Eigenvalues, Eigenvectors and the Characteristic Polynomial

If $\lambda$ is an eigenvalue, [L1] and [L2] show that $T-\lambda I_V$ is not injective; it therefore cannot be invertible by [L4].

Conversely, suppose $\lambda$ is not an eigenvalue. Then [L1] and [L2] make $S=T-\lambda I_V$ injective, so $\ker S=\{0\}$. By [L3], $\dim \mathrm{im}S=\dim V$, hence $\mathrm{im}S=V$ and $S$ is bijective. Its inverse function is linear: applying $S$ to $S^{-1}(au+bv)$ and to $a{S}^{-1}(u)+b{S}^{-1}(v)$ gives $au+bv$ in both cases, and injectivity makes the two inputs equal. Thus $S$ is invertible; contraposition gives that noninvertibility forces $\lambda$ to be an eigenvalue.

The two implications establish the equivalence. If $V=\{0_V\}$, [L1] makes the left side false and the unique endomorphism is the identity, so the same argument includes that case.

For $m=0$ the empty family is independent, and for $m=1$ the equation $a_1{v}_1=0$ forces $a_1=0$ because $v_1\mathrm{\ne }0$.

Assume $m\ge 2$ and that the claim holds for $m-1$ eigenvectors. Suppose ${\sum }_{i=1}^m{a}_i{v}_i=0$. Applying $T-{\lambda }_m{I}_V$ and using [L1] gives ${\sum }_{i=1}^{m-1}{a}_i({\lambda }_i-{\lambda }_m)v_i=0$.

The first $m-1$ eigenvalues remain pairwise distinct, so the induction hypothesis gives $a_i({\lambda }_i-{\lambda }_m)=0$ for $i<m$. Since ${\lambda }_i-{\lambda }_m\mathrm{\ne }0$, [F1] gives $a_i=0$ for every $i<m$.

The original relation now reads $a_m{v}_m=0$, so $a_m=0$ by [L1] and [F1]. Thus the family is independent by [L2].

The base cases and induction step prove the claim for every finite family.

From any finite set of distinct eigenvalues, choose one eigenvector belonging to each; this is a finite sequence of individual choices.

The chosen vectors are linearly independent by [L1], so [L2] extends them to a basis of $V$. Every basis of the $n$-dimensional space has $n$ elements, so the chosen family, and hence the set of chosen eigenvalues, has at most $n$ elements.

There cannot be $n+1$ distinct eigenvalues. Equivalently, $T$ has at most $n$ distinct eigenvalues; when $n=0$, no eigenvector exists and the bound is still valid.

If $n=0$, [L1] gives ${\chi }_A=1$, which is monic of degree $0$; the additional coefficient assertions are restricted to $n\ge 1$.

Assume $n\ge 1$. In [L2] for $x{I}_n-A$, the identity permutation contributes ${\prod }_{i<n}(x-a_{ii})$. Every nonidentity permutation moves at least two indices, so its term contains at most $n-2$ diagonal factors and has degree at most $n-2$.

Evaluating at $x=0$ gives the constant coefficient $\det (-A)=(-1{)}^n\det (A)$ directly from [L2].

Hence the coefficient of $x^n$ is $1$, the coefficient of $x^{n-1}$ comes only from the identity term and is $-{\sum }_{i<n}{a}_{ii}=-\mathrm{tr}(A)$, and there are no terms of degree above $n$. Thus ${\chi }_A$ is monic of degree $n$.

Steps 1.1–2.1 establish every asserted dimension and coefficient case.

Suppose first that $M$ is block upper triangular. In a nonzero Leibniz term for $xI-M$, a permutation cannot send an index from the second block into the first: if it did, bijectivity would force some first-block index into the second, where the lower-left block is zero. Thus every surviving permutation preserves both index blocks.

The surviving permutation splits uniquely into one permutation of each block, and its sign and product split accordingly. The determinant sum therefore factors as $\det (x{I}_r-A)\det (x{I}_s-D)={\chi }_A(x){\chi }_D(x)$.

The lower-triangular case is identical with the two block directions interchanged. If either block has size zero, its characteristic polynomial is $1$ and the identity reduces to the other block.

Hence both block-triangular forms have the claimed characteristic polynomial.

If $n=0$, [L4] gives ${\chi }_A=1={\chi }_B$.

Suppose $n\ge 1$ and choose $P$ from [L1]. Over $F[x]$, $xI-B=P^{-1}(xI-A)P$.

By [L2], $\det (xI-B)=\det (P^{-1})\det (xI-A)\det (P)$. Since $P^{-1}P=I$, multiplicativity also gives $\det (P^{-1})\det (P)=1$.

Using [L4] in step 2.1 gives ${\chi }_B(x)={\chi }_A(x)$, and step 1.1 supplies the remaining size.

If $\dim V=0$, the spectrum is empty because there is no nonzero eigenvector, while [L1] gives ${\chi }_T=1$, which has no root.

Suppose $\dim V=n\ge 1$ and choose a basis with matrix $A=[T]$. Then ${\chi }_T(\lambda )=\det (\lambda I_n-A)=(-1{)}^n\det (A-\lambda I_n)$.

The scalar $(-1{)}^n$ is nonzero. Thus step 1.2, [L3], and [L2] give ${\chi }_T(\lambda )=0$ if and only if $T-\lambda I_V$ is not invertible if and only if $\lambda \in {\sigma }_F(T)$.

Steps 1.1 and 2.1 prove the set equality in every finite dimension and prove both directions of the equivalence.

Since $V\mathrm{\ne }\{0\}$, its finite dimension $n$ is positive, so [L2] makes ${\chi }_T$ a nonconstant polynomial.

By [L1], ${\chi }_T$ has a root $\lambda \in F$, and [L3] makes $\lambda$ an eigenvalue of $T$.

Hence every such endomorphism has an eigenvalue; the excluded zero-dimensional case would have characteristic polynomial $1$ and no eigenvalue.

If $n=0$, [L3] gives $\mathrm{tr}(T)=0$, while the sum indexed by the empty set is $0$.

Suppose $n\ge 1$. By [L1]–[L3], the coefficient of $x^{n-1}$ in ${\chi }_T$ is $-\mathrm{tr}(T)$. In the given product, obtaining degree $n-1$ means choosing $-{\lambda }_i$ from exactly one factor, so the same coefficient is $-{\sum }_{i<n}{\lambda }_i$.

Equality of coefficients and additive cancellation give $\mathrm{tr}(T)={\sum }_{i<n}{\lambda }_i$. By [L4], the factors list exactly the eigenvalues with their algebraic multiplicities.

Steps 1.1 and 2.1 prove the formula in every finite dimension.

If $n=0$, [L3] gives $\det (T)=1$, while the product indexed by the empty set is $1$.

Suppose $n\ge 1$. By [L1]–[L3], the constant coefficient of ${\chi }_T$ is $(-1{)}^n\det (T)$. The constant coefficient of the given product is ${\prod }_{i<n}(-{\lambda }_i)=(-1{)}^n{\prod }_{i<n}{\lambda }_i$.

Equality of coefficients and cancellation of the nonzero scalar $(-1{)}^n$ give $\det (T)={\prod }_{i<n}{\lambda }_i$. By [L4], the factors list exactly the eigenvalues with their algebraic multiplicities.

Steps 1.1 and 2.1 prove the formula in every finite dimension.

Put $g=\dim E_{\lambda }(T)$. Choose a basis $e_1,\dots ,e_g$ of the eigenspace and extend it by [L2] to a basis of $V$.

Since $T(e_i)=\lambda e_i$ for $i\le g$, [L3] makes the matrix of $T$ in this basis block upper triangular with leading diagonal block $\lambda I_g$.

By [L4], ${\chi }_T(x)=\det (x{I}_g-\lambda I_g)q(x)=(x-\lambda {)}^{g}q(x)$ for the characteristic polynomial $q$ of the other diagonal block.

Thus $(x-\lambda {)}^g$ divides ${\chi }_T$, so the maximal exponent in [L1] is at least $g$. This is the claimed inequality; $g\ge 1$ because $\lambda$ is an eigenvalue.

Since $E_{\lambda }(T)$ contains an eigenvector, it is nonzero and its finite dimension is at least $1$.

By [L2] and the given algebraic multiplicity, $\dim E_{\lambda }(T)\le 1$.

Combining steps 1.1 and 1.2 gives $\dim E_{\lambda }(T)=1$.

If $n=0$, both characteristic polynomials are $1$ by [L1]. Assume henceforth that $n\ge 1$, and work over $F[t]$.

Put $N=\left(\begin{array}{cc}I& tA\\ -B& I\end{array}\right)$. Left multiplication by $\left(\begin{array}{cc}I& -tA\\ 0& I\end{array}\right)$ produces $\left(\begin{array}{cc}I+tAB& 0\\ -B& I\end{array}\right)$, while left multiplication by $\left(\begin{array}{cc}I& 0\\ B& I\end{array}\right)$ produces $\left(\begin{array}{cc}I& tA\\ 0& I+tBA\end{array}\right)$.

In the Leibniz sum [L2] for a block-triangular matrix, every nonzero term preserves the two index blocks, so its determinant is the product of the two diagonal-block determinants. Both multiplying matrices in step 1.2 consequently have determinant $1$, and [L3] yields $\det (I+tAB)=\det (N)=\det (I+tBA)$ in $F[t]$.

Replacing $t$ by $-t$, the coefficient of $t^k$ in $\det (I-tC)$ is the coefficient of $x^{n-k}$ in $\det (xI-C)$, directly from the Leibniz formula. Equality in step 2.1 therefore gives equality of every coefficient of ${\chi }_{AB}$ and ${\chi }_{BA}$.

Together with the zero-sized case, ${\chi }_{AB}={\chi }_{BA}$ for all $n$.

If $n=0$, both sides are the empty product $1$ by [L2].

Assume $n>0$ and the result in dimension $n-1$. The factor $x-{\lambda }_0$ shows that ${\lambda }_0$ is a root of ${\chi }_T$, so [L3] supplies a nonzero eigenvector $v$. Extend $(v)$ by [L4] to a basis. In that basis [L4] gives $[T]=\left(\begin{array}{cc}{\lambda }_0& \ast \\ 0& B\end{array}\right)$.

By [L5], ${\chi }_T(x)=(x-{\lambda }_0){\chi }_B(x)$. Comparing with the given factorization and cancelling $x-{\lambda }_0$ in the domain [L6] gives ${\chi }_B(x)={\prod }_{0<i<n}(x-{\lambda }_i)$.

Powers, linear combinations, and [L1] preserve this block upper-triangular shape, so $[p(T)]=\left(\begin{array}{cc}p({\lambda }_0)& \ast \\ 0& p(B)\end{array}\right)$. The induction hypothesis applied to $B$ and step 2.1 gives ${\chi }_{p(B)}(y)={\prod }_{0<i<n}(y-p({\lambda }_i))$.