This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Now we examine the characteristic polynomial of a matrix and a linear map, and through it we define eigenvalues.
Characteristic Polynomial and Eigenvalues
Definition 1 For any \(n\times n\) square matrix \(A\), the characteristic polynomial of \(A\) is defined as the polynomial in \(\x\) given by \(\det(\x I-A)\).
From the formula
\[\det(\x I-A)=\sum_{\tau\in S_n}\sgn(\tau)(\x I-A)_{\tau(1),1}\cdots(\x I-A)_{\tau(n),n}\tag{1}\]we see that the degree of the characteristic polynomial of \(A\) is at most \(n\). Each summand on the right-hand side is a product of \(n\) terms, and each factor \((\x I-A)_{\tau(k),k}\) is linear in \(x\) only when \(\tau(k)=k\), and is constant otherwise. Hence, if the characteristic polynomial is actually of degree \(n\), the degree-\(n\) term must appear only for the permutation \(\tau\) satisfying \(\tau(k)=k\) for all \(k\), that is, for \(\tau=\id_{S_n}\). In this case, the corresponding term is
\[(\x I-A)_{1,1}\cdots(\x I-A)_{n,n}=(\x-A_{11})\cdots(\x-A_{nn})\tag{2}\]and expanding this shows that the coefficient of \(\x\) is \(1\); thus the characteristic polynomial always has degree \(n\).
If there exists some \(i\) with \(\tau(i)\neq i\), then by the pigeonhole principle there must also be another \(j\) with \(\tau(j)\neq j\). From this we see that no term of degree \(n-1\) appears on the right-hand side of (1). That is, the degree-(\(n-1\)) term of the characteristic polynomial arises solely from (2), and its coefficient is
\[-(A_{1,1}+\cdots+A_{n,n}).\]Finally, let us find the constant term of the characteristic polynomial. Substituting \(\x=0\) gives
\[\det(0I-A)=\det(-A)=(-1)^n\det(A).\]Therefore we have proved the following proposition.
Proposition 2 The characteristic polynomial of an \(n\times n\) square matrix is always a polynomial of degree \(n\); the coefficient of its degree-(\(n-1\)) term equals \(-\tr A\), and its constant term equals \((-1)^n\det A\).
The roots of the characteristic polynomial provide important information when studying a matrix.
Definition 3 For an \(n\times n\) matrix \(A\), the roots of the characteristic equation \(\det(\x I-A)=0\) are called the eigenvalues of \(A\). The set of eigenvalues of \(A\) is called the spectrum of \(A\) and is denoted by \(\Spec(A)\).
Consider two similar \(n\times n\) matrices \(A\) and \(B\). Then from \(A=PBP^{-1}\) we obtain
\[\det(\x I-A)=\det(\x I-PBP^{-1})=\det(P(\x I-B)P^{-1})=\det P\det(\x I-B)\det P^{-1}=\det(\x I-B).\]Hence \(A\) and \(B\) have the same characteristic polynomial. From this we obtain the following corollaries.
Corollary 4 For any linear map \(L:V\rightarrow V\), defining the characteristic polynomial of \(L\) as the characteristic polynomial of the matrix \([L]_\mathcal{B}^\mathcal{B}\) is well-defined.
Proof
That is, we must show that the characteristic polynomial of \(L\) does not change if we choose a basis \(\mathcal{C}\) of \(V\) instead of \(\mathcal{B}\). By the preceding argument, it suffices to observe from the formula after §Change of Basis, ⁋Proposition 5 that the two matrix representations \([L]_\mathcal{B}^\mathcal{B}\) and \([L]_\mathcal{C}^\mathcal{C}\) are similar matrices.
For convenience, all subsequent discussion will be unified in terms of matrices, but by the above corollary we can prove the same results for any linear map \(L\) as well.
Corollary 5 Similar matrices have the same trace and determinant.
Proof
From the preceding argument we know that \(A\) and \(B\) have the same characteristic polynomial, and by Proposition 2 the trace and determinant of a matrix are determined by its characteristic polynomial.
In particular, when any linear map \(L:V\rightarrow V\) is given, by decomposing \([L]_\mathcal{B}^\mathcal{B}\) via diagonalization, which we will cover in the next post, we can decompose \(V\) into eigenspaces of \(L\).
Algebraic Multiplicity
Eigenvalues are all roots of the characteristic polynomial, but some eigenvalues may have greater multiplicity than others. This is defined as follows.
Definition 6 Let \(p(\x)\) be an arbitrary polynomial in \(\mathbb{K}[\x]\), and let \(a\in\mathbb{K}\) be a root of \(p(\x)=0\). If \((\x-a)^k\) divides \(p(\x)\) but \((\x-a)^{k+1}\) does not, then the multiplicity of \(a\) is defined to be \(k\).
Let \(p_A(\x)\) be the characteristic polynomial of an \(n\times n\) matrix \(A\), and let \(\lambda\) be an eigenvalue of \(A\). Then the multiplicity of \(\lambda\) as a root of \(p_A\) is called the algebraic multiplicity of \(\lambda\). This terminology serves to distinguish it from the geometric multiplicity, which we will define shortly.
Let \(p(\x)\) be an arbitrary element of \(\mathbb{K}[\x]\). If \(p\) is a polynomial of degree \(n\), then \(p\) has at most \(n\) roots. However, it need not have exactly \(n\) roots.
Example 7 For example, let \(\mathbb{K}=\mathbb{R}\) and consider the \(2\times 2\) matrix
\[J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.\]Then the characteristic polynomial of \(J\) is \(\x^2+1\), which has no roots in \(\mathbb{R}\).
A field in which this does not happen is called an algebraically closed field. The following Fundamental Theorem of Algebra can be proved algebraically or analytically, but any approach is beyond our current level, so we accept it as a fact and move on.
Theorem 8 (Fundamental Theorem of Algebra) The set of complex numbers \(\mathbb{C}\) is an algebraically closed field.
The characteristic equation of the matrix \(J\) in Example 7 above has no roots in \(\mathbb{R}\), but has two roots in \(\mathbb{C}\). Henceforth, it will be important to distinguish over which field the roots of a polynomial are defined.
Eigenvectors and Geometric Multiplicity
Let \(A\) be an \(n\times n\) matrix and let \(\lambda\) be an eigenvalue of \(A\). Then by definition the matrix \(\lambda I-A\) is singular, and therefore there exists a nonzero vector \(v\) satisfying
\[(\lambda I-A)v=0.\]Definition 9 For an \(n\times n\) matrix \(A\) and an eigenvalue \(\lambda\), a vector \(v\) satisfying \(Av=\lambda v\) is called an eigenvector of \(A\) corresponding to \(\lambda\).
For a matrix \(A\), let \(E_\lambda\) denote the set of all eigenvectors corresponding to \(\lambda\). Then one easily verifies that \(E_\lambda\) forms a vector space. It is called the eigenspace corresponding to \(\lambda\). Since \(E_\lambda\) always contains at least one nonzero vector, \(\dim E_\lambda\) is always greater than \(0\).
Definition 10 For an \(n\times n\) matrix \(A\) and an eigenvalue \(\lambda\), the dimension \(\dim E_\lambda\) of \(E_\lambda\) is called the geometric multiplicity of \(\lambda\).
References
[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.
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