Practice Set-2 — Complete Worked Solutions

Section A — Eigenvalues & Eigenvectors (detailed)

2×2 Matrices

1. Problem: \(A=\begin{bmatrix}2&1\\1&2\end{bmatrix}\).
   1. Compute characteristic polynomial: \(\det(A-\lambda I)=\det\begin{bmatrix}2-\lambda&1\\1&2-\lambda\end{bmatrix}=(2-\lambda)^2-1.\)
   Expand: \((2-\lambda)^2-1 = \lambda^2-4\lambda+3\). Solve \(\lambda^2-4\lambda+3=0\) → \((\lambda-3)(\lambda-1)=0\).
   
   2. Eigenvalues: \(\lambda_1=3,\ \lambda_2=1\).
   
   3. Eigenvectors:
   For \(\lambda=3\): solve \((A-3I)v=0\): \[ \begin{bmatrix}-1&1\\1&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=0 \Rightarrow -x+y=0 \Rightarrow y=x. \] Choose \(v_1=\begin{bmatrix}1\\1\end{bmatrix}\).
   
   For \(\lambda=1\): solve \((A-I)v=0\): \[ \begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=0 \Rightarrow x+y=0 \Rightarrow y=-x. \] Choose \(v_2=\begin{bmatrix}1\\-1\end{bmatrix}\) (or \([-1,1]^T\)).
   
   Final: eigenpairs \((3,[1,1]^T)\) and \((1,[1,-1]^T)\).
2. Problem: \(B=\begin{bmatrix}4&-2\\1&1\end{bmatrix}\).
   1. \(\det(B-\lambda I)=\det\begin{bmatrix}4-\lambda&-2\\1&1-\lambda\end{bmatrix}=(4-\lambda)(1-\lambda)+2\).
   Expand: \((4-\lambda)(1-\lambda)+2 = 4-5\lambda+\lambda^2+2 = \lambda^2-5\lambda+6\).
   Solve \(\lambda^2-5\lambda+6=0\) → \((\lambda-2)(\lambda-3)=0\).
   
   2. Eigenvalues: \(\lambda=2,3\).
   
   3. Eigenvectors:
   For \(\lambda=2\): \((B-2I)=\begin{bmatrix}2&-2\\1&-1\end{bmatrix}\). Equation \(2x-2y=0\Rightarrow x=y\). Choose \(v=\begin{bmatrix}1\\1\end{bmatrix}\).
   For \(\lambda=3\): \((B-3I)=\begin{bmatrix}1&-2\\1&-2\end{bmatrix}\). Equation \(x-2y=0\Rightarrow x=2y\). Choose \(v=\begin{bmatrix}2\\1\end{bmatrix}\).
3. Problem: \(C=\begin{bmatrix}0&1\\-4&-5\end{bmatrix}\).
   1. \(\det(C-\lambda I)=\det\begin{bmatrix}-\lambda&1\\-4&-5-\lambda\end{bmatrix} = \lambda(\lambda+5)+4=\lambda^2+5\lambda+4.\)
   Solve \(\lambda^2+5\lambda+4=0\) → \((\lambda+4)(\lambda+1)=0\) so \(\lambda=-4,-1\).
   
   2. Eigenvectors:
   For \(\lambda=-4\): \(C+4I=\begin{bmatrix}4&1\\-4&-1\end{bmatrix}\). Equation \(4x+y=0\Rightarrow y=-4x\). Choose \(v=\begin{bmatrix}1\\-4\end{bmatrix}\) (or scaled \([-1/4,1]^T\) as earlier).
   For \(\lambda=-1\): \(C+I=\begin{bmatrix}1&1\\-4&-4\end{bmatrix}\). Equation \(x+y=0\Rightarrow y=-x\). Choose \(v=\begin{bmatrix}1\\-1\end{bmatrix}\).
4. Problem: \(D=\begin{bmatrix}3&2\\0&3\end{bmatrix}\).
   1. Char. poly \(\det(D-\lambda I)=(3-\lambda)^2\) → eigenvalue \(\lambda=3\) (algebraic multiplicity 2).
   
   2. Solve \((D-3I)v=0\): \(\begin{bmatrix}0&2\\0&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=0\Rightarrow 2y=0\Rightarrow y=0\). No restriction on \(x\) → eigenvectors proportional to \(\begin{bmatrix}1\\0\end{bmatrix}\).
   
   Only one independent eigenvector exists though algebraic multiplicity is 2 → geometric multiplicity 1 → matrix is defective (not diagonalizable).
5. Problem: \(E=\begin{bmatrix}1&2\\2&1\end{bmatrix}\).
   1. Characteristic polynomial: \(\det\begin{bmatrix}1-\lambda&2\\2&1-\lambda\end{bmatrix}=(1-\lambda)^2-4=\lambda^2-2\lambda-3\).
   Solve \(\lambda^2-2\lambda-3=0\) → \((\lambda-3)(\lambda+1)=0\) → \(\lambda=3,-1\).
   
   2. Eigenvectors:
   \(\lambda=3:\ (E-3I)=\begin{bmatrix}-2&2\\2&-2\end{bmatrix}\Rightarrow -2x+2y=0 \Rightarrow y=x.\) Choose \([1,1]^T\).
   \(\lambda=-1:\ (E+I)=\begin{bmatrix}2&2\\2&2\end{bmatrix}\Rightarrow 2x+2y=0\Rightarrow y=-x.\) Choose \([1,-1]^T\).

3×3 Matrices — full steps

6. Problem: \(F=\begin{bmatrix}1&0&0\\0&2&1\\0&0&2\end{bmatrix}\).
   This is upper-triangular, so eigenvalues are diagonal entries: \(1,2,2\).
   
   For \(\lambda=1\): solve \((F-I)v=0\) → matrix \(\begin{bmatrix}0&0&0\\0&1&1\\0&0&1\end{bmatrix}\). The equations give \(v_3=0\), \(v_2\) free? Check carefully: From last row: \(v_3=0\). From second row: \(v_2+v_3=0\Rightarrow v_2=0\). So \(v=[1,0,0]^T\) (free \(v_1\)).
   
   For \(\lambda=2\): \((F-2I)=\begin{bmatrix}-1&0&0\\0&0&1\\0&0&0\end{bmatrix}\). Equations: \(-x_1=0\Rightarrow x_1=0\); second row: \(x_3=0\); third row: 0. So eigenvectors have \(x_2\) free and \(x_1=x_3=0\). So eigenvector \( [0,1,0]^T\).
   
   There is only one independent eigenvector for \(\lambda=2\) (though algebraic mult. 2). Therefore \(F\) is not diagonalizable (has a Jordan block of size 2).
7. Problem: \(G=\begin{bmatrix}2&0&0\\0&3&4\\0&0&3\end{bmatrix}\).
   Upper-triangular → eigenvalues \(2,3,3\).
   
   For \(\lambda=2\): eigenvector \([1,0,0]^T\).
   For \(\lambda=3\): solve \(\begin{bmatrix}-1&0&0\\0&0&4\\0&0&0\end{bmatrix}v=0\). Equations give \( -x_1=0\Rightarrow x_1=0\); \(4x_3=0\Rightarrow x_3=0\). So eigenvector with \(x_2\) free: \([0,1,0]^T\). Only one independent eigenvector for \(\lambda=3\) → not diagonalizable over \(\mathbb{R}\).
8. Problem: \(H=\begin{bmatrix}4&0&0\\0&2&1\\0&1&2\end{bmatrix}\).
   The matrix is block diagonal with scalar 4 and the \(2\times2\) symmetric block \(S=\begin{bmatrix}2&1\\1&2\end{bmatrix}\).
   
   Eigenvalues: from block: \(4\) and eigenvalues of \(S\): solve \(\det(S-\lambda I)=(2-\lambda)^2-1=\lambda^2-4\lambda+3\Rightarrow \lambda=3,1\).
   
   Eigenvectors: \(\lambda=4:\ [1,0,0]^T\). \(\lambda=3:\) eigenvector for \(S\) is \([1,1]^T\) → for full matrix \([0,1,1]^T\). \(\lambda=1:\) eigenvector for \(S\) is \([1,-1]^T\) → for full matrix \([0,1,-1]^T\).
9. Problem: \(I=\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}\) (permutation/cyclic).
   Compute characteristic polynomial: \(\det(I-\lambda I)= -\lambda^3+1\) (or compute \(\lambda^3-1=0\)). So eigenvalues are cube roots of unity: \(\lambda=1,\ e^{2\pi i/3},\ e^{4\pi i/3}\). The latter two are complex: \(-\tfrac12\pm i\tfrac{\sqrt3}{2}\).
   
   Real eigenvector for \(\lambda=1\): solve \((I-I)v=0\Rightarrow v_1=v_2=v_3\). Choose \([1,1,1]^T\).
   
   Complex eigenvectors correspond to complex eigenvalues; over \(\mathbb{R}\) the real Jordan/real canonical form is a block with a \(2\times2\) rotation block.
10. Problem: \(J=\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}\) (symmetric tridiagonal).
    This is well-known: eigenvalues for this 3×3 are: \[ \lambda_k = 2-2\cos\left(\frac{k\pi}{4}\right)\quad (k=1,2,3) \] which numerically give approximately \(2-2\cos(\pi/4)=2-\sqrt2\), \(2\) and \(2+ \sqrt2\).
    
    Analytical factorization gives eigenvalues \(\lambda_1=2-\sqrt2,\ \lambda_2=2,\ \lambda_3=2+\sqrt2\).
    
    Eigenvectors (up to scaling) can be chosen as: \[ \lambda=2:\ v=[-1,0,1]^T,\quad \lambda=2-\sqrt2:\ v=[1,\sqrt2,1]^T,\quad \lambda=2+\sqrt2:\ v=[1,-\sqrt2,1]^T. \] (You can confirm by substitution \(Jv=\lambda v\).)

Section B — Cayley–Hamilton (with explicit calculations)

11. Problem: Verify Cayley–Hamilton for \(A=\begin{bmatrix}2&1\\1&2\end{bmatrix}\).
    Char. poly \(p(\lambda)=\lambda^2-4\lambda+3\). CH says \(p(A)=A^2-4A+3I=0\).
    
    Compute \(A^2=\begin{bmatrix}2&1\\1&2\end{bmatrix}^2=\begin{bmatrix}5&4\\4&5\end{bmatrix}\).
    Then \(A^2-4A+3I=\begin{bmatrix}5&4\\4&5\end{bmatrix}-4\begin{bmatrix}2&1\\1&2\end{bmatrix}+3\begin{bmatrix}1&0\\0&1\end{bmatrix}\) \[ =\begin{bmatrix}5-8+3 & 4-4+0 \\ 4-4+0 & 5-8+3 \end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}. \] Verified.
12. Problem: Use CH to compute \(A^3\) for \(A=\begin{bmatrix}1&1\\0&1\end{bmatrix}\).
    For this matrix direct multiplication is easiest: \[ A^2=\begin{bmatrix}1&1\\0&1\end{bmatrix}^2=\begin{bmatrix}1&2\\0&1\end{bmatrix},\quad A^3=A^2A=\begin{bmatrix}1&2\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}1&3\\0&1\end{bmatrix}. \] (Cayley–Hamilton would say \(p(\lambda)=(\lambda-1)^2\) so \(A\) satisfies \((A-I)^2=0\). That identity can be used to reduce powers, but direct is simplest here.)
13. Problem: Use CH to find \(A^{-1}\) for \(A=\begin{bmatrix}3&2\\1&1\end{bmatrix}\).
    Char. poly: \(\lambda^2-4\lambda+1\). CH: \(A^2-4A+I=0\). Multiply on right by \(A^{-1}\): \[ A-4I + A^{-1} = 0 \quad\Rightarrow\quad A^{-1} = 4I - A. \] Thus \[ A^{-1} = 4\begin{bmatrix}1&0\\0&1\end{bmatrix} - \begin{bmatrix}3&2\\1&1\end{bmatrix} = \begin{bmatrix}1&-2\\-1&3\end{bmatrix}. \] (Confirm by \(AA^{-1}=I\): compute product to check — it equals \(I\).)
14. Problem: Verify CH for \(A=\begin{bmatrix}0&1&0\\0&0&1\\-6&11&-6\end{bmatrix}\).
    This is a companion matrix of polynomial \(\lambda^3-6\lambda^2+11\lambda-6\). The polynomial factors \((\lambda-1)(\lambda-2)(\lambda-3)\).
    
    CH states \(A^3-6A^2+11A-6I=0\). Compute \(A^2\), \(A^3\) explicitly (lengthy but straightforward) and verify the combination equals zero matrix. (Because companion matrix was constructed from that polynomial, CH holds identically.)
15. Problem: Use CH to compute \(A^4\) for \(A=\begin{bmatrix}4&3\\3&2\end{bmatrix}\).
    First compute characteristic polynomial: \(\det(A-\lambda I)=\lambda^2-6\lambda-? \) — compute precisely: \[ \det\begin{bmatrix}4-\lambda&3\\3&2-\lambda\end{bmatrix}=(4-\lambda)(2-\lambda)-9=\lambda^2-6\lambda-1. \] So \(p(\lambda)=\lambda^2-6\lambda-1\). CH: \(A^2 -6A - I =0\Rightarrow A^2 =6A + I\).
    
    Then \(A^3=A\cdot A^2 = A(6A+I)=6A^2 + A = 6(6A+I)+A = 37A + 6I\).
    And \(A^4 = A\cdot A^3 = A(37A+6I)=37A^2 + 6A = 37(6A+I)+6A = (222A + 37I) + 6A = 228A +37I.\)
    
    Now substitute numeric matrix: \[ 228A = 228\begin{bmatrix}4&3\\3&2\end{bmatrix}=\begin{bmatrix}912&684\\684&456\end{bmatrix},\quad 37I=\begin{bmatrix}37&0\\0&37\end{bmatrix}. \] So \(A^4=\begin{bmatrix}949&684\\684&493\end{bmatrix}\) (same result as direct multiplication).

Section C — Diagonalization (explicit P, inverse where real)

16. Problem: Diagonalize \(A=\begin{bmatrix}5&4\\1&2\end{bmatrix}\).
    1. Char. poly \(\det(A-\lambda I)=(5-\lambda)(2-\lambda)-4=\lambda^2-7\lambda+6=(\lambda-6)(\lambda-1)\).
    
    2. Eigenvalues \(\lambda_1=6,\ \lambda_2=1\).
    
    3. Eigenvectors: For \(\lambda=6\): \((A-6I)=\begin{bmatrix}-1&4\\1&-4\end{bmatrix}\Rightarrow -x+4y=0\Rightarrow x=4y.\) Choose \(v_1=\begin{bmatrix}4\\1\end{bmatrix}\) (or scaled \([2,0.5]\) etc.). I'll use \(v_1=[2,1]^T\) (scaling by 1/2 from [4,1]).
    For \(\lambda=1\): \((A-I)=\begin{bmatrix}4&4\\1&1\end{bmatrix}\Rightarrow 4x+4y=0\Rightarrow x=-y\). Choose \(v_2=[-1,1]^T\).
    
    4. Form \(P=[v_1\ v_2]=\begin{bmatrix}2&-1\\1&1\end{bmatrix}\). Compute \(P^{-1}\) (determinant \(=2\cdot1 -(-1)\cdot1 =3\)): \[ P^{-1}=\frac{1}{3}\begin{bmatrix}1&1\\-1&2\end{bmatrix}. \] Then \(P^{-1}AP=\operatorname{diag}(6,1)\). (You can multiply out to verify.)
17. Problem: \(B=\begin{bmatrix}2&0\\0&3\end{bmatrix}\) — find \(P\) s.t. \(P^{-1}BP\) diagonal.
    \(B\) already diagonal: eigenvalues 2 and 3, eigenvectors \(e_1=[1,0]^T\) and \(e_2=[0,1]^T\). So choose \(P=I\) or any matrix with eigenvectors as columns. \(P^{-1}BP=B\).
18. Problem: Is \(C=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) diagonalizable?
    Char. poly \((\lambda-1)^2\). Solve \((C-I)v=0\Rightarrow\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=0\Rightarrow y=0\). So eigenvectors spanned by \([1,0]^T\). Geometric multiplicity 1 < algebraic multiplicity 2 → not diagonalizable (Jordan block).
19. Problem: Diagonalize (over \(\mathbb{C}\)) \(D=\begin{bmatrix}6&-2\\2&3\end{bmatrix}\).
    Char. poly \(\lambda^2-9\lambda+22\). Discriminant \(81-88=-7\) → complex eigenvalues \(\frac{9\pm i\sqrt7}{2}\) (distinct). Distinct eigenvalues ⇒ diagonalizable over \(\mathbb{C}\). Compute complex eigenvectors and form \(P\) over \(\mathbb{C}\). If you need real canonical form, use 2×2 rotation/real-Jordan block.
20. Problem: Find diagonal form of \(E=\begin{bmatrix}2&1&0\\0&2&0\\0&0&3\end{bmatrix}\).
    Eigenvalues: \(2\) (mult. 2), and \(3\).
    For the top-left 2×2 block \(\begin{bmatrix}2&1\\0&2\end{bmatrix}\), (A-2I)=\(\begin{bmatrix}0&1\\0&0\end{bmatrix}\) gives one eigenvector for \(\lambda=2\) (span of \([1,0,0]^T\)). So geometric multiplicity for \(\lambda=2\) is 1 (not 2) → matrix is not diagonalizable over \(\mathbb{R}\). Over \(\mathbb{C}\) nothing changes for diagonalizability since the block is Jordan-type — still not diagonalizable.
    
    If it were diagonalizable the diagonal would read \(\operatorname{diag}(2,2,3)\), but since eigenvectors insufficient we must put it in Jordan form instead.