Eigenvalues and Orthogonality: math test solutions, College – download the PDF

1 Eigenvalues and eigenvectors / 4 pts

(a) Trace \(15\), determinant \(56-6=50\): \(\lambda^2-15\lambda+50\). (1.5 pts)

(b) \(\lambda^2-15\lambda+50=(\lambda-5)(\lambda-10)\), so the eigenvalues are \(5\) and \(10\). (1 pt)

(c) \(\lambda=5\): \(A-5I=\begin{pmatrix}2&2\\3&3\end{pmatrix}\) gives \((1,-1)\). \(\lambda=10\): \(A-10I=\begin{pmatrix}-3&2\\3&-2\end{pmatrix}\) gives \((2,3)\). (1.5 pts)

2 Diagonalization / 3 pts

(a) \(\det(M-\lambda I)=(3-\lambda)\big[(1-\lambda)^2-4\big]=(3-\lambda)(\lambda-3)(\lambda+1)\). The eigenvalues are \(3\) (twice) and \(-1\). Check: trace \(5=3+3-1\). (2 pts)

(b) Yes. The vectors \((1,0,0)\) and \((0,1,1)\) are independent eigenvectors for \(3\), and \((0,1,-1)\) is an eigenvector for \(-1\). That gives three independent eigenvectors. (1 pt)

3 Dot product and angle / 4 pts

(a) \(\|\mathbf{u}\|=\sqrt{9+16+144}=13\). (1 pt)

(b) \(\mathbf{u}\cdot\mathbf{w}=6-4+60=62\). \(\|\mathbf{w}\|=\sqrt{30}\). \(\cos\theta=\dfrac{62}{13\sqrt{30}}\approx0.8707\), so \(\theta\approx29.5^\circ\). (2 pts)

(c) \(3t+12+24=0\), so \(t=-12\). (1 pt)

4 Orthogonal projection / 3 pts

(a) \(\mathbf{b}\cdot\mathbf{a}=13\), \(\mathbf{a}\cdot\mathbf{a}=5\), so \(\mathbf{p}=\tfrac{13}{5}(2,-1)=(5.2,\,-2.6)\). (1.5 pts)

(b) \(\mathbf{b}-\mathbf{p}=(1.8,\,3.6)\) and \((1.8)(2)+(3.6)(-1)=0\). (0.5 pt) The distance is \(\sqrt{3.24+12.96}=\sqrt{16.2}\approx4.02\). (1 pt)

5 Gram-Schmidt / 3 pts

\(\mathbf{u}_1=(2,1,2)\), \(\mathbf{u}_1\cdot\mathbf{u}_1=9\) and \(\mathbf{v}_2\cdot\mathbf{u}_1=9\). (1 pt)

\(\mathbf{u}_2=(3,3,0)-1\cdot(2,1,2)=(1,2,-2)\). Check: \(2+2-4=0\). (1 pt)

Both vectors have length \(3\), so the orthonormal pair is \(\left(\tfrac23,\tfrac13,\tfrac23\right)\) and \(\left(\tfrac13,\tfrac23,-\tfrac23\right)\). (1 pt)

6 Least squares / 3 pts

\(n=4\), \(\sum x=6\), \(\sum y=24\), \(\sum x^2=14\), \(\sum xy=0+5+12+30=47\). (1 pt)

\(b=\dfrac{4\cdot47-6\cdot24}{4\cdot14-36}=\dfrac{44}{20}=2.2\) and \(a=\dfrac{24-13.2}{4}=2.7\). The line is \(y=2.7+2.2x\). (1.5 pts)

For day \(4\): \(2.7+8.8=11.5\), about \(115\) bikes. (0.5 pt)