Proof. By definition \(\lambda \) is an eigenvalue of \(A\) exactly when there is a non-zero vector \(x\) with \(Ax=\lambda x\), that is \[(A-\lambda I)x=0.\] So \(\lambda \) is an eigenvalue if and only if the homogeneous system with coefficient matrix \(A-\lambda I\) has a non-trivial solution. By the \(n\) by \(n\) determinant theorem, part (4), that happens if and only if \(\det (A-\lambda I)=0\).

The insistence on \(x\neq 0\) is what makes the statement have content: \(x=0\) satisfies \(Ax=\lambda x\) for every \(\lambda \) whatsoever, so admitting it would make every scalar an eigenvalue. □

Proof. The characteristic polynomial is defined by \(X_A(x)=\det (A-xI)\). So \[X_A(\lambda )=0 \iff \det (A-\lambda I)=0 \iff \lambda \text { is an eigenvalue of } A,\] the second equivalence being Theorem 8.1.3.

This converts the search for eigenvalues, which as stated is a search over all of \(\mathbb {F}\) for scalars admitting a non-zero solution, into the finite problem of factoring a polynomial of degree \(n\). In particular an \(n\) by \(n\) matrix has at most \(n\) distinct eigenvalues. □

Solution. \(\lambda \) is eigenvalue if and only if it is root of \(|A-x\lambda |=0\).

Now \(|A-xI|=0\) \begin {align*} \begin {vmatrix} 4-x&2\\3&-1-x\\ \end {vmatrix} &=0\\\\ (4-x)(-1-x)-6 &=0\\ -4-4x+x+x^2-6 &=0\\\\ x^2-3x-10 &=0\\ x^2-5x+2x-10 &=0\\ (x-5)(x+2) &=0\\\\ \implies \hspace {0.4cm} x&=5, -2 \end {align*}

i.e. eigenvalues are \(\lambda _1=5\) and \(\lambda _2=-2\).

If \(\lambda =5\) then \((A-\lambda I)=0\) \begin {align*} \begin {pmatrix} 4-\lambda _1 &2\\3&-1-\lambda _1\\ \end {pmatrix} \begin {pmatrix} x_1\\x_2\\ \end {pmatrix} &= \begin {pmatrix} 0\\0\\ \end {pmatrix}\\\\ \begin {pmatrix} -1&2\\3&-6\\ \end {pmatrix} \begin {pmatrix} x_1\\x_2\\ \end {pmatrix} &= \begin {pmatrix} 0\\0\\ \end {pmatrix} \end {align*}

\(-x_1+2x_2=0\implies x_1=2x_2\implies x_1=2\) if \(x_2=1\).

Therefore, if \(\lambda =5\) then \( \begin {pmatrix} x_1\\x_2\\ \end {pmatrix} = \begin {pmatrix} 2\\1\\ \end {pmatrix} \).

Now if \(\lambda =-2\) then \begin {align*} \begin {pmatrix} 4+2&2\\3&-1+2\\ \end {pmatrix} \begin {pmatrix} x_1\\x_2\\ \end {pmatrix} &= \begin {pmatrix} 0\\0\\ \end {pmatrix}\\\\ \begin {pmatrix} 6&2\\3&1\\ \end {pmatrix} \begin {pmatrix} x_1\\x_2\\ \end {pmatrix} &= \begin {pmatrix} 0\\0\\ \end {pmatrix} \end {align*}

\(6x_1+2x_2=0\implies x_2=-3x_1\implies x_2=-3\) if \(x_1=1\).

Therefore, if \(\lambda =-2\) then \( \begin {pmatrix} x_1\\x_2\\ \end {pmatrix} = \begin {pmatrix} 1\\-3\\ \end {pmatrix} \).

Recall that if \(A,B\in M_n(\mathbb {F})\), then \(A\) is similar to matrix \(B\) if there exists an invertible matrix \(P\in M_n(\mathbb {F})\ni B=P^{-1}AP\).

Proof. Suppose \(B=P^{-1}AP\) with \(P\) invertible. Then, using \(P^{-1}IP=I\), \[B-xI=P^{-1}AP-xP^{-1}IP=P^{-1}(A-xI)P.\] Taking determinants and using the multiplicative property, \[X_B(x)=\det (B-xI)=\det (P^{-1})\det (A-xI)\det (P) =\frac {1}{\det P}\det (A-xI)\det P=\det (A-xI)=X_A(x).\] So the two characteristic polynomials are identical, and by Theorem 8.1.5 their roots — the eigenvalues — coincide.

Note the order of the statement: equality of the characteristic polynomials is the stronger fact, and the equality of eigenvalues follows from it, not the other way round. □