Proof.

i.

Suppose \(A\) is diagonalisable, so \(P^{-1}AP=D\) with \(D\) diagonal, say with diagonal entries \(d_1,\dots ,d_n\). Rewriting, \(AP=PD\). Comparing the \(j\)-th columns, and writing \(p_j\) for the \(j\)-th column of \(P\), \[Ap_j=d_jp_j.\] Each \(p_j\) is non-zero because \(P\) is invertible, so each \(p_j\) is an eigenvector with eigenvalue \(d_j\); and the columns of an invertible matrix are linearly independent, so \(\{p_1,\dots ,p_n\}\) is a basis.

Conversely, suppose there is a basis \(\{p_1,\dots ,p_n\}\) consisting of eigenvectors, \(Ap_j=d_jp_j\). Let \(P\) be the matrix with these as columns. It is invertible, its columns being a basis, and the computation above run backwards gives \(AP=PD\), that is \(P^{-1}AP=D\).

ii.

It suffices, by (i), to show that eigenvectors belonging to distinct eigenvalues are linearly independent; \(n\) of them then form a basis, since \(\dim V_n(\mathbb {F})=n\).

Let \(v_1,\dots ,v_k\) be eigenvectors for distinct eigenvalues \(\lambda _1,\dots ,\lambda _k\) and suppose they are dependent. Among all non-trivial relations choose one with the fewest non-zero coefficients, say \[c_1v_1+c_2v_2+\cdots +c_mv_m=0\] after renumbering, with every \(c_i\neq 0\) and \(m\geq 2\) (a single term \(c_1v_1=0\) with \(c_1\neq 0\) would force \(v_1=0\), which no eigenvector is). Apply \(A-\lambda _mI\) to both sides. Since \((A-\lambda _mI)v_i =(\lambda _i-\lambda _m)v_i\), this gives \[c_1(\lambda _1-\lambda _m)v_1+\cdots +c_{m-1}(\lambda _{m-1}-\lambda _m)v_{m-1}=0.\] The eigenvalues are distinct, so every coefficient \(c_i(\lambda _i-\lambda _m)\) with \(i<m\) is non-zero — yet this relation has one fewer term, contradicting minimality. Hence the eigenvectors are independent.

Solution. From Example 8.1.6 the eigenvalues are \(\lambda _1=5\), \(\lambda _2=-2\). i.e. two distinct eigenvalues for \(2\times 2\) matrix.

\(\implies A\) is diagonalisable.

Now, \(P^{-1}AP=\) diag\((\lambda _1,\lambda _2)=\) diag\((5,-2)= \begin {pmatrix} 5&0\\0&-2\\ \end {pmatrix} \).

We can find \(P\), since \(\{(2,1)\}\) is a basis for \(V(5)\) and \(\{(1,-3)\}\) is a basis for \(V(-2)\) then \(P= \begin {pmatrix} 2&1\\1&-3\\ \end {pmatrix} \) and indeed \[ P^{-1}AP= \begin {pmatrix} 2&1\\1&-3\\ \end {pmatrix}^{-1} \begin {pmatrix} 4&2\\3&-1\\ \end {pmatrix} \begin {pmatrix} 2&1\\1&-3\\ \end {pmatrix} =\begin {pmatrix} 5&0\\0&2\\ \end {pmatrix} =\text {diag}(\lambda _1,\lambda _2) \]

Proof. Let \(M(x)\) and \(M'(x)\) be minimum polynomial of \(A\).

\(\implies \deg (M(x))=\deg (M'(x))\).
Define \(f(x)=M(x)-M'(x)\) then \(\deg (f(x))<\deg (M(x))\) or \(\deg (M'(x))\). Since \(M(x)\) and \(M'(x)\) are Monic.
Now \(f(A)=M(A)-M'(A)=0-0=0\) and so \(f(x)\) is monic.
Therefore \(f(x)\) is a minimum polynomial with \(\deg (f(x))<\deg (M(x))\) a contradiction. So \(f\) does not exist and \(M(x)=M'(x)\). □

Proof. By division algorithm the exist polynomials \(q(x)\) and \(r(x)\) such that \(f(x)=q(x)M(x)+r(x)\) where \(\deg (r(x))<\deg (M(x))\) or \(r(x)=0\). Now, \begin {align*} r(A) &=f(A)-q(A)M(A)\\ &=0-q(A)(0)\\ &=0-0\\ &=0 \end {align*}

Suppose \(\deg (r(x))\geq 1\), then \(r(x)\) is minimum with \(\deg (r(x))<\deg (M(x))\) a contradiction, since minimum polynomial is unique. Thus \(r(x)=0\) and \(f(x)=q(x)M(x)\) \[\implies M(x)=\frac {f(x)}{q(x)}\] □

Proof. Let \(B=P^{-1}AP\). First note that powers behave well under similarity: \[B^k=(P^{-1}AP)^k=P^{-1}A^kP,\] because the interior factors \(PP^{-1}\) cancel in pairs. Consequently, for any polynomial \(f(x)=\sum _k a_kx^k\), \[f(B)=\sum _k a_kB^k=\sum _k a_kP^{-1}A^kP=P^{-1}\left (\sum _k a_kA^k\right )P =P^{-1}f(A)P.\] Since \(P\) is invertible, \(f(B)=0\) if and only if \(f(A)=0\). So \(A\) and \(B\) are annihilated by exactly the same polynomials, and in particular the monic polynomial of least degree annihilating one annihilates the other. Hence their minimum polynomials agree. □