CHARACTERISTIC POLYNOMIAL

In linear algebra, one associates a polynomial to every square matrix, its 'characteristic polynomial'. This polynomial encodes several important properties of the matrix, most notably its eigenvalues, its determinant and its trace.

Contents

Motivation

Formal definition

Example

Properties

Characteristic polynomial of a product of two matrices

See also

Motivation

Given a square matrix ''A'', we want to find a polynomial whose roots are precisely the eigenvalues of ''A''.
For a diagonal matrix ''A'', the characteristic polynomial is easy to define: if the diagonal entries are ''a'', ''b'', ''c'' the characteristic polynomial will be:
(''t'' − ''a'')(''t'' − ''b'')(''t'' − ''c'')...
This works because the diagonal entries are also the eigenvalues of this matrix.
For a general matrix ''A'', one can proceed as follows. If λ is an eigenvalue of ''A'', then there is an eigenvector 'v'≠'0' such that
:''A'' 'v' = λ'v',
or
:(λ'''I''' - ''A'')'v' = 0
(where '''I''' is the identity matrix). Since 'v' is non-zero, this means that the matrix λ'''I''' − ''A'' is singular, which in turn means that its determinant is 0. We have just shown that the roots of the function det(''t'' '''I''' − ''A'') are the eigenvalues of ''A''. Since this function is a polynomial in ''t'', we're done.

Formal definition

We start with a field ''K'' (such as the real or complex numbers) and an ''n''×''n'' matrix ''A'' over ''K''. The characteristic polynomial of ''A'', denoted by ''p''_''A''(''t''), is the polynomial defined by
:''p''_''A''(''t'') = det(''t'' '''I''' − ''A'')
where '''I''' denotes the ''n''-by-''n'' identity matrix and the determinant is being taken in ''K(t)'', the field of rational functions in ''t''. This is indeed a polynomial, since determinants are defined in terms of sums of products.
(Some authors define the characteristic polynomial to be det(''A'' − ''t'' '''I'''); the difference is immaterial since the two polynomials differ at most by a sign - when ''n'' is odd .)

Example

Suppose we want to compute the characteristic polynomial of the matrix
:
2 & 1\
-1& 0
end{pmatrix}.

We have to compute the determinant of
:
t-2&-1\
1&t
end{pmatrix}

and this determinant is
:$(t-2)t\; -\; 1(-1)\; =\; t^2-2t+1.,!$
The latter is the characteristic polynomial of ''A''.

Properties

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''). The constant coefficient ''p''_''A''(0) is equal to (−1)^''n'' times the determinant of ''A'', and the coefficient of ''t'' ^{''n'' − 1} is equal to the negative of tr(''A''), the matrix trace of ''A''. For a 2×2 matrix ''A'', the characteristic polynomial is nicely expressed then as
: ''t'' ² − tr(''A'')''t'' + det(''A'').
All real polynomials of odd degree have a real number as a root, so for odd ''n'', every real matrix has at least one real eigenvalue. Many real polynomials of even degree do not have a real root, but the fundamental theorem of algebra states that every polynomial of degree n has n complex roots, counted with their multiplicities. The non-real roots of real polynomials, hence the non-real eigenvalues, come in conjugate pairs.
The Cayley-Hamilton theorem states that replacing ''t'' by ''A'' in the expression for ''p''_''A''(''t'') yields the zero matrix: ''p''_''A''(''A'') = 0. Simply, every matrix satisfies its own characteristic equation. As a consequence of this, one can show 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''.
In fact, ''A'' is even similar to a matrix in Jordan normal form in this case.

Characteristic polynomial of a product of two matrices

If ''A'' and ''B'' are two square ''n×n'' matrices then characterictic polynomials of ''AB'' and ''BA'' coincide:
:$p\_\{AB\}(t)=p\_\{BA\}(t).,$
More generally, if ''A'' is ''m×n''-matrix and ''B'' is ''n×m'' matrices such that ''m''<''n'', then ''AB'' is ''m×m'' and ''BA'' is ''n×n'' matrix.
One has
: $p\_\{A\; B\}(t)\; =\; t^\{n-m\}\; p\_\{B\; A\}(t).,$
To prove the first result, recognize that the equation to be proved, as a polynomial in t and in the entries of ''A'' and ''B'' is a universal polynomial identity. It therefore suffices to check it on an open set of parameter values in the complex numbers. The tuples (''A'',''B'',''t'') where ''A'' is an invertible complex ''n'' by ''n'' matrix, ''B'' is any complex ''n'' by ''n'' matrix, and ''t'' is any complex number form an open set in complex space of dimension 2''n''² + 1.
When ''A'' is non-singular our result follows from the fact that ''AB'' and ''BA'' are similar:
:$BA\; =\; A^\{-1\}\; (AB)\; A.,$

See also

★ Invariants of tensors

★ Characteristic equation

★ Linear differential equation