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

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1. Answer to Question 1 (4 marks)

   \(U\): contains \(\mathbf 0\) (1); if \(x=2y\) and \(x'=2y'\) then \(x+x'=2(y+y')\) and \(\lambda x=2\lambda y\) (1): subspace (1). \(V\): \((1,0)\in V\) but \(-1\cdot(1,0)=(-1,0)\notin V\) (1): not closed under scalar multiplication, so not a subspace.

   For \(U\), the three subspace conditions are checked: the zero vector satisfies \(0=2\times0\), and the set is closed under addition and under scalar multiplication because the condition \(x=2y\) is linear. Geometrically, \(U\) is a plane through the origin in \(\mathbb R^3\). For \(V\), one counterexample is enough to show it is not a subspace; \(V\) is a closed half-plane, closed under addition but not under multiplication by negative scalars.

2. Answer to Question 2 (4 marks)

   The determinant of the matrix with these rows is \(1(18-12)-1(9-3)+1(4-2)=6-6+2=2\ne0\) (2): a basis. Solve \(a+b+c=3\), \(a+2b+4c=7\), \(a+3b+9c=13\) (one equation for each coordinate): subtract: \(b+3c=4\), \(b+5c=6\) so \(c=1\), \(b=1\), \(a=1\) (2). Coordinates \((1,1,1)\).

   Three vectors in the three-dimensional space \(\mathbb R^3\) form a basis exactly when they are linearly independent, which is equivalent to a non-zero determinant. In the elimination, subtracting the first equation from the second and third gives \(b+3c=4\) and \(2b+8c=10\), that is \(b+4c=5\); a further subtraction gives \(c=1\). Check: \(1\times(1,1,1)+1\times(1,2,3)+1\times(1,4,9)=(3,7,13)\).

3. Answer to Question 3 (4 marks)

   1. \(\begin{pmatrix}1&1&0\\0&1&1\end{pmatrix}\) (1).
   2. Kernel: \(x=-y\), \(z=-y\), spanned by \((-1,1,-1)\) (1). Image: the columns \((1,0)\), \((1,1)\) span \(\mathbb R^2\), basis \(\{(1,0),(0,1)\}\) (1).
   3. Rank 2 + nullity 1 \(=3\) (1).

   In (a), the columns of the matrix are the images of the standard basis vectors: \(T(1,0,0)=(1,0)\), \(T(0,1,0)=(1,1)\) and \(T(0,0,1)=(0,1)\). In (b), the kernel is the solution set of \(x+y=0\) and \(y+z=0\); taking \(y=1\) as the free variable gives the spanning vector. Check: \(T(-1,1,-1)=(0,0)\). In (c), the dimension of the domain \(\mathbb R^3\) is 3, which equals rank plus nullity.

4. Answer to Question 4 (4 marks)

   1. \(5-2=3\) (1).
   2. The image has dimension 2 (1), less than 3, so it cannot cover \(\mathbb R^3\) (1).
   3. Not injective since the kernel is non-trivial (nullity 3) (1).

   In (a), the rank-nullity theorem says rank plus nullity equals the dimension of the domain, which is 5 because \(A\) acts on \(\mathbb R^5\). In (b), the rank is the dimension of the image, a plane inside \(\mathbb R^3\), so the map is not surjective. In (c), a linear map is injective exactly when its kernel is \(\{\mathbf 0\}\); here three independent directions are sent to zero, so different inputs can give the same measurement.

5. Answer to Question 5 (4 marks)

   1. Coordinates \((1,1,0)\), \((0,1,1)\), \((1,0,1)\) (1); determinant \(2\ne0\) (1).
   2. \(a+c=2\), \(a+b=3\), \(b+c=1\) (1): sum \(a+b+c=3\), so \(b=1\), \(c=0\), \(a=2\). \(q=2p_1+p_2\) (1). Check: \(2+2x+x+x^2=2+3x+x^2\).

   In (a), the coordinate vector lists the coefficients of \(1\), \(x\) and \(x^2\): for example \(p_2=0+1x+1x^2\). Three vectors in the three-dimensional space \(P_2\) form a basis when the determinant of their coordinates is non-zero; here it equals \(1(1-0)-0+1(1-0)=2\). In (b), compare the coefficients of \(1\), \(x\) and \(x^2\) in \(ap_1+bp_2+cp_3=q\). Adding the three equations gives \(2(a+b+c)=6\).

Total : 20 marks