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 define eigenvalues.
Characteristic Polynomial and Eigenvalues
Definition 1 For any \(n\times n\) square matrix \(A\), we define the characteristic polynomial of \(A\) as the polynomial \(\det(\x I-A)\) in \(\x\).
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 can see that the degree of the characteristic polynomial of \(A\) is at most \(n\). This is because each summand on the right-hand side is a product of \(n\) terms, and each \((\x I-A)_{\tau(k),k}\) is a linear expression in \(x\) only when \(\tau(k)=k\), and a constant otherwise. From this we know that if the characteristic polynomial is actually of degree \(n\), then the degree-\(n\) term necessarily appears only for \(\tau\) satisfying \(\tau(k)=k\) for all \(k\), that is, only when \(\tau=\id_{S_n}\). In this case, the corresponding term becomes
\[(\x I-A)_{1,1}\cdots(\x I-A)_{n,n}=(\x-A_{11})\cdots(\x-A_{nn})\tag{2}\]and expanding this shows that the coefficient of \(\x\) is \(1\), so we see that the degree of the characteristic polynomial is always \(n\).
If there exists an \(i\) with \(\tau(i)\neq i\), then by the pigeonhole principle there must also exist another \(j\) with \(\tau(j)\neq j\). From this we know that no term of degree \(n-1\) appears among the summands on the right-hand side of (1). That is, the degree-\((n-1)\) term of the characteristic polynomial necessarily arises solely from (2), and its coefficient is
\[-(A_{1,1}+\cdots+A_{n,n})\]as can be seen.
Finally, let us find the constant term of the characteristic polynomial. To do this, we simply substitute \(\x=0\) into the characteristic polynomial. Then the result is
\[\det(0I-A)=\det(-A)=(-1)^n\det(A)\].
Thus we have proved the following proposition.
Proposition 2 The characteristic polynomial of an \(n\times n\) square matrix is necessarily 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 matrices.
Definition 3 For an \(n\times n\) matrix \(A\), the roots of the characteristic polynomial \(\det(\x I-A)=0\) of \(A\) are called the eigenvalues of \(A\). The collection of eigenvalues of \(A\) is called the spectrum of \(A\), and this set is denoted \(\Spec(A)\).
Consider two similar \(n\times n\) matrices \(A\) and \(B\). Then, from \(A=PBP^{-1}\),
\[\det(\x I-A)=\det(\x I-PBP^{-1})=\det(P(\x I-B)P^{-1})=\det P\det(\x I-B)\det P^{-1}=\det(\x I-B)\]we obtain. Therefore, \(A\) 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
Proof
That is, we must show that the characteristic polynomial of \(L\) does not change even 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, ⁋Definition 2 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 to be about matrices, but through 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, given any linear map \(L:V\rightarrow V\), if we decompose \([L]_\mathcal{B}^\mathcal{B}\) via the diagonalization of matrices to be discussed in the next post, we can decompose \(V\) into the eigenspaces of \(L\).
Algebraic Multiplicity
Eigenvalues are all roots of the characteristic polynomial, but some eigenvalues may have a 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 divide \(p(\x)\), then we define the multiplicity of \(a\) to be \(k\).
Let \(p_A(\x)\) denote 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 to be defined 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, \(p\) need not have exactly \(n\) roots.
Example 7 For example, suppose \(\mathbb{K}=\mathbb{R}\), and consider the \(2\times 2\) matrix
\[J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\]Then the characteristic polynomial of \(J\) is \(\x^2+1\), and this polynomial 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 through analysis, but either way is beyond our current level, so we accept it as 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 remain important to distinguish over which field the roots of a polynomial are defined.
Eigenvectors and Geometric Multiplicity
Consider an \(n\times n\) matrix \(A\) and its eigenvalue \(\lambda\). Then by definition the matrix \(\lambda I-A\) is singular, and therefore
\[(\lambda I-A)v=0\]there exists a nonzero vector \(v\) satisfying this equation.
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 can easily verify that \(E_\lambda\) forms a vector space. This 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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