polynomial in λ.

We consider an n×n matrix A. The characteristic polynomial of A, denoted by p_A(t), is the polynomial defined by[2]

${\displaystyle p_{A}(t)=\det \left(tI-A\right)}$

where I denotes the n-by-n identity matrix.

Some authors define the characteristic polynomial to be det(A - t I). That polynomial differs from the one defined here by a sign (−1)ⁿ, so it makes no difference for properties like having as roots the eigenvalues of A; however the definition above always gives a monic polynomial, whereas the alternative definition is monic only when ${\displaystyle n}$ is even.

Suppose we want to compute the characteristic polynomial of the matrix

${\displaystyle A={\begin{pmatrix}2&1\\-1&0\end{pmatrix}}.}$

We now compute the determinant of

${\displaystyle tI-A={\begin{pmatrix}t-2&-1\\1&t-0\end{pmatrix}}}$

which is ${\displaystyle (t-2)t-1(-1)=t^{2}-2t+1\,\!,}$ the characteristic polynomial of A.

Another example uses hyperbolic functions of a hyperbolic angle φ. For the matrix take

${\displaystyle A={\begin{pmatrix}\cosh(\varphi )&\sinh(\varphi )\\\sinh(\varphi )&\cosh(\varphi )\end{pmatrix}}.}$

Its characteristic polynomial is

${\displaystyle \det(tI-A)=(t-\cosh(\varphi ))^{2}-\sinh ^{2}(\varphi )=t^{2}-2t\ \cosh(\varphi )+1=(t-e^{\varphi })(t-e^{-\varphi }).}$

The polynomial p_A(t) is monic (its leading coefficient is 1) and its degree is n. The most important fact about the characteristic polynomial was already mentioned in the motivational paragraph: the eigenvalues of A are precisely the roots of p_A(t) (this also holds for the minimal polynomial of A, but its degree may be less than n). The coefficients of the characteristic polynomial are all polynomial expressions in the entries of the matrix. In particular its constant coefficient p_A (0)  is det(−A) = (−1)ⁿ det(A), the coefficient of tⁿ is one, and the coefficient of tⁿ⁻¹ is tr(−A) = −tr(A), where tr(A) is the trace of A. (The signs given here correspond to the formal definition given in the previous section;[3] for the alternative definition these would instead be det(A) and (−1)^n − 1 tr(A) respectively.[4])

For a 2×2 matrix A, the characteristic polynomial is thus given by

${\displaystyle t^{2}-\operatorname {tr} (A)t+\det(A).}$

Using the language of exterior algebra, one may compactly express the characteristic polynomial of an n×n matrix A as

${\displaystyle p_{A}(t)=\sum _{k=0}^{n}t^{n-k}(-1)^{k}\operatorname {tr} (\Lambda ^{k}A)}$

where tr(Λ^kA) is the trace of the k^th exterior power of A, which has dimension ${\displaystyle {\tbinom {n}{k}}}$. This trace may be computed as the sum of all principal minors of A of size k. The recursive Faddeev–LeVerrier algorithm computes these coefficients more efficiently.

When the characteristic is 0 it may alternatively be computed as a single determinant, that of the k×k matrix,

${\displaystyle \operatorname {tr} (\Lambda ^{k}A)={\frac {1}{k!}}{\begin{vmatrix}\operatorname {tr} A&k-1&0&\cdots &\\\operatorname {tr} A^{2}&\operatorname {tr} A&k-2&\cdots &\\\vdots &\vdots &&\ddots &\vdots \\\operatorname {tr} A^{k-1}&\operatorname {tr} A^{k-2}&&\cdots &1\\\operatorname {tr} A^{k}&\operatorname {tr} A^{k-1}&&\cdots &\operatorname {tr} A\end{vmatrix}}~.}$

The Cayley–Hamilton theorem states that replacing t by A in the characteristic polynomial (interpreting the resulting powers as matrix powers, and the constant term c as c times the identity matrix) yields the zero matrix. Informally speaking, every matrix satisfies its own characteristic equation. This statement is equivalent to saying that the minimal polynomial of A divides the characteristic polynomial of A.

Two similar matrices have the same characteristic polynomial. The converse however is not true in general: two matrices with the same characteristic polynomial need not be similar.

The matrix A and its transpose have the same characteristic polynomial. A is similar to a triangular matrix if and only if its characteristic polynomial can be completely factored into linear factors over K (the same is true with the minimal polynomial instead of the characteristic polynomial). In this case A is similar to a matrix in Jordan normal form.

If A and B are two square n×n matrices then characteristic polynomials of AB and BA coincide:

${\displaystyle p_{AB}(t)=p_{BA}(t).\,}$

When A is non-singular this result follows from the fact that AB and BA are similar:

${\displaystyle BA=A^{-1}(AB)A.}$

For the case where both A and B are singular, one may remark that the desired identity is an equality between polynomials in t and the coefficients of the matrices. Thus, to prove this equality, it suffices to prove that it is verified on a non-empty open subset (for the usual topology, or, more generally, for the Zariski topology) of the space of all the coefficients. As the non-singular matrices form such an open subset of the space of all matrices, this proves the result.

More generally, if A is a matrix of order m×n and B is a matrix of order n×m, then AB is m×m and BA is n×n matrix, and one has

${\displaystyle p_{BA}(t)=t^{n-m}p_{AB}(t).\,}$

To prove this, one may suppose n > m, by exchanging, if needed, A and B. Then, by bordering A on the bottom by n – m rows of zeros, and B on the right, by, n – m columns of zeros, one gets two n×n matrices A' and B' such that B'A' = BA, and A'B' is equal to AB bordered by n – m rows and columns of zeros. The result follows from the case of square matrices, by comparing the characteristic polynomials of A'B' and AB.

The above definition of the characteristic polynomial of a matrix ${\displaystyle A\in M_{n}(F)}$ with entries in a field F generalizes without any changes to the case when F is just a commutative ring. Garibaldi (2004) defines the characteristic polynomial for elements of an arbitrary finite-dimensional (associative, but not necessarily commutative) algebra over a field F and proves the standard properties of the characteristic polynomial in this generality.

• Characteristic equation (disambiguation)
• Minimal polynomial (linear algebra)
• Invariants of tensors
• Companion matrix
• Faddeev–LeVerrier algorithm
• Cayley–Hamilton theorem
• Samuelson–Berkowitz algorithm

• T.S. Blyth & E.F. Robertson (1998) Basic Linear Algebra, p 149, Springer ISBN 3-540-76122-5 .
• John B. Fraleigh & Raymond A. Beauregard (1990) Linear Algebra 2nd edition, p 246, Addison-Wesley ISBN 0-201-11949-8 .
• Garibaldi, Skip (2004), "The characteristic polynomial and determinant are not ad hoc constructions", American Mathematical Monthly, 111 (9): 761–778, arXiv:math/0203276, doi:10.2307/4145188, MR 2104048
• Werner Greub (1974) Linear Algebra 4th edition, pp 120–5, Springer, ISBN 0-387-90110-8 .
• Paul C. Shields (1980) Elementary Linear Algebra 3rd edition, p 274, Worth Publishers ISBN 0-87901-121-1 .
• Gilbert Strang (1988) Linear Algebra and Its Applications 3rd edition, p 246, Brooks/Cole ISBN 0-15-551005-3 .