Vector spaces and linear maps – university Year 1 maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(\mathbf 0=(0,0,0)\) satisfies the equation. If \((x,y,z)\) and \((x',y',z')\) are in \(W_1\) then the sum has coordinate total \((x+y+z)+(x'+y'+z')=0\); and \(\lambda(x,y,z)\) has total \(\lambda\cdot0=0\). So \(W_1\) is a subspace (a plane through the origin).
   2. \((1,0)\) and \((0,1)\) are in \(W_2\) (since \(1\cdot0=0\) and \(0\cdot1=0\)), but their sum \((1,1)\) has \(xy=1\ne0\). Closure under addition fails: not a subspace.
   3. \((0,0,0)\) gives total \(0\ne1\), so the zero vector is missing: not a subspace (it is a plane not through the origin).

2. Answer to Question 2

   1. Solve \(a(1,1)+b(1,2)=(3,5)\): \(a+b=3\), \(a+2b=5\). Subtracting, \(b=2\) and \(a=1\). So yes: \((3,5)=1(1,1)+2(1,2)\).
   2. The determinant of the matrix with these rows is \(1(0-24)-2(0-20)+3(0-5)=1\ne0\), so they are independent.
   3. Try \((7,8,9)=a(1,2,3)+b(4,5,6)\): \(a+4b=7\), \(2a+5b=8\) give \(b=2\), \(a=-1\), and the third coordinate \(-3+12=9\) works. So \((1,2,3)-2(4,5,6)+(7,8,9)=\mathbf 0\): dependent.

3. Answer to Question 3

   1. The determinant of the matrix with rows \(\mathbf b_i\) is \(1(1\cdot1-0)-1(0\cdot1-0)+0=\,\)... computing: \(\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}=1(1-0)-1(0-1)+0=2\ne0\). Three independent vectors in \(\mathbb R^3\) form a basis.
   2. Solve \(a\mathbf b_1+b\mathbf b_2+c\mathbf b_3=(2,3,4)\): \(a+c=2\), \(a+b=3\), \(b+c=4\). Adding: \(2(a+b+c)=9\), so \(a+b+c=4.5\). Then \(b=4.5-2=2.5\), \(c=4.5-3=1.5\), \(a=4.5-4=0.5\). Coordinates: \((0.5,\,2.5,\,1.5)\).
   3. The dimension is the number of vectors in a basis: 3.

4. Answer to Question 4

   1. Basis \(1,x,x^2,x^3\): dimension 4.
   2. One basis matrix for each of the 6 positions (a 1 in that position, 0 elsewhere): dimension 6.
   3. \((x,y,z)=(-y-z,y,z)=y(-1,1,0)+z(-1,0,1)\), so the basis \(\{(-1,1,0),(-1,0,1)\}\) gives dimension 2.
   4. A symmetric matrix is \(\begin{pmatrix}a&b\\b&d\end{pmatrix}\)\(=a\)\(\begin{pmatrix}1&0\\0&0\end{pmatrix}\)\(+b\)\(\begin{pmatrix}0&1\\1&0\end{pmatrix}\)\(+d\)\(\begin{pmatrix}0&0\\0&1\end{pmatrix}\), so the dimension is 3.

5. Answer to Question 5

   1. \(T(\lambda\mathbf u+\mu\mathbf v)\) has components \(\lambda(u_1+u_2)+\mu(v_1+v_2)\) and \(\lambda(2u_1-u_2)+\mu(2v_1-v_2)\), which equals \(\lambda T(\mathbf u)+\mu T(\mathbf v)\). The matrix columns are \(T(1,0)=(1,2)\) and \(T(0,1)=(1,-1)\), so the matrix is \(\begin{pmatrix}1&1\\2&-1\end{pmatrix}\).
   2. \(\det=-1-2=-3\ne0\), so \(T\) is invertible.
   3. \(S(0,0)=(1,0)\ne(0,0)\). A linear map must send \(\mathbf 0\) to \(\mathbf 0\), so \(S\) is not linear (it is a translation).

6. Answer to Question 6

   1. Solve \(A\mathbf x=\mathbf 0\). Row 2 is twice row 1, so we need \(x+2y+3z=0\) and \(x+y+z=0\). Subtracting: \(y+2z=0\), so \(y=-2z\) and \(x=z\). The kernel is spanned by \((1,-2,1)\) (nullity 1).
   2. The image is the span of the columns \((1,2,1)\), \((2,4,1)\) and \((3,6,1)\). The third column is \((3,6,1)=2(2,4,1)-(1,2,1)\), so a basis is \(\{(1,2,1),(2,4,1)\}\) (rank 2).
   3. \(\operatorname{rank}+\operatorname{nullity}=2+1=3=\dim\mathbb R^3\).

7. Answer to Question 7

   1. The determinant of the matrix with columns \(\mathbf p_i\) is \(2(1-0)-0+1(1-0)=3\ne0\), so they are independent.
   2. Solve \(2a+c=5\), \(a+b=4\), \(b+c=3\). From the second, \(b=4-a\); from the third \(c=3-b=a-1\); first: \(2a+a-1=5\) gives \(a=2\), so \(b=2\), \(c=1\). The recipe is 2 units of \(\mathbf p_1\), 2 of \(\mathbf p_2\), 1 of \(\mathbf p_3\).
   3. Three independent vectors in \(\mathbb R^3\) form a basis, so every vector has exactly one coordinate triple.
   4. For instance \((0,0,1)\): solve \(2a+c=0\), \(a+b=0\), \(b+c=1\): \(c=-2a\), \(b=-a\), so \(-a-2a=1\) and \(a=-\tfrac13\), which is negative. Such colours are outside the range the three inks can produce.

8. Answer to Question 8

   1. The determinant is \(\begin{vmatrix}1&2&0\\0&1&2\\1&0&1\end{vmatrix}=1(1-0)-2(0-2)+0=5\ne0\), so the vectors are independent and form a basis.
   2. Solve \(a+c=3\), \(2a+b=7\), \(2b+c=7\). From the first \(c=3-a\); second \(b=7-2a\); third: \(14-4a+3-a=7\), so \(a=2\), \(b=3\), \(c=1\). Servings: 2, 3 and 1.
   3. Two independent vectors span a plane through the origin in \(\mathbb R^3\).
   4. Only if \((3,7,7)\) lies in that plane. By (b) the coordinates are unique, and the coefficient of \(\mathbf f_3\) is \(1\ne0\), so the answer is no.

9. Answer to Question 9

   1. The image (column space) has dimension equal to the rank: 3.
   2. \(\operatorname{rank}+\operatorname{nullity}=6\) (the number of columns), so the nullity is \(6-3=\) 3.
   3. Three independent changes of the readings leave all four summaries unchanged, so they cannot be detected from the summaries.
   4. No. The image has dimension 3, but \(\mathbb R^4\) has dimension 4, so the summaries always lie in a three-dimensional subspace.

10. Answer to Question 10

    1. \(D(1)=0\), \(D(t)=1\), \(D(t^2)=2t\). In the basis \(\{1,t,t^2\}\) the columns are \((0,0,0)\), \((1,0,0)\), \((0,2,0)\): \(\begin{pmatrix}0&1&0\\0&0&2\\0&0&0\end{pmatrix}\).
    2. The kernel is the constant polynomials (dimension 1). The image is \(\{a+bt\}\), spanned by \(1\) and \(t\) (dimension 2). Check: \(1+2=3=\dim P_2\).
    3. The coordinate vector of \(h\) is \((2,20,-4.9)\). Then \(\begin{pmatrix}0&1&0\\0&0&2\\0&0&0\end{pmatrix}\)\(\begin{pmatrix}2\\20\\-4.9\end{pmatrix}\)\(=\) \(\begin{pmatrix}20\\-9.8\\0\end{pmatrix}\), that is \(D(h)=20-9.8t\).
    4. \(D(h)\) is the velocity of the ball. \(D^2h=-9.8\) is the (constant) acceleration due to gravity.

11. Answer to Question 11

    1. \(\begin{pmatrix}1&0.5\\0&1\end{pmatrix}\), since \(S(1,0)=(1,0)\) and \(S(0,1)=(0.5,1)\).
    2. \((0,0)\mapsto(0,0)\); \((1,0)\mapsto(1,0)\); \((0,2)\mapsto(1,2)\); \((1,2)\mapsto(2,2)\). The rectangle becomes a parallelogram with the top edge shifted one unit to the right.
    3. \(\det S=1\): the area of every shape is unchanged by the shear, so letters keep the same ink area.
    4. \(S^{-1}=\) \(\begin{pmatrix}1&-0.5\\0&1\end{pmatrix}\), which shears back by the same amount in the opposite direction.

12. Answer to Question 12

    1. The determinant of the matrix with rows \((1,1)\), \((1,-1)\) is \(-2\ne0\), so the vectors are independent; two independent vectors in \(\mathbb R^2\) form a basis.
    2. Solve \(a+b=3\), \(a-b=1\): \(a=2\), \(b=1\). Coordinates \((2,1)\).
    3. \(a+b=2\), \(a-b=6\): \(a=4\), \(b=-2\). Coordinates \((4,-2)\).
    4. The reflection fixes \(\mathbf d_1=(1,1)\) and sends \(\mathbf d_2=(1,-1)\) to \((-1,1)=-\mathbf d_2\). In the basis \(\mathcal B\) its matrix is \(\begin{pmatrix}1&0\\0&-1\end{pmatrix}\).

13. Answer to Question 13

    1. \(R(1,0)=(0,1)\) and \(R(0,1)=(-1,0)\), so the matrix is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\).
    2. No non-zero vector is sent to zero, so the kernel is \(\{\mathbf 0\}\). By rank-nullity the rank is 2, so the image is all of \(\mathbb R^2\). Hence \(R\) is injective and surjective: an isomorphism.
    3. After 1 minute: \((-1,3)\). After 2: \((-3,-1)\). After 3: \((1,-3)\). After 4: \((3,1)\).
    4. Four quarter turns make one full turn, which returns every vector to its starting position. In matrices, \(R^2=-I\), so \(R^4=(-I)^2=I\).

14. Answer to Question 14

    1. Carbon: \(3a=c\). Hydrogen: \(8a=2d\). Oxygen: \(2b=2c+d\).
    2. \(A=\) \(\begin{pmatrix}3&0&-1&0\\8&0&0&-2\\0&2&-2&-1\end{pmatrix}\). From the first two equations \(c=3a\) and \(d=4a\). The third gives \(2b=6a+4a=10a\), so \(b=5a\). The kernel is spanned by \((1,5,3,4)\).
    3. The kernel has dimension 1. There are four unknowns and three independent equations, so \(4-3=1\) by rank-nullity; the single free parameter is the overall scale of the reaction.
    4. Taking \(a=1\): \(\mathrm{C_3H_8}+5\,\mathrm{O_2}\to3\,\mathrm{CO_2}+4\,\mathrm{H_2O}\).

15. Answer to Question 15

    1. \(U\): \(\{(-1,1,0),(-1,0,1)\}\) (dimension 2). \(W\): \((x,x,z)=x(1,1,0)+z(0,0,1)\), basis \(\{(1,1,0),(0,0,1)\}\) (dimension 2).
    2. Points with \(x=y\) and \(x+y+z=0\) have \(z=-2x\): \((x,x,-2x)=x(1,1,-2)\). So \(U\cap W\) is the line spanned by \((1,1,-2)\), dimension 1.
    3. \(\dim(U+W)=2+2-1=\) 3, so \(U+W=\mathbb R^3\).
    4. \((1,-1,0)\in U\) and \((0,0,1)\in W\), but their sum \((1,-1,1)\) has \(x+y+z=1\ne0\) and \(x\ne y\), so it is in neither: \(U\cup W\) is not closed under addition.

16. Answer to Question 16

    1. \(T(\lambda p+\mu q)\) has entries \(\lambda p(t)+\mu q(t)\), so \(T\) is linear. The images are \(T(1)=(1,1,1)\), \(T(t)=(0,1,2)\), \(T(t^2)=(0,1,4)\), so the matrix is \(\begin{pmatrix}1&0&0\\1&1&1\\1&2&4\end{pmatrix}\).
    2. \(\det=1(4-2)=2\ne0\), so \(T\) is invertible: for any three values there is exactly one polynomial of degree at most 2.
    3. Solve \(a=1\), \(a+b+c=3\), \(a+2b+4c=7\): \(b+c=2\), \(2b+4c=6\) so \(c=1\), \(b=1\). Hence \(p(t)=1+t+t^2\).
    4. \(p(3)=1+3+9=13\). Three data points fit a quadratic exactly, but a perfect fit does not mean the prediction is reliable: seasonal or random effects are ignored.

17. Answer to Question 17

    1. Row 1 + row 2 \(=(1,3,1,2)=\) row 3. The first two rows are independent, so the rank is 2.
    2. Solve \(x_1+2x_2+x_4=0\) and \(x_2+x_3+x_4=0\). Put \(x_3=s\), \(x_4=t\): \(x_2=-s-t\) and \(x_1=-2x_2-x_4=2s+2t-t=2s+t\). So \((x_1,\dots,x_4)=s(2,-1,1,0)+t(1,-1,0,1)\). A basis: \(\{(2,-1,1,0),(1,-1,0,1)\}\).
    3. The columns are \((1,0,1)\), \((2,1,3)\), \((0,1,1)\), \((1,1,2)\). The first two are independent and the others are combinations of them, so a basis of the image is \(\{(1,0,1),(2,1,3)\}\).
    4. Rank \(2\) + nullity \(2\) \(=4\), the number of columns. The kernel consists of signal patterns that the monitors cannot detect.

18. Answer to Question 18

    1. \(t=0\): \(a+b=0\). \(t=\frac\pi2\): \(a+c=0\). \(t=\pi\): \(a-b=0\). From the first and third, \(a=b=0\) (adding gives \(2a=0\)), then \(c=0\).
    2. The only linear combination that equals the zero function is the trivial one, so the three functions are linearly independent.
    3. \(g=1\cdot f_2+1\cdot f_3\), so the coordinates with respect to \((f_2,f_3)\) are \((1,1)\).
    4. \(\{\cos t,\sin t\}\) is a basis: both solve \(y''+y=0\), and they are independent. The constant function 1 does not solve it.

19. Answer to Question 19

    1. \(T(1,1)=(4,4)=4(1,1)\) and \(T(1,-1)=(2,-2)=2(1,-1)\).
    2. The matrix is diagonal: \(\begin{pmatrix}4&0\\0&2\end{pmatrix}\).
    3. \((2,0)=1(1,1)+1(1,-1)\). Applying \(T^5\) multiplies the first component by \(4^5=1024\) and the second by \(2^5=32\): \(T^5(2,0)=1024(1,1)+32(1,-1)=(1056,992)\).
    4. The direction \((1,1)\) (equal amounts in each tank) is multiplied by 4 each step, while the direction \((1,-1)\) (difference between the tanks) is multiplied by 2. The total grows faster than the imbalance.

20. Answer to Question 20

    1. \((\alpha+\gamma)\mathbf u+(\alpha+\beta)\mathbf v+(\beta+\gamma)\mathbf w=\mathbf 0\).
    2. By independence all three coefficients vanish: \(\alpha+\gamma=0\), \(\alpha+\beta=0\), \(\beta+\gamma=0\).
    3. Adding all three: \(2(\alpha+\beta+\gamma)=0\), so \(\alpha+\beta+\gamma=0\). Subtracting each equation: \(\beta=0\), \(\gamma=0\), \(\alpha=0\). Hence only the trivial combination works.
    4. The matrix is \(\begin{pmatrix}1&0&1\\1&1&0\\0&1&1\end{pmatrix}\) with determinant \(2\). A non-zero determinant means the new vectors are independent. Modulo 2 the determinant is \(0\) (and in fact \((\mathbf u+\mathbf v)+(\mathbf v+\mathbf w)+(\mathbf w+\mathbf u)=2(\mathbf u+\mathbf v+\mathbf w)=\mathbf 0\)), so the vectors are dependent: the step "divide by 2" is impossible.