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

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1. Answer to Question 1

   1. Subtract \(\lambda\) from the diagonal and take the determinant: \(\det(A-\lambda I)=(4-\lambda)(3-\lambda)-2\times1=\lambda^2-7\lambda+10\). This factorises as \((\lambda-2)(\lambda-5)=0\), so the eigenvalues are \(\lambda=2\) and \(\lambda=5\).
   2. For \(\lambda=2\): \((A-2I)\mathbf v=\mathbf 0\) gives \(\begin{pmatrix}2&1\\2&1\end{pmatrix}\)\(\mathbf v=\mathbf 0\). Both rows say \(2x+y=0\), so \(y=-2x\) and we may take \(\mathbf v=(1,-2)\). For \(\lambda=5\): \(\begin{pmatrix}-1&1\\2&-2\end{pmatrix}\)\(\mathbf v=\mathbf 0\) gives \(x=y\), so \(\mathbf v=(1,1)\). Any non-zero multiple is also correct.
   3. The trace is \(4+3=7\), which equals \(2+5\). The determinant is \(12-2=10\), which equals \(2\times5\). Both checks confirm the eigenvalues.

2. Answer to Question 2

   1. The matrix is upper triangular, so the eigenvalues are the diagonal entries: 1 and 3.
   2. Expanding along the first row, \(\det(A-\lambda I)=(2-\lambda)\left[(3-\lambda)(9-\lambda)-16\right]=(2-\lambda)(\lambda^2-12\lambda+11)=(2-\lambda)(\lambda-1)(\lambda-11)\). The eigenvalues are 1, 2 and 11.
   3. For a \(2\times2\) matrix the characteristic equation is \(\lambda^2-(\operatorname{tr})\lambda+\det=0\), here \(\lambda^2-7\lambda+10=0\), so the eigenvalues are 2 and 5.

3. Answer to Question 3

   1. \(\det(A-\lambda I)=\lambda^2+1=0\), so \(\lambda=\pm i\). For \(\lambda=i\): \(-ix-y=0\), so \(y=-ix\) and \(\mathbf v=(1,-i)\). A real rotation has no real direction left unchanged, hence the complex eigenvalues.
   2. \((1-\lambda)^2+4=0\) gives \(1-\lambda=\pm2i\), so \(\lambda=1\mp2i\), that is \(1+2i\) and \(1-2i\). For \(\lambda=1+2i\): \(-2ix-2y=0\) so \(y=-ix\) and \(\mathbf v=(1,-i)\). The eigenvalues come in a conjugate pair, with modulus \(\sqrt{1+4}=\sqrt5\): the matrix is a rotation combined with an enlargement by \(\sqrt5\).

4. Answer to Question 4

   1. The characteristic equation is \((2-\lambda)^2-1=0\), so \(2-\lambda=\pm1\) and \(\lambda=1\) or \(3\). For \(\lambda=3\): \(-x+y=0\), so \(x=y\) and \(\mathbf v=(1,1)\). For \(\lambda=1\): \(x+y=0\), so \(\mathbf v=(1,-1)\).
   2. Place the eigenvectors as columns of \(P\) in the same order as the eigenvalues in \(D\): \(P=\) \(\begin{pmatrix}1&1\\1&-1\end{pmatrix}\), \(D=\) \(\begin{pmatrix}3&0\\0&1\end{pmatrix}\). Since \(\det P=-2\), \(P^{-1}=\dfrac12\)\(\begin{pmatrix}1&1\\1&-1\end{pmatrix}\).
   3. Powers of a diagonal matrix are easy: \(A^5=PD^5P^{-1}\) with \(D^5=\) \(\begin{pmatrix}243&0\\0&1\end{pmatrix}\). Multiplying out, \(A^5=\dfrac12\)\(\begin{pmatrix}244&242\\242&244\end{pmatrix}\)\(=\) \(\begin{pmatrix}122&121\\121&122\end{pmatrix}\). Check: the trace \(244=3^5+1^5\).

5. Answer to Question 5

   1. \((2-\lambda)^2=0\), so \(\lambda=2\) with algebraic multiplicity 2.
   2. \((N-2I)\mathbf v=\mathbf 0\): \(\begin{pmatrix}0&1\\0&0\end{pmatrix}\)\(\mathbf v=\mathbf 0\) gives \(y=0\), so all eigenvectors are multiples of \((1,0)\). Only one independent eigenvector: geometric multiplicity 1.
   3. Diagonalisation needs two independent eigenvectors in \(\mathbb R^2\). \(N\) has only one, so it is not diagonalisable.

6. Answer to Question 6

   1. Let \(A\mathbf v=\lambda\mathbf v\) with \(\mathbf v\ne\mathbf 0\). Applying \(A\) again, \(A^2\mathbf v=\lambda A\mathbf v=\lambda^2\mathbf v\). Multiplying \(A\mathbf v=\lambda\mathbf v\) by \(A^{-1}\) and dividing by \(\lambda\ne0\) gives \(A^{-1}\mathbf v=\lambda^{-1}\mathbf v\). Finally \((A+3I)\mathbf v=(\lambda+3)\mathbf v\). So \(A^2\) has eigenvalues 4 and 25; \(A^{-1}\) has \(\frac12\) and \(\frac15\); \(A+3I\) has 5 and 8.
   2. The determinant is the product of the eigenvalues and the trace is their sum, so \(\det(A^2)=4\times25=100\) and \(\operatorname{tr}(A^2)=4+25=29\). Directly, \(A^2=\) \(\begin{pmatrix}18&7\\14&11\end{pmatrix}\), with trace \(29\) and determinant \(198-98=100\). The two methods agree.

7. Answer to Question 7

   1. Trace \(=1.7\) and determinant \(=0.72-0.02=0.70\). The characteristic equation \(\lambda^2-1.7\lambda+0.7=0\) gives \((\lambda-1)(\lambda-0.7)=0\): eigenvalues 1 and 0.7.
   2. For \(\lambda=1\): \(-0.1a+0.2b=0\) gives \(a=2b\), \(\mathbf v=(2,1)\). For \(\lambda=0.7\): \(0.2a+0.2b=0\) gives \(\mathbf v=(1,-1)\).
   3. The eigenvalue 1 eigenvector scaled to total 1 is \(\left(\frac23,\frac13\right)\): in the long run A has 66.7% and B has 33.3%.
   4. Write \((1,0)=\frac13(2,1)+\frac13(1,-1)\). Then \(\mathbf v_n=\frac13(2,1)+\frac13(0.7)^n(1,-1)\). For \(n=5\): \(0.7^5=0.16807\), so \(a_5=\frac23+\frac13(0.16807)=0.7227\): about 72.3%.

8. Answer to Question 8

   1. The characteristic equation is \(\det(L-\lambda I)=-\lambda(0.8-\lambda)-2\times0.5=\lambda^2-0.8\lambda-1=0\). By the quadratic formula, \(\lambda=\dfrac{0.8\pm\sqrt{0.64+4}}2=0.4\pm1.077\), so the eigenvalues are 1.477 and \(-0.677\).
   2. The eigenvalue with the largest modulus dominates, so long-term behaviour is governed by \(1.477\). Each period the population is multiplied by about 1.477, a growth of about 47.7% per period.
   3. For \(\lambda=1.477\), the first row of \((L-\lambda I)\mathbf v=\mathbf 0\) gives \(-1.477J+2A=0\). So in the long run \(A/J=1.477/2\approx\) 0.739: about 74 adults for every 100 juveniles.
   4. Writing the starting vector in terms of the two eigenvectors, the second part is multiplied by \((-0.677)^n\). Since \(|-0.677|\lt 1\), this tends to zero, so only the dominant behaviour remains.

9. Answer to Question 9

   1. \((2-\mu)^2-1=0\) gives \(\mu=1\) or \(\mu=3\).
   2. For \(\mu=1\): \(\mathbf v=(1,1)\) — the carriages move together, so the middle spring is not stretched. For \(\mu=3\): \(\mathbf v=(1,-1)\) — they move in opposite directions.
   3. \(\dfrac km=100\). So \(\omega^2=100\mu\): \(\omega=10\) rad/s for \(\mu=1\) and \(\omega=\sqrt{300}\approx17.3\) rad/s for \(\mu=3\).
   4. The opposite-direction mode is faster because the middle spring is stretched and compressed as well, so the restoring force is larger.

10. Answer to Question 10

    1. The characteristic equation is \((3-\lambda)^2-1=0\), so \(3-\lambda=\pm1\), which gives \(\lambda=2\) or \(4\).
    2. For \(\lambda=4\): \(-x+y=0\), so \(\mathbf v=(1,1)\). For \(\lambda=2\): \(x+y=0\), so \(\mathbf v=(1,-1)\). Vectors along these lines are mapped to multiples of themselves; they are the two diagonals of the picture.
    3. Points on the diagonal \(y=x\) move 4 times further from the centre, and points on \(y=-x\) move 2 times further. Every other direction is both stretched and turned.
    4. The area scale factor is \(|\det S|=9-1=8\), and \(4\times2=8\). Any shape in the photo has its area multiplied by 8.

11. Answer to Question 11

    1. \((-3-\lambda)^2-1=0\) gives \(\lambda=-2\) and \(\lambda=-4\). Eigenvectors: \((1,1)\) for \(-2\), and \((1,-1)\) for \(-4\).
    2. \(\mathbf x(t)=c_1e^{-2t}(1,1)+c_2e^{-4t}(1,-1)\).
    3. At \(t=0\): \(c_1+c_2=4\) and \(c_1-c_2=0\), so \(c_1=c_2=2\). After 1 hour: \(x_1=2e^{-2}+2e^{-4}\approx0.307\) and \(x_2=2e^{-2}-2e^{-4}\approx0.234\).
    4. Both eigenvalues are negative, so \(e^{-2t}\) and \(e^{-4t}\) tend to 0: the deviations vanish and the system is stable.

12. Answer to Question 12

    1. Column A: \(0+\frac12+\frac12\); column B: \(0+0+1\); column C: \(1+0+0\). All equal 1.
    2. The equations are \(r_A=r_C\), \(r_B=\frac12r_A\), \(r_C=\frac12r_A+r_B\). So \(r_B=\frac12r_A\) and \(r_C=r_A\). With the sum equal to 1: \(r_A\left(1+\frac12+1\right)=1\), so \(r_A=0.4\), \(r_B=0.2\), \(r_C=0.4\).
    3. Pages A and C tie with 0.4 each; B has 0.2.
    4. Page C receives all of B's importance as well as half of A's; B receives only half of A's.

13. Answer to Question 13

    1. The characteristic equation is \((1-\lambda)(-\lambda)-1=\lambda^2-\lambda-1=0\). By the quadratic formula, \(\lambda=\dfrac{1\pm\sqrt5}2\), so the eigenvalues are \(\varphi\) and \(\psi\).
    2. The first row of \((Q-\lambda I)\mathbf v=\mathbf 0\) is \((1-\lambda)x+y=0\). Using \(\lambda^2=\lambda+1\), the vector \(\mathbf v=(\lambda,1)\) satisfies both rows, so the eigenvectors are \((\varphi,1)\) and \((\psi,1)\).
    3. Since \(Q=PDP^{-1}\) with \(D=\operatorname{diag}(\varphi,\psi)\), we get \(Q^n=PD^nP^{-1}\). Multiplying out, the off-diagonal entry is \(\dfrac{\varphi^n-\psi^n}{\varphi-\psi}\), and \(\varphi-\psi=\sqrt5\). This entry is \(F_n\), which proves Binet's formula.
    4. \(\varphi^{10}\approx122.99\), and \(122.99/2.236\approx55.0\), matching \(F_{10}=55\). Because \(|\psi|\lt 1\), \(\psi^n\to0\), so \(F_n\) is the nearest integer to \(\varphi^n/\sqrt5\).

14. Answer to Question 14

    1. Trace \(1.55\), determinant \(0.57-0.02=0.55\): \(\lambda^2-1.55\lambda+0.55=0\) gives \((\lambda-1)(\lambda-0.55)=0\).
    2. For \(\lambda=1\): \(-0.05w+0.4b=0\), so \(w=8b\). With \(w+b=1\): \(w=\frac89\approx\) 0.889.
    3. \(\mathbf v_1=(0.95,0.05)\), \(\mathbf v_2=T\mathbf v_1=(0.9025+0.02,\;0.0475+0.03)=(0.9225,0.0775)\).
    4. The second eigenvalue is \(0.55\), so the distance from the steady state shrinks by a factor of \(0.55\) each day: after 5 days it is \(0.55^5\approx5\%\) of its original size.

15. Answer to Question 15

    1. Using the recurrence: \(a_2=5(1)-6(0)=5\), \(a_3=5(5)-6(1)=19\) and \(a_4=5(19)-6(5)=65\). The table is correct.
    2. The top row must give \(a_{n+1}=5a_n-6a_{n-1}\) and the bottom row simply copies \(a_n\), so \(A=\) \(\begin{pmatrix}5&-6\\1&0\end{pmatrix}\).
    3. The characteristic equation is \(\lambda^2-5\lambda+6=(\lambda-2)(\lambda-3)=0\), giving eigenvalues \(2\) and \(3\), so \(a_n=\alpha2^n+\beta3^n\). From \(a_0=\alpha+\beta=0\) and \(a_1=2\alpha+3\beta=1\) we find \(\beta=1\) and \(\alpha=-1\). Hence \(a_n=3^n-2^n\). Check: \(a_4=81-16=65\).
    4. \(a_{10}=59049-1024=58\,025\) million. Since \(a_{n+1}/a_n=\dfrac{3^{n+1}-2^{n+1}}{3^n-2^n}\) and \((2/3)^n\to0\), the ratio tends to \(3\), the dominant eigenvalue.

16. Answer to Question 16

    1. \(\lambda^2-7\lambda+6=0\) gives \(\lambda=1\) and \(\lambda=6\).
    2. For \(\lambda=6\): \(-x+2y=0\), so \((2,1)/\sqrt5\). For \(\lambda=1\): \(4x+2y=0\), so \((1,-2)/\sqrt5\). The dot product is \(2-2=0\), so they are perpendicular (a general property of symmetric matrices).
    3. \(Q=6u^2+v^2\), where \(u\) is measured along \((2,1)/\sqrt5\) and \(v\) along \((1,-2)/\sqrt5\).
    4. \(Q=6\) becomes \(u^2+\dfrac{v^2}{6}=1\) — an ellipse. The semi-axis along the \(u\)-direction is \(1\) and along the \(v\)-direction is \(\sqrt6\approx2.45\).

17. Answer to Question 17

    1. Multiplying, \(A^2=\) \(\begin{pmatrix}5&5\\5&10\end{pmatrix}\); then \(A^3=A^2A=\) \(\begin{pmatrix}15&20\\20&35\end{pmatrix}\). The top-left entries are \(5\) and \(15\), as in the table.
    2. With trace \(5\) and determinant \(6-1=5\), the characteristic equation is \(\lambda^2-5\lambda+5=0\), so \(\lambda=\dfrac{5\pm\sqrt5}2\), about \(3.618\) and \(1.382\).
    3. By Cayley-Hamilton, \(A\) satisfies its own characteristic equation: \(A^2-5A+5I=0\), so \(A^2=5A-5I\). Rearranging, \(A(5I-A)=5I\), hence \(A^{-1}=\dfrac15(5I-A)=\) \(\begin{pmatrix}3/5&-1/5\\-1/5&2/5\end{pmatrix}\).
    4. Diagonalising, \(A^n=\lambda_1^nP_1+\lambda_2^nP_2\) for fixed matrices \(P_1,P_2\). As \(n\) grows, \(\lambda_1^n\) dominates, so \(\dfrac{(A^{n+1})_{11}}{(A^n)_{11}}\to\lambda_1\approx3.618\). The table gives \(50/15\approx3.33\), already close.

18. Answer to Question 18

    1. \((k-\lambda)^2-1=0\) gives \(\lambda=k\pm1\). The eigenvector for \(k+1\) is \((1,1)\) and for \(k-1\) is \((1,-1)\), independent of \(k\).
    2. \(\det=k^2-1\ne0\), so \(A_k\) is invertible exactly when \(k\ne\pm1\) (the eigenvalues are non-zero).
    3. Solutions behave like \(e^{(k+1)t}\) and \(e^{(k-1)t}\), which both tend to zero only if \(k+1<0\), so \(k<-1\).
    4. \(A_0=\) \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\), the reflection in \(y=x\). Its eigenvalues are \(\pm1\): it fixes \((1,1)\) and reverses \((1,-1)\).

19. Answer to Question 19

    1. P1: \(\sqrt{25-24}=1\), so \(\lambda=-2,-3\). P2: \(\sqrt{4-20}=4i\), so \(\lambda=-1\pm2i\). P3: \(\sqrt{1+8}=3\), so \(\lambda=2,-1\). P4: \(\sqrt{16-12}=2\), so \(\lambda=3,1\).
    2. P1 stable node; P2 stable spiral; P3 saddle (\(\Delta<0\)); P4 unstable node.
    3. P1 and P2: all eigenvalues have negative real part.
    4. P2: \(\mathbf x=e^{-t}(\mathbf a\cos2t+\mathbf b\sin2t)\), a decaying oscillation with angular frequency 2 (period \(\pi\approx3.14\)).

20. Answer to Question 20

    1. Applying \(A\): \(\alpha\lambda_1\mathbf v_1+\beta\lambda_2\mathbf v_2=\mathbf 0\).
    2. Multiply the original equation by \(\lambda_2\) and subtract: \(\alpha(\lambda_1-\lambda_2)\mathbf v_1=\mathbf 0\). Since \(\mathbf v_1\ne\mathbf 0\) and \(\lambda_1\ne\lambda_2\), \(\alpha=0\).
    3. Then \(\beta\mathbf v_2=\mathbf 0\) with \(\mathbf v_2\ne\mathbf 0\) gives \(\beta=0\). So the vectors are independent.
    4. \(\det(A^T-\lambda I)=\det\left((A-\lambda I)^T\right)=\det(A-\lambda I)\), so the characteristic polynomials agree. For \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\) the characteristic polynomial is \(\lambda^2-(a+d)\lambda+(ad-bc)\), so the sum of its roots is \(a+d=\operatorname{tr}A\).