Eigenvalues and eigenvectors – university Year 1 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   The characteristic equation is \(\det(A-\lambda I)=(1-\lambda)^2-4=0\) (1). So \(1-\lambda=\pm2\), which gives \(\lambda=3\) or \(\lambda=-1\) (1). For \(\lambda=3\): the system \(-2x+2y=0\) gives \(x=y\), so \(\mathbf v=(1,1)\) (1). For \(\lambda=-1\): \(2x+2y=0\) gives \(y=-x\), so \(\mathbf v=(1,-1)\) (1). Check: trace \(2=3+(-1)\) and determinant \(-3=3\times(-1)\).

   Eigenvectors are only determined up to a non-zero multiple, so \((2,2)\) or \((-1,1)\) also score full marks. The matrix is symmetric, and its two eigenvectors are perpendicular: \((1,1)\cdot(1,-1)=0\).

2. Answer to Question 2 (4 marks)

   The matrix is upper triangular, so its eigenvalues are the diagonal entries \(3\) and \(2\) (1). For \(\lambda=3\): \((A-3I)\mathbf v=\mathbf 0\) gives \(y=0\), so \(\mathbf v=(1,0)\); for \(\lambda=2\): \(x+y=0\), so \(\mathbf v=(1,-1)\) (1). Hence \(P=\) \(\begin{pmatrix}1&1\\0&-1\end{pmatrix}\), \(D=\operatorname{diag}(3,2)\), and here \(P^{-1}=P\) (1). Therefore \(A^n=PD^nP^{-1}=\) \(\begin{pmatrix}3^n&3^n-2^n\\0&2^n\end{pmatrix}\) (1). For \(n=1\) this gives \(A\) back, as a check.

   Check \(P^{-1}=P\): \(P^2=I\). The diagonalisation works because the eigenvalues are distinct.

3. Answer to Question 3 (4 marks)

   1. The characteristic equation is \((-\lambda)^2-(-4)(1)=\lambda^2+4=0\), so \(\lambda^2=-4\) and \(\lambda=\pm2i\) (2). A real matrix can have complex eigenvalues; they come in conjugate pairs and here describe a rotation combined with a stretch.
   2. The matrix is triangular, so \(\lambda=1\) is a repeated eigenvalue (1). Solving \((A-I)\mathbf v=\mathbf 0\), the only condition is \(y=0\), so every eigenvector is a multiple of \((1,0)\) (1). There is only one independent eigenvector, so the matrix cannot be diagonalised.

4. Answer to Question 4 (4 marks)

   1. The trace is \(0.8+0.9=1.7\) and the determinant is \(0.72-0.02=0.70\) (1). The eigenvalues satisfy \(\lambda^2-1.7\lambda+0.7=(\lambda-1)(\lambda-0.7)=0\), so they are \(1\) and \(0.7\) (1). A transition matrix always has eigenvalue \(1\).
   2. The long-term state is the eigenvector for \(\lambda=1\): \(-0.2c+0.1s=0\), so \(s=2c\) (1). Since \(c+s=1\), \(3c=1\) and the cinema share tends to \(\frac13\), about 33% (1). The other part decays like \(0.7^n\).

   Check: \((\tfrac13,\tfrac23)\) is unchanged by the matrix.

5. Answer to Question 5 (4 marks)

   For a \(2\times2\) matrix the characteristic equation is \(\lambda^2-(\operatorname{tr}A)\lambda+\det A=0\). Here \(\operatorname{tr}A=1\) and \(\det A=0-6=-6\), so \(\lambda^2-\lambda-6=0\) (1). By Cayley-Hamilton, \(A^2-A-6I=0\), that is \(A^2=A+6I\) (1). Factorising, \(A(A-I)=6I\) (1), so \(A^{-1}=\dfrac16(A-I)=\dfrac16\)\(\begin{pmatrix}0&2\\3&-1\end{pmatrix}\) (1). Check: multiplying by \(A\) gives \(I\).

   Check explicitly: \(A^2=\begin{pmatrix}7&2\\3&6\end{pmatrix}\), and \(A+6I=\begin{pmatrix}7&2\\3&6\end{pmatrix}\). The usual formula for a \(2\times2\) inverse, \(\tfrac1{-6}\begin{pmatrix}0&-2\\-3&1\end{pmatrix}\), gives the same matrix. Since \(\det A\ne0\), the inverse exists, consistent with 0 not being an eigenvalue.

Total : 20 marks