MA20008 Algebra 1, 2004, Sheet 8 Solutions

  Geoff Smith, http://www.bath.ac.uk/∼masgcs

1. Suppose that V is a finite dimensional vector space over a field F, and that α : V ⟶ V is a linear map. Suppose that e_{1}, e_{2}, …, e_{n} is a basis of V such that each subspace ⟨e_{i}⟩ is α-invariant. Prove that the matrix of α with respect to this basis (in both domain and codomain) is diagonal. Solution For each i we have α(e_{i}) = λ_{i}e_{i} for some scalar λ_{i} ∈ F. It follows that the matrix of α with respect to this basis is

( \begin{matrix}λ_{1} & 0 & 0 & . & . & . \\ 0 & λ_{2} & 0 & . & . & . \\ 0 & 0 & λ_{3} & . & . & . \\ . & . & . & . & . & . \\ . & . & . & . & λ_{n−1} & 0 \\ . & . & . & . & 0 & λ_{n} \end{matrix} ) .

2. We consider the ordinary inner product on ℝ2 or ℝ3 defined as the “dot” or “scalar” product. Let the vertices of triangle ABC have position vectors a, b and c respectively.

(a) Show that the point G with position vector \frac{1}{3}(a + b + c) lies on each median (a line joining a vertex to the midpoint of the opposite side). Conclude that the medians are concurrent (at a point G which is called the centroid of △ABC). Solution The position vector of the point \frac{2}{3} of the way down the median from A is a+ \frac{2}{3}(\frac{1}{2}(b+c)−a) = \frac{1}{3}(a+b+c). By symmetry this point G is on all three medians, which must therefore be concurrent.

(b) The altitude of △ABC through A is the straight line through A which is perpendicular to BC. Show that a point P is on this altitude if and only if the position vector r of P satisfies (r − a) · (b − c) = 0. Solution The equation specifies that the lines AP and BC are perpendicular, or that A = P. This is the required altitude.

(c) Now we insist that the origin is the circumcentre O of △ABC. Thus there is a quantity R such that

a · a = b · b = c · c = R2.

Show that the point H with position vector a+b+c is on all three altitudes. Deduce that the three altitudes of △ABC are concurrent at H. The point H is called the orthocentre of △ABC. Solution H is on the altitude from A because (a+b+c−a)·(b− c) = (b + c) · (b − c) = b · b − c · c = R2 − R2 = 0. By symmetry H is on all three altitudes which are therefore concurrent.

(d) Deduce that the three points O, G and H are colinear, and that the distances are such that |OH| = 3|OG|. The line through O and H is called the Euler line of △ABC. Solution We know that \overline{OH} = 3\overline{OG} because \overline{OH} = a + b + c and \overline{OG} = \frac{1}{3}(a + b + c).

(e) Let L be the midpoint of BC and M be the midpoint of AH. Let N be the midpoint of LM. Find the position vector of N (with the origin still at O). Solution \overline{ON} = \frac{1}{4}(a + a + b + c) + \frac{1}{4}(b + c) = \frac{1}{2}(a + b + c).

(f) Deduce that N is the midpoint of OH so that O, G, N, H are colinear and the ratios of lengths are

|OG| : |GN| : |NH| = 2 : 1 : 3.

Solution This is because

\overline{OG} = \frac{1}{3}(a + b + c),

\overline{ON} = \frac{1}{2}(a + b + c)

and

\overline{OH} = a + b + c.

(g) Show that |LM| = R. \overline{LM} = \frac{1}{2}(a + a + b + c) − \frac{1}{2}(b + c) = \frac{1}{2}a. Now the length of a is R, so |LM| = \frac{1}{2}R.

(h) Deduce that the circle with centre N and radius R/2 goes through the following nine interesting points: the midpoints of the sides of △ABC, the feet of the altitudes of △ABC and the three points which are midway between H and each of the three vertices A, B and C. This is the ‘nine-point circle’ or ‘Feuerbach circle’. Solution We have shown that the circle with centre N and radius \frac{1}{2}R has LM as a diameter. Let the line AL intersect the line BC at D. Now ∠LDM is a right angle, so by the converse of the “angle in a semicircle” theorem, this circle passes through D. The position vector of N reveals that this is the same circle if we cyclically permute A, B and C, and the result follows.

3. Let ABCD be a cyclic quadrilateral. A maltitude is a straight line through the midpoint of a side which is perpendicular to the opposite side. Show that the four maltitudes are concurrent. Solution The maltitude from the midpoint of AB which is perpendicular to CD has equation

(r − \frac{1}{2}(a + b)) · (c − d).

Now the point P with position vector

\frac{1}{2}(a + b + c + d).

The symmetry of this position vector ensures that this point is on all four maltitudes.

4. Suppose that V is an inner product space, with inner product denoted by ⟨ , ⟩. Suppose that U and W are subspaces of V.

(a) Show that (U + W)^{⊥} = (U ∪ W)^{⊥} = U^{⊥} ∩ W^{⊥}.

Solution The second equality is a formality. As for the first, both inclusions are formailities.

(b) Suppose that U ≤ W. Show that W^{⊥} ≤ U^{⊥}.

Solution This is a formality.

5. Suppose that V is an inner product space of dimension n, and 0 ≠ v ∈ V. Prove that dim({v}^{⊥}) = n − 1. Solution Consider the map θ : V ⟶ F defined by x ↦ (x, v). This is a linear map, and is not the zero map since θ(v) = (v, v) ≠ 0. It is linear, and F is 1-dimensional, so θ is surjective. The rank-nullity theorem applies and we are done.