Practice: Diagonalization

(a) $A$ is diagonalizable over $\mathbb{R}$. $\mathrm{tr}A=5$, $detA=4-4=0$, so $p(\lambda )={\lambda }^2-5\lambda =\lambda (\lambda -5)$: eigenvalues $0$ and $5$, distinct, each of algebraic multiplicity 1. Two distinct real eigenvalues in a $2\times 2$, so by the corollary $A$ is diagonalizable. $E_0=\ker A$: from $v_1+2v_2=0$, take $(2,-1)$. $E_5$: , giving $2v_1-v_2=0$ and the vector $(1,2)$. Taking the eigenvectors in the order $(2,-1)$ then $(1,2)$:

Verification.

Column 1 says $A(2,-1{)}^{\mathsf{T}}=0$, and column 2 says $A(1,2{)}^{\mathsf{T}}=5(1,2{)}^{\mathsf{T}}$. Invariant check. $\mathrm{tr}D=0+5=5=\mathrm{tr}A$ ✓, and the diagonal product $0\times 5=0=detA$ ✓. (b) $A^3$ by the factorisation. $detP=2(2)-1(-1)=5$, so . $D^3=\mathrm{diag}(0^3,5^3)=\mathrm{diag}(0,125)$.

Direct check. , so ✓. (That $A^2=5A$ is the eigenvalue structure showing through: $A$ acts as 0 on one direction and 5 on the other, so $A^k=5^{k-1}A$.) (c) $R$: not over $\mathbb{R}$, yes over $\mathbb{C}$. $\mathrm{tr}R=0$, $detR=1$, so $p(\lambda )={\lambda }^2+1$. Over $\mathbb{R}$ this has no roots, so $R$ has no real eigenvalues and no real basis of eigenvectors: not diagonalizable over $\mathbb{R}$. Over $\mathbb{C}$ the roots are $i$ and $-i$, distinct, so $R$ is diagonalizable. For $\lambda =i$, $(R-iI)v=0$ gives $-i{v}_1-v_2=0$, so $v=(1,-i)$; conjugating, $\lambda =-i$ has $v=(1,i)$. Thus

Verification of column 1. $R(1,-i{)}^{\mathsf{T}}=(0(1)+(-1)(-i),1(1)+0(-i){)}^{\mathsf{T}}=(i,1{)}^{\mathsf{T}}$, and $i(1,-i{)}^{\mathsf{T}}=(i,-i^2{)}^{\mathsf{T}}=(i,1{)}^{\mathsf{T}}$ ✓. (d) $Q$: not diagonalizable over any field. $p(\lambda )=(\lambda -3{)}^2$, which already splits over $\mathbb{R}$, and over $\mathbb{C}$, and over every field containing 3. So there is no field to move to. has rank 1, so $\dim E_3=2-1=1$. The geometric multiplicities sum to 1, short of 2, so no basis of eigenvectors exists and $Q$ is not diagonalizable over $\mathbb{C}$ either. The rank of $Q-3I$ is computed by row reduction over whatever field contains the entries, and extending the field does not change it: a matrix of rank 1 over $\mathbb{R}$ has rank 1 over $\mathbb{C}$. So the shortfall is permanent. (e) The two obstructions. For $R$ the obstruction is that the characteristic polynomial does not split over the field in question. That is a property of the field, not of the matrix: ${\lambda }^2+1$ has roots somewhere, and moving to a field containing them produces distinct eigenvalues and a full basis of eigenvectors. The obstruction is removable, and what removes it is enlarging the field. For $Q$ the polynomial splits and the obstruction is that the geometric multiplicity falls short of the algebraic. That is a property of the matrix, expressed as the rank of $Q-3I$, and rank is invariant under field extension. There is nowhere to move to, and the obstruction is permanent. What can be done about each: for $R$, work over $\mathbb{C}$ and the factorisation exists. For $Q$, accept that no diagonal representative exists in its similarity class and settle for the nearest normal form. The Jordan form, which for $Q$ is $Q$ itself, a single $2\times 2$ block with 3 on the diagonal and 1 above it. Reporting "not diagonalizable" without naming which obstruction applies leaves out exactly the part that decides whether anything can be done.