Practice: Eigenvalues and Eigenvectors

(a) $P$: singular, diagonalisable. $\mathrm{tr}P=1+4=5$, $detP=4-4=0$, so $p(\lambda )={\lambda }^2-5\lambda =\lambda (\lambda -5)$. Eigenvalues $0$ and $5$, each of algebraic multiplicity 1. Invariant check. $0+5=5=\mathrm{tr}P$ ✓, $0\times 5=0=detP$ ✓. $E_0=\ker P$. $P-0I=P$, and row reducing leaves $v_1+2v_2=0$, so $v_1=-2v_2$ and $E_0=\mathrm{span}\{(2,-1)\}$ (taking $v_2=-1$). Geometric multiplicity 1. $E_5$. , determinant $4-4=0$ ✓. Reducing gives $2v_1-v_2=0$, so $v_2=2v_1$ and $E_5=\mathrm{span}\{(1,2)\}$. Geometric multiplicity 1. Verification. $P(1,2{)}^{\mathsf{T}}=(1+4,2+8{)}^{\mathsf{T}}=(5,10{)}^{\mathsf{T}}=5(1,2{)}^{\mathsf{T}}$ ✓. And $P(2,-1{)}^{\mathsf{T}}=(2-2,4-4{)}^{\mathsf{T}}=(0,0{)}^{\mathsf{T}}=0\cdot (2,-1{)}^{\mathsf{T}}$ ✓. Verdict. Geometric equals algebraic for both, and $\{(2,-1),(1,2)\}$ is a basis of ${\mathbb{R}}^2$ of eigenvectors. $P$ is diagonalisable. That $0$ is an eigenvalue simply records that $P$ is singular. It is not an obstruction. (b) $Q$: defective. Triangular, so the eigenvalues are the diagonal entries: $\lambda =3$ twice, algebraic multiplicity 2. Equivalently $\mathrm{tr}Q=6$, $detQ=9$, $p(\lambda )={\lambda }^2-6\lambda +9=(\lambda -3{)}^2$. Invariant check. $3+3=6$ ✓, $3\times 3=9$ ✓. $E_3$. , which has rank 1, so its kernel has dimension $2-1=1$. The condition is $v_2=0$, giving $E_3=\mathrm{span}\{(1,0)\}$. Geometric multiplicity 1. Verification. $Q(1,0{)}^{\mathsf{T}}=(3,0{)}^{\mathsf{T}}=3(1,0{)}^{\mathsf{T}}$ ✓. Verdict. Geometric algebraic $2$, so $Q$ is defective. There are not two independent eigenvectors, so no basis of eigenvectors exists and $Q$ is not diagonalisable over any field. (c) $R$: no real eigenvalues, diagonalisable over $\mathbb{C}$. $\mathrm{tr}R=0$, $detR=0-(-1)=1$, so $p(\lambda )={\lambda }^2+1$. Over $\mathbb{R}$ this has no roots, so $R$ has no real eigenvalues, which the geometry predicts, since $R$ is a quarter-turn rotation and no real direction survives it. Over $\mathbb{C}$ the roots are $\lambda =i$ and $\lambda =-i$, a conjugate pair, each of algebraic multiplicity 1. Invariant check. $i+(-i)=0=\mathrm{tr}R$ ✓, $i(-i)=-i^2=1=detR$ ✓. $E_i$. . The first row gives $-i{v}_1-v_2=0$, so $v_2=-i{v}_1$ and $E_i=\mathrm{span}\{(1,-i)\}$. Verification. $R(1,-i{)}^{\mathsf{T}}=(0(1)+(-1)(-i),1(1)+0(-i){)}^{\mathsf{T}}=(i,1{)}^{\mathsf{T}}$, and $i(1,-i{)}^{\mathsf{T}}=(i,-i^2{)}^{\mathsf{T}}=(i,1{)}^{\mathsf{T}}$ ✓. $E_{-i}$. By conjugation, $\mathrm{span}\{(1,i)\}$. Verdict. Over $\mathbb{R}$: not diagonalisable, indeed no eigenvalues at all. Over $\mathbb{C}$: two distinct eigenvalues, geometric equals algebraic for both, so diagonalisable. (d) Three different situations. $P$ does not fail. It is diagonalisable over $\mathbb{R}$ with two distinct real eigenvalues. Its only peculiarity is that one eigenvalue is zero, which reports singularity rather than any obstruction to diagonalisation. $R$ fails over $\mathbb{R}$, and the obstruction is removable. Its characteristic polynomial has no real roots, but every polynomial splits over $\mathbb{C}$, and there it has two distinct eigenvalues and a basis of eigenvectors. Enlarging the field fixes it completely. $Q$ fails, and the obstruction cannot be removed. Its characteristic polynomial already splits over $\mathbb{R}$, the root is 3, twice, so there is no larger field to move to. The shortfall is in the geometric multiplicity, which is a statement about the rank of $Q-3I$ and is unchanged by enlarging the field. A defective matrix is defective over every field containing its eigenvalues. That is the distinction worth carrying: missing roots are a property of the field; missing eigenvectors are a property of the matrix.