Proof. We will give an explicit expression for the polynomial \(g.\) If one is not interested in such an expression, a proof using induction can also be given. Write \(f(x)=\sum_{k=0}^n a_kx^k\) for coefficients \((a_0,\ldots,a_n) \in \mathbb{K}^{n+1}.\) For \(0\leqslant j\leqslant n-1\) consider \[\tag{12.1} b_j=\sum_{k=0}^{n-j-1}a_{k+j+1}x_0^k\] and the polynomial \[g(x)=\sum_{j=0}^{n-1}b_jx^j\] of degree \(n-1.\) We have \[\begin{aligned} g(x)(x-x_0)&=\sum_{j=0}^{n-1}\sum_{k=0}^{n-j-1}\left(a_{k+j+1}x_0^kx^{j+1}\right)-\sum_{j=0}^{n-1}\sum_{k=0}^{n-j-1}\left(a_{k+j+1}x_0^{k+1}x^{j}\right)\\ &=\sum_{j=1}^n\sum_{k=0}^{n-j}\left(a_{k+j}x_0^kx^j\right)-\sum_{j=0}^{n-1}\sum_{k=1}^{n-j}\left(a_{k+j}x_0^kx^j\right)\\ &=a_nx^n+\sum_{j=1}^{n-1}a_jx^j+a_0-a_0-\sum_{k=1}^na_kx_0^k=f(x)-f(x_0). \end{aligned}\]

Proof. We use induction. The case \(n=0\) is clear, hence the statement is anchored.

Inductive step: Suppose \(f \in \mathsf{P}(\mathbb{K})\) is a polynomial of degree \(n\) with \(n+1\) distinct zeros \(\lambda_1,\ldots,\lambda_{n+1}.\) Since \(f(\lambda_{n+1})=0,\) Theorem 12.1 implies that \[f(x)=(x-\lambda_{n+1})g(x)\] for some polynomial \(g\) of degree \(n-1.\) For \(1\leqslant i\leqslant n,\) we thus have \[0=f(\lambda_i)=(\lambda_i-\lambda_{n+1})g(\lambda_i).\] Since \(\lambda_i\neq \lambda_{n+1}\) it follows that \(g(\lambda_i)=0.\) Therefore, \(g\) has \(n\) distinct zeros and must be the zero polynomial by the induction hypothesis. It follows that \(f\) is the zero polynomial as well.

Proof. By Lemma 11.38 and Lemma 11.43, the eigenvalues of \(g\) are the zeros of the characteristic polynomial. The characteristic polynomial of \(g\) has degree \(n.\) The claim follows by applying Proposition 12.2.

Proof. We use induction on the number \(m\) of distinct eigenvalues of \(g.\) Let \(\{\lambda_1,\ldots,\lambda_m\}\) be distinct eigenvalues of \(g.\) For \(m=1\) the statement is trivially true, so the statement is anchored.

Inductive step: Assume \(m-1\) eigenspaces are in direct sum. We want to show that then \(m\) eigenspaces are also in direct sum. Let \(v_i,v_i^{\prime} \in \operatorname{Eig}_{g}(\lambda_i)\) be eigenvectors such that \[\tag{12.2} v_1+v_2+\cdots+v_m=v_1^{\prime}+v_2^{\prime}+\cdots+v_{\tilde{m}}.\] Applying \(g\) to this last equation gives \[\tag{12.3} \lambda_1v_1+\lambda_2 v_2+\cdots+\lambda_m v_m=\lambda_1 v_1^{\prime}+\lambda_2 v_2^{\prime}+\cdots+\lambda_m v_{\tilde{m}}.\] Subtracting \(\lambda_m\) times (12.2) from (12.3) gives \[(\lambda_1-\lambda_m)v_1+\cdots+(\lambda_{m-1}-\lambda_m)v_{m-1}=(\lambda_1-\lambda_m)v_1^{\prime}+\cdots+(\lambda_{m-1}-\lambda_m)v_{m-1}^{\prime}.\] Since \(m-1\) eigenspaces are in direct sum, this implies that \((\lambda_i-\lambda_m)v_i=(\lambda_i-\lambda_m)v_i^{\prime}\) for \(1\leqslant i\leqslant m-1.\) Since the eigenvalues are distinct, we have \(\lambda_i-\lambda_m\neq 0\) for all \(1\leqslant i\leqslant m-1\) and hence \(v_i=v_i^{\prime}\) for all \(1\leqslant i\leqslant m-1.\) Now (12.3) implies that \(v_m=v_{\tilde{m}}\) as well and the inductive step is complete.

Since the eigenspaces are in direct sum, the linear independence of eigenvectors with respect to distinct eigenvalues follows from Remark 11.19 (ii).

Proof. Let \(n=\dim(V).\) Let \(\lambda_1,\ldots,\lambda_n\) denote the distinct eigenvalues of \(g.\) Let \(0_V\neq v_i \in \operatorname{Eig}_g(\lambda_i)\) for \(i=1,\ldots,n.\) Then, by Proposition 12.4, the eigenvectors are linearly independent, it follows that \((v_1,\ldots,v_n)\) is an ordered basis of \(V\) consisting of eigenvectors, hence \(g\) is diagonalisable.

Proof. By Lemma 11.38 and Lemma 11.43, the eigenvalues of \(g\) are the zeros of the characteristic polynomial and this is an element of \(\mathsf{P}(\mathbb{C}).\) The first statement thus follows by applying the fundamental theorem of algebra to the characteristic polynomial of \(g.\)

Applying Theorem 12.1 and the fundamental theorem of algebra repeatedly, we find \(k \in \mathbb{N}\) and multiplicities \(m_1,\ldots,m_k \in \mathbb{N}\) such that \[\operatorname{char}_g(x)=(x-\lambda_1)^{m_1}(x-\lambda_2)^{m_2}\cdots(x-\lambda_k)^{m_k}\] where \(\lambda_1,\ldots,\lambda_k\) are zeros of \(\operatorname{char}_g.\) Since \(\operatorname{char}_g\) has degree \(n,\) it follows that \(\sum_{i=1}^k m_i=n.\)

Proof. (i) Let \(\dim \operatorname{Eig}_{g}(\lambda)=m\) and \(\mathbf{b}\) be an ordered basis of \(\operatorname{Eig}_{g}(\lambda).\) Furthermore, let \(\mathbf{b}^{\prime}\) be an ordered tuple of vectors such that \(\mathbf{c}=(\mathbf{b},\mathbf{b}^{\prime})\) is an ordered basis of \(V.\) The eigenspace \(\operatorname{Eig}_{g}(\lambda)\) is stable under \(g\) and \[\mathbf{M}(g|_{\operatorname{Eig}_{g}(\lambda)},\mathbf{b},\mathbf{b})=\lambda\mathbf{1}_{m}.\] By Proposition 11.36, the matrix representation of \(g\) with respect to the basis \(\mathbf{c}\) takes the form \[\mathbf{M}(g,\mathbf{c},\mathbf{c})=\begin{pmatrix} \lambda\mathbf{1}_{m} & * \\ \mathbf{0}_{n-m,m} & \mathbf{B}\end{pmatrix}\] for some matrix \(\mathbf{B}\in M_{n-m,n-m}(\mathbb{K}).\) We thus obtain \[\operatorname{char}_g(x)=\det\begin{pmatrix} (x-\lambda)\mathbf{1}_{m} & * \\ \mathbf{0}_{n-m,m} & x\mathbf{1}_{n-m}-\mathbf{B}\end{pmatrix}\] Applying the Laplace expansion (9.5) with respect to the first column, we have \[\operatorname{char}_g(x)=(x-\lambda)\det\begin{pmatrix} (x-\lambda)\mathbf{1}_{m-1} & * \\ \mathbf{0}_{n-m,m-1} & x\mathbf{1}_{n-m}-\mathbf{B}\end{pmatrix}\] Applying the Laplace expansion again with respect to the first column, \(m\)-times in total, we get \[\operatorname{char}_g(x)=(x-\lambda)^m\det(x\mathbf{1}_{n-m}-\mathbf{B})=(x-\lambda)^m\operatorname{char}_\mathbf{B}(x).\] The algebraic multiplicity of \(\lambda\) is thus at least \(m.\)

(ii) Suppose \(\mathbb{K}=\mathbb{C}\) and that \(g :V \to V\) is diagonalisable. Hence we have an ordered basis \((v_1,\ldots,v_n)\) of \(V\) consisting of eigenvectors of \(g.\) Therefore, \[\operatorname{char}_g(x)=\prod_{i=1}^n(x-\lambda_i)\] where \(\lambda_i\) is the eigenvalue of the eigenvector \(v_i,\) \(1\leqslant i \leqslant n.\) For any eigenvalue \(\lambda_j,\) its algebraic multiplicity is the number of indices \(i\) with \(\lambda_i=\lambda_j.\) For each such index \(i,\) the eigenvector \(v_i\) satisfies \(g(v_i)=\lambda_iv_i=\lambda_j v_i\) and hence is an element of the eigenspace \(\operatorname{Eig}_{g}(\lambda_j).\) The geometric multiplicity of each eigenvalue is thus at least as big as the algebraic multiplicity, but by the previous statement, the latter cannot be bigger than the former, hence they are equal.

Conversely, suppose that for all eigenvalues of \(g,\) the algebraic and geometric multiplicity are the same. Since \(\mathbb{K}=\mathbb{C},\) by Theorem 12.7, the sum of the algebraic multiplicities is \(n.\) The sum of the geometric multiplicities is by assumption also \(n.\) Since, by Proposition 12.4, the eigenspaces with respect to different eigenvalues are in direct sum, we obtain a basis of \(V\) consisting of eigenvectors of \(g.\)

Proof. Suppose \(\lambda \in \mathbb{K}\) is an eigenvalue of \(\iota\) so that \(\iota(v)=\lambda v\) for some non-zero vector \(v \in V.\) Then \(\iota(\iota(v))=v=\lambda \iota(v)=\lambda^2 v.\) Hence \((1-\lambda^2)v=0_V\) and since \(v\) is non-zero, we conclude that \(\lambda=\pm 1.\)

By Proposition 12.4, the eigenspaces \(\operatorname{Eig}_{\iota}(1)\) and \(\operatorname{Eig}_{\iota}(-1)\) are in direct sum (interpreting \(\operatorname{Eig}_{\iota}(1)\) as \(\{0\}\) if \(1\) is not an eigenvalue, and similarly for \(-1\)). What we need to show is that their sum is all of \(V.\) For \(v \in V\) we write \[v= \frac{1}{2}(v+\iota(v))+\frac{1}{2}(v-\iota(v)).\] We claim that \(\frac{1}{2}(v+\iota(v)) \in \operatorname{Eig}_{\iota}(1),\) and \(\frac{1}{2}(v-\iota(v)) \in \operatorname{Eig}_{\iota}(-1).\)

For the first half of the claim, we compute \[\iota\left(\frac{1}{2}(v+\iota(v))\right)= \frac{1}{2}(\iota(v) + \iota(\iota(v))) = \frac{1}{2}(\iota(v) + v).\] The second half of the claim is similar.

It follows that the sum of \(\operatorname{Eig}_{\iota}(1)\) and \(\operatorname{Eig}_\iota(-1)\) is \(V,\) so \(\iota\) is diagonalisable as required.

Proof. Let \(v \in V\) be an eigenvector of the projection \(\Pi\) with eigenvalue \(\lambda.\) Hence we obtain \(\Pi(\Pi(v))=\lambda^2 v=\Pi(v)=\lambda v,\) equivalently, \(\lambda(\lambda-1)v=0_V.\) Since \(v\) is non zero, it follows that \(\lambda=0\) or \(\lambda=1.\)

Using an argument similar to Proposition 12.12, one can show that \(V = \operatorname{Ker}\Pi + \operatorname{Im}\Pi.\) Since \(\operatorname{Ker}\Pi=\operatorname{Eig}_\Pi(0),\) the theorem will follow if we can show that \(\operatorname{Im}\Pi=\operatorname{Eig}_{\Pi}(1).\) Let \(v \in \operatorname{Im}\Pi\) so that \(v=\Pi(\hat{v})\) for some vector \(\hat{v} \in V.\) Hence \(\Pi(v)=\Pi(\Pi(\hat{v}))=\Pi(\hat{v})=v\) and \(v\) is an eigenvector with eigenvalue \(1.\) Conversely, suppose \(v \in V\) is an eigenvector of \(\Pi\) with eigenvalue \(1.\) Then \(\Pi(v)=v=\Pi(\Pi(v))\) and hence \(v \in \operatorname{Im}\Pi.\) We thus conclude that \(\operatorname{Im}\Pi=\operatorname{Eig}_{\Pi}(1).\) Choosing an ordered basis of \(\operatorname{Ker}\Pi\) and an ordered basis of \(\operatorname{Im}\Pi\) gives a basis of \(V\) consisting of eigenvectors, hence \(\Pi\) is diagonalisable.