Eigenvalues and Orthogonality: practice solutions, College – download the PDF

1 Is it an eigenvector? ★★★

(a) \(A\mathbf{v}=(1+2,\,3+0)=(3,3)=3\mathbf{v}\). Yes, it is an eigenvector with eigenvalue \(3\).

(b) \(A\mathbf{w}=(1+4,\,3)=(5,3)\). This is not a multiple of \((1,2)\), because \(5/1\neq 3/2\). So \(\mathbf{w}\) is not an eigenvector.

2 Dot product and length ★★★

(a) \(\mathbf{u}\cdot\mathbf{w}=2\cdot4+(-1)\cdot5+3\cdot(-1)=8-5-3=0\). The vectors are orthogonal.

(b) \(\|\mathbf{u}\|=\sqrt{4+1+9}=\sqrt{14}\approx 3.74\).

3 A first characteristic polynomial ★★★

The trace is \(9\) and the determinant is \(20-2=18\). So \(\det(A-\lambda I)=\lambda^2-9\lambda+18=(\lambda-3)(\lambda-6)\). The eigenvalues are \(3\) and \(6\). Check: \(3+6=9\) and \(3\cdot 6=18\).

4 Triangular matrix ★★★

\(T\) is upper triangular, so its eigenvalues are the diagonal entries: \(2,\,-3,\,4\). Check with the trace: \(2+(-3)+4=3\), which equals the sum of the diagonal entries \(2-3+4=3\).

5 Angle between two vectors ★★★

\(\mathbf{u}\cdot\mathbf{w}=0+0+1=1\) and \(\|\mathbf{u}\|=\|\mathbf{w}\|=\sqrt2\). So \(\cos\theta=\dfrac{1}{\sqrt2\cdot\sqrt2}=\dfrac12\), and \(\theta=60^\circ\).

6 A quick projection ★★★

\(\mathbf{b}\cdot\mathbf{a}=12+2=14\) and \(\mathbf{a}\cdot\mathbf{a}=5\). So \(\operatorname{proj}_{\mathbf{a}}\mathbf{b}=\tfrac{14}{5}(2,1)=(5.6,\,2.8)\).

7 True or false: powers and shifts ★★★

\(A^2\mathbf{v}=A(4\mathbf{v})=4A\mathbf{v}=16\mathbf{v}\). Then \(A^3\mathbf{v}=4\cdot16\mathbf{v}=64\mathbf{v}\). And \((A+3I)\mathbf{v}=4\mathbf{v}+3\mathbf{v}=7\mathbf{v}\).

True: if \(A\mathbf{v}=0\mathbf{v}=\mathbf{0}\) with \(\mathbf{v}\neq\mathbf{0}\), then \(A\) sends a nonzero vector to zero, so it cannot be invertible (equivalently, \(\det A=0\)).

8 Finding eigenvectors ★★★

The trace is \(7\) and the determinant is \(12-2=10\). So \(\lambda^2-7\lambda+10=(\lambda-5)(\lambda-2)\).

For \(\lambda=5\): \(A-5I=\begin{pmatrix}-1&1\\2&-2\end{pmatrix}\) gives \(y=x\), so \((1,1)\).

For \(\lambda=2\): \(A-2I=\begin{pmatrix}2&1\\2&1\end{pmatrix}\) gives \(y=-2x\), so \((1,-2)\).

Check: \(A(1,-2)=(4-2,\,2-6)=(2,-4)=2(1,-2)\).

9 Diagonalize and take a power ★★★

(a) Trace \(7\), determinant \(6\): \(\lambda^2-7\lambda+6=(\lambda-6)(\lambda-1)\). For \(\lambda=6\): \(A-6I=\begin{pmatrix}-1&2\\2&-4\end{pmatrix}\) gives \((2,1)\). For \(\lambda=1\): \(A-I=\begin{pmatrix}4&2\\2&1\end{pmatrix}\) gives \((1,-2)\).

(b) \(P=\begin{pmatrix}2&1\\1&-2\end{pmatrix}\), \(D=\begin{pmatrix}6&0\\0&1\end{pmatrix}\).

(c) \(A^3=PD^3P^{-1}\), so its eigenvalues are \(6^3=216\) and \(1^3=1\). Directly, \(A^2=\begin{pmatrix}29&14\\14&8\end{pmatrix}\) and \(A^3=A^2A=\begin{pmatrix}173&86\\86&44\end{pmatrix}\). Check: trace \(217=216+1\) and determinant \(7612-7396=216\).

10 A 3 by 3 characteristic polynomial ★★★

Expanding along the third row, \(\det(M-\lambda I)=(4-\lambda)\big[(1-\lambda)^2-4\big]=(4-\lambda)(\lambda-3)(\lambda+1)\). The eigenvalues are \(4,\,3,\,-1\).

\(\lambda=3\): \((1,1,0)\). \(\lambda=-1\): \((1,-1,0)\). \(\lambda=4\): \((0,0,1)\).

Check with the trace: \(4+3-1=6=1+1+4\).

11 A rotation has no real eigenvalues ★★★

\(\det(R-\lambda I)=\lambda^2+1\). This equals \(0\) only for \(\lambda=\pm i\), which are not real. So there is no real eigenvalue.

Geometrically, rotating by \(90^\circ\) turns every nonzero vector off its line, so no direction is sent to a multiple of itself.

12 Gram-Schmidt in the plane ★★★

\(\mathbf{u}_1=(1,2)\). \(\mathbf{v}_2\cdot\mathbf{u}_1=6\), \(\mathbf{u}_1\cdot\mathbf{u}_1=5\), so \(\mathbf{u}_2=(4,1)-\tfrac65(1,2)=\left(\tfrac{14}{5},-\tfrac75\right)=\tfrac75(2,-1)\).

Check: \((1,2)\cdot(2,-1)=0\). Both \((1,2)\) and \((2,-1)\) have length \(\sqrt5\), so an orthonormal basis is \(\tfrac{1}{\sqrt5}(1,2)\) and \(\tfrac{1}{\sqrt5}(2,-1)\).

13 Distance from a point to a line ★★★

\(\mathbf{b}=(1,7)\), \(\mathbf{a}=(1,2)\): \(\mathbf{b}\cdot\mathbf{a}=15\), \(\mathbf{a}\cdot\mathbf{a}=5\). The closest point is \(3\mathbf{a}=(3,6)\).

The error is \((1,7)-(3,6)=(-2,1)\); check \((-2,1)\cdot(1,2)=0\). The distance is \(\sqrt{4+1}=\sqrt5\approx 2.24\), that is about \(224\) feet (about \(68\) meters).

14 A three-point fit ★★★

\(A=\begin{pmatrix}1&-1\\1&0\\1&1\end{pmatrix}\), \(\mathbf{b}=(1,2,6)\). Then \(A^TA=\begin{pmatrix}3&0\\0&2\end{pmatrix}\) and \(A^T\mathbf{b}=\begin{pmatrix}9\\5\end{pmatrix}\).

So \(3a=9\) and \(2b=5\): \(a=3\), \(b=2.5\). The line is \(y=3+2.5x\).

15 Diagonalizable or not? ★★★

(a) The characteristic polynomial is \((\lambda-3)^2\), so \(3\) is the only eigenvalue. \(A-3I=\begin{pmatrix}0&1\\0&0\end{pmatrix}\) forces \(y=0\), so all eigenvectors are multiples of \((1,0)\). There is only one independent eigenvector, so \(A\) is not diagonalizable.

(b) \(B\) is already diagonal (\(B=3I\)), so it is diagonalizable: every nonzero vector is an eigenvector. Repeated eigenvalues alone do not decide the question; the number of independent eigenvectors does.

16 Finding a missing entry ★★★

\(5\) is an eigenvalue exactly when \(\det(A-5I)=0\): \((1-5)(4-5)-2k=4-2k=0\), so \(k=2\).

The trace is \(5\), so the other eigenvalue is \(5-5=0\). Check: \(\det A=4-4=0\) and \(5\cdot0=0\).

17 Orthogonal diagonalization of a symmetric matrix ★★★

(a) \(S(1,1,1)=(4,4,4)=4(1,1,1)\), eigenvalue \(4\).

(b) \(S(1,-1,0)=(2-1,\,1-2,\,0)=(1,-1,0)\). \(S(1,1,-2)=(2+1-2,\,1+2-2,\,1+1-4)=(1,1,-2)\). Both have eigenvalue \(1\). Dot products: \((1,1,1)\cdot(1,-1,0)=0\), \((1,1,1)\cdot(1,1,-2)=0\), \((1,-1,0)\cdot(1,1,-2)=0\).

(c) The eigenvalues of \(S\) are \(4,1,1\), so those of \(S^5\) are \(4^5,1,1\) and the trace is \(1024+1+1=1026\).

18 Gram-Schmidt in three dimensions ★★★

\(\mathbf{u}_1=(1,1,0)\). Since \(\mathbf{v}_2\cdot\mathbf{u}_1=1\) and \(\mathbf{u}_1\cdot\mathbf{u}_1=2\), \(\mathbf{u}_2=(1,0,1)-\tfrac12(1,1,0)=\left(\tfrac12,-\tfrac12,1\right)\).

For \(\mathbf{v}_3\): \(\mathbf{v}_3\cdot\mathbf{u}_1=1\) and \(\mathbf{v}_3\cdot\mathbf{u}_2=\tfrac12\) with \(\mathbf{u}_2\cdot\mathbf{u}_2=\tfrac32\). So
\[\mathbf{u}_3=(0,1,1)-\tfrac12(1,1,0)-\tfrac13\left(\tfrac12,-\tfrac12,1\right)=\left(-\tfrac23,\tfrac23,\tfrac23\right).\]

Check: \(\mathbf{u}_3\cdot\mathbf{u}_1=0\) and \(\mathbf{u}_3\cdot\mathbf{u}_2=-\tfrac13-\tfrac13+\tfrac23=0\).

19 Least squares for plant growth ★★★

(a) \(n=4\), \(\sum x=6\), \(\sum y=9\), \(\sum x^2=14\), \(\sum xy=0+2+4+12=18\). The normal equations \(4a+6b=9\), \(6a+14b=18\) give \(b=\dfrac{4\cdot18-6\cdot9}{4\cdot14-36}=\dfrac{18}{20}=0.9\) and \(a=\dfrac{9-5.4}{4}=0.9\). So \(y=0.9+0.9x\).

(b) The predicted values are \(0.9,\,1.8,\,2.7,\,3.6\), so the residuals are \(0.1,\,0.2,\,-0.7,\,0.4\) and the sum of squares is \(0.01+0.04+0.49+0.16=0.70\).

(c) At \(x=5\): \(0.9+4.5=5.4\) inches (about \(13.7\) cm).

20 Eigenvectors of symmetric matrices ★★★

Since \(A\) is symmetric, \((A\mathbf{u})\cdot\mathbf{w}=\mathbf{u}\cdot(A\mathbf{w})\). The left side is \(\lambda(\mathbf{u}\cdot\mathbf{w})\) and the right side is \(\mu(\mathbf{u}\cdot\mathbf{w})\). Hence \((\lambda-\mu)(\mathbf{u}\cdot\mathbf{w})=0\). Because \(\lambda\neq\mu\), we must have \(\mathbf{u}\cdot\mathbf{w}=0\).

21 A projection matrix ★★★

(a) \(\mathbf{a}\cdot\mathbf{a}=9\), so \(P=\dfrac19\begin{pmatrix}1&2&2\\2&4&4\\2&4&4\end{pmatrix}\). Since \(\mathbf{b}\cdot\mathbf{a}=3+0+6=9\), \(P\mathbf{b}=\tfrac99\mathbf{a}=(1,2,2)\).

(b) \(P\mathbf{a}=\mathbf{a}\), so \(1\) is an eigenvalue. Every vector orthogonal to \(\mathbf{a}\) is sent to \(\mathbf{0}\), a plane of eigenvectors for \(0\). The eigenvalues are \(1,0,0\); the trace \(\tfrac{1+4+4}{9}=1\) agrees.

22 Two apps, long-term behavior ★★★

(a) Trace \(1.7\), determinant \(0.72-0.02=0.70\): \(\lambda^2-1.7\lambda+0.7=(\lambda-1)(\lambda-0.7)\). For \(\lambda=1\): \((M-I)\mathbf{v}=\mathbf{0}\) gives \(\mathbf{v}_1=(2,1)\). For \(\lambda=0.7\): \(M-0.7I=\begin{pmatrix}0.2&0.2\\0.1&0.1\end{pmatrix}\) gives \(\mathbf{v}_2=(1,-1)\).

(b) Solving \(c_1(2,1)+c_2(1,-1)=(150,150)\): \(2c_1+c_2=150\) and \(c_1-c_2=150\), so \(c_1=100\), \(c_2=-50\). Then \(\mathbf{x}_n=100(2,1)-50(0.7)^n(1,-1)\). Check \(n=1\): \((200-35,\,100+35)=(165,135)\), the same as \(M\mathbf{x}_0\).

(c) Since \((0.7)^n\to0\), \(\mathbf{x}_n\to(200,100)\): about \(200{,}000\) users on A and \(100{,}000\) on B, whatever the split was at the start, with total \(300{,}000\) unchanged.