Solution. The characteristic polynomial is \[X_A(x)=\det (A-xI)=\begin {vmatrix}2-x&5\\4&1-x\end {vmatrix} =(2-x)(1-x)-20=x^2-3x-18=(x-6)(x+3),\] so the eigenvalues are \(\lambda _1=6\) and \(\lambda _2=-3\).

For \(\lambda =6\), solve \((A-6I)v=0\): \[\begin {pmatrix}-4&5\\4&-5\end {pmatrix} \begin {pmatrix}v_1\\v_2\end {pmatrix}=0 \implies 4v_1=5v_2,\] the second row being \(-1\) times the first, as it must be for a non-trivial solution to exist. Taking \(v_2=4\) gives the eigenvector \((5,4)\).

For \(\lambda =-3\), solve \((A+3I)v=0\): \[\begin {pmatrix}5&5\\4&4\end {pmatrix}v=0\implies v_1=-v_2,\] giving the eigenvector \((-1,1)\).

Two useful checks: the eigenvalues sum to \(6+(-3)=3=\operatorname {tr}A\), and their product is \(-18=\det A\).

Solution. \(A\) is lower triangular, and the determinant of a triangular matrix is the product of its diagonal entries. So \[X_A(x)=\det (A-xI)=(-5-x)(7-x)(3-x),\] giving eigenvalues \(-5\), \(7\) and \(3\) immediately — the eigenvalues of a triangular matrix are its diagonal entries, with no computation needed.

For the eigenvectors, solve \((A-\lambda I)v=0\) in each case:

• \(\lambda =3\): the equations force \(v_1=v_2=0\), leaving \(v=(0,0,1)\).
• \(\lambda =7\): solving gives \(v=(0,-2,1)\).
• \(\lambda =-5\): solving gives \(v=(-16,4,9)\), or any multiple.

Since the three eigenvalues are distinct, Theorem 8.2.3(ii) guarantees \(A\) is diagonalisable, and the three eigenvectors form a basis.

Solution. From the first problem the eigenvalues are \(6\) and \(-3\) with eigenvectors \((5,4)\) and \((-1,1)\). Since the eigenvalues are distinct these are independent, so put them in as columns: \[P=\begin {pmatrix}5&-1\\4&1\end {pmatrix},\qquad D=\begin {pmatrix}6&0\\0&-3\end {pmatrix}.\] Here \(\det P=5\cdot 1-(-1)\cdot 4=9\neq 0\), confirming invertibility, and \[P^{-1}=\frac {1}{9}\begin {pmatrix}1&1\\-4&5\end {pmatrix}.\] Checking, \[P^{-1}AP=\frac {1}{9}\begin {pmatrix}1&1\\-4&5\end {pmatrix} \begin {pmatrix}2&5\\4&1\end {pmatrix} \begin {pmatrix}5&-1\\4&1\end {pmatrix} =\begin {pmatrix}6&0\\0&-3\end {pmatrix}=D.\] The order matters: the \(j\)-th diagonal entry of \(D\) must be the eigenvalue of the eigenvector placed in the \(j\)-th column of \(P\). Swapping the columns of \(P\) swaps the diagonal entries of \(D\).

Solution. The characteristic polynomial is \(X_A(x)=x^2-3x-18\), so by Theorem 8.3.1, \[A^2-3A-18I=0.\] Rearrange to isolate a multiple of \(I\): \[A^2-3A=18I\implies A(A-3I)=18I.\] Since \(18\neq 0\) we may divide, and the equation exhibits an explicit inverse: \[A^{-1}=\frac {1}{18}(A-3I) =\frac {1}{18}\begin {pmatrix}-1&5\\4&-2\end {pmatrix} =\begin {pmatrix}-1/18&5/18\\2/9&-1/9\end {pmatrix}.\] This matches the usual formula \(A^{-1}=\frac {1}{\det A}\begin {pmatrix}1&-5\\-4&2\end {pmatrix}\) with \(\det A=-18\).

The step that makes this work in general is that the constant term of \(X_A\) is \(\pm \det A\), which is non-zero exactly when \(A\) is invertible — so the rearrangement is always available for an invertible matrix, of any size.

Solution.

Powers

Let \(B=P^{-1}AP\). Then \[B^k=(P^{-1}AP)(P^{-1}AP)\cdots (P^{-1}AP)=P^{-1}A^kP,\] since each interior \(PP^{-1}\) cancels. So \(B^k\) is similar to \(A^k\) via the same \(P\).

Trace

Use the fact that \(\operatorname {tr}(XY)=\operatorname {tr}(YX)\) for any matrices for which both products are defined — both traces equal \(\sum _i\sum _j x_{ij}y_{ji}\). Applying it with \(X=P^{-1}\) and \(Y=AP\), \[\operatorname {tr}B=\operatorname {tr}(P^{-1}AP)=\operatorname {tr}(APP^{-1}) =\operatorname {tr}A.\]

Both facts also follow from Lemma 8.1.7: similar matrices have the same characteristic polynomial, and the trace is (up to sign) one of its coefficients.

Solution. \(A\) is triangular, so \(X_A(x)=(x-3)^2\) and the only eigenvalue is \(3\), repeated.

For the minimum polynomial, the candidates are the monic divisors of \(X_A\), by Theorem 8.3.1 and Lemma 8.2.7: either \(x-3\) or \((x-3)^2\). Test the smaller: \[A-3I=\begin {pmatrix}0&1\\0&0\end {pmatrix}\neq 0,\] so \(x-3\) does not annihilate \(A\) and therefore \[M(x)=(x-3)^2.\]

\(A\) is not diagonalisable. By Corollary 8.3.5 a matrix is diagonalisable exactly when its minimum polynomial is a product of distinct linear factors, and here the factor \(x-3\) is repeated.

The direct reason is visible too: solving \((A-3I)v=0\) gives \(v_2=0\), so all eigenvectors are multiples of \((1,0)\). The eigenvectors span only a one-dimensional space and cannot form a basis for \(\mathbb {R}^2\), which by Theorem 8.2.3(i) is exactly the obstruction.