Tutorial: Vectors and Linear Algebra
Francis Wright, June 2018
Click on a problem header to show/hide my REDUCE solution.
Let \(A = (1,3,-1)\) and \(B = (4,1,1)\), and let
\(\mathbf{a}\) and \(\mathbf{b}\) be the position vectors
of \(A\) and \(B\) respectively. Determine:
- the length of \(\mathbf{a}\);
- the vector having length \(3\) in the same direction as \(\mathbf{a}\);
- the vector represented by \(\overrightarrow{AB}\);
- parametric equations for the line through \(A\) and \(B\);
- the cosine of the angle between \(\mathbf{a}\) and \(\mathbf{b}\);
- the vector product \(\mathbf{a}\times\mathbf{b}\) of \(\mathbf{a}\) and \(\mathbf{b}\).
load_package avector; a := avec(1,3,-1); b := avec(4,1,1); % (a) vmod a; % (b) 3 a/ws; % (c) ab := b - a; % (d) line := a + alpha*ab; % Or as scalar equations: {x = line(0), y = line(1), z = line(2)}; % (e) (a dot b)/(vmod a * vmod b); % (f) a cross b;
Let \(\mathbf{a} =
\left(\begin{array}{c}3\\-1\\-6\end{array}\right)\),
and let \(\mathbf{b}\) and \(\mathbf{c}\) be position
vectors of points in the \((x,y)\)-plane such that the
area of the parallelogram with sides \(\mathbf{b}\) and
\(\mathbf{c}\) is \(7\). Let the parallelepiped with sides
corresponding to \(\mathbf{a}\), \(\mathbf{b}\) and
\(\mathbf{c}\) have volume \(V\).
- Is there sufficient information to calculate \(V\)?
- If so, determine \(V\).
load_package avector; a := avec(3,-1,-6); k := avec(0,0,1); % b cross c = |b cross c| k = 7 k % V = |a dot (b cross c)| = 7 |a dot k| % (a) Yes % (b) a dot k; V := 7 abs ws;
Determine all solutions to the following system of linear
equations in \(x, y, z\) defined over \(\mathbb{R}\):
\[
\left.\begin{eqnarray*}
-x-2y+5z &=& 8 \\
3x+2y+ z &=& -4 \\
x+ y- z &=& -3
\end{eqnarray*}\right\}.
\]
eqns := {-x-2y+5z = 8, 3x+2y+z = -4, x+y-z = -3}; solve ws; % regarding arbcomplex() as a real parameter
Calculate the distance between the parallel
lines with vector equations
\[
r = \left(\begin{array}{c}2\\1\\2\end{array}\right) + \lambda
\left(\begin{array}{c}1\\2\\3\end{array}\right)
\quad\mbox{and}\quad
r =\left(\begin{array}{c}2\\-1\\3\end{array}\right) + \mu
\left(\begin{array}{c}-2\\-4\\-6\end{array}\right)
\]
respectively. (The parameters \(\lambda\) and \(\mu\) both range over
the whole of \(\mathbb{R}\).)
load_package avector; % This is a vector from one line to the other: avec(2,1,2) - avec(2,-1,3); % and this is the magnitude of its projection perpendicular to the lines: vmod(ws cross avec(1,2,3));
Let
\[
A = \left(\begin{array}{rrr}1 & 2 & 0\\-1 & 1 & 2\\1 & -2 & -1\end{array}\right)
\quad\mbox{and}\quad
B = \left(\begin{array}{rrr}2 & -1 & 0\\-1 & 2 & 0\\1 & -2 & -3\end{array}\right).
\]
Calculate the following:
- \(-2A+B\);
- \(A^2\);
- the characteristic polynomial of \(B\);
- \(\det B\) [Hint: the answer to the previous part should help].
a := mat((1,2,0),(-1,1,2),(1,-2,-1)); b := mat((2,-1,0),(-1,2,0),(1,-2,-3)); -2a+b; a^2; det(lam*mat((1,0,0),(0,1,0),(0,0,1))-b); -sub(lam=0,ws); % using the hint, or directly: det b;
Let \(S_\theta\) denote the \(2\times2\) matrix representing the
reflexion (in the \((x,y)\)-plane) in the line through the origin at
anticlockwise angle \(\theta/2\) to the \(x\)-axis. Then
\[
S_\theta = \left(\begin{array}{cc}
\cos\theta & \sin\theta \\
\sin\theta\ & -\cos\theta
\end{array}\right).
\]
- Determine the eigenvalues of \(S_\theta\).
- Now let \(S\) be the matrix of the reflexion in the line \(y = x\). Write down \(S\), and write down an eigenvector of \(S\) and its corresponding eigenvalue.
S_theta := mat((cos theta, sin theta),(sin theta, -cos theta)); % (a) let cos(~th)^2 + sin(~th)^2 => 1; E_theta := mateigen(S_theta, lam); % Hence eigenvalues are +/- 1. % (b) theta := pi/2; S := S_theta; E_theta; third first ws; % with corresponding eigenvalue 1.
Compute the determinant of the matrix
\[ \left(\begin{array}{ccc}
1 & -1 & 1 \\
2 & 0 & 1 \\
-1 & 2 & -2
\end{array}\right).
\]
Is this matrix invertible? If so, find its inverse.
Hence or otherwise solve the linear system
\[ \begin{eqnarray}
x_1 - x_2 + x_3 &=& 1 \\
2x_1 + x_3 &=& 2 \\
-x_1 + 2x_2 - 2x_3 &=& -3.
\end{eqnarray}
\]
M := mat((1,-1,1),(2,0,1),(-1,2,-2)); det M; M^-1; % Hence: ws * mat((1),(2),(-3)); % or otherwise: eqns := {x1 - x2 + x3 = 1, 2x1 + x3 = 2, -x1 + 2x2 - 2x3 = -3}; solve ws;
Let
\[
A = \left(\begin{array}{cc}
0 & 3 \\
-1 & 2
\end{array}\right), \quad
B =\left(\begin{array}{cc}
-1 & 2 \\
0 & 3 \\
1 & -2
\end{array}\right).
\]
For each of the products \(A^2\), \(AB\), \(BA\), \(B^2\), state
whether or not it exists; if it exists then evaluate it.
A := mat((0,3),(-1,2)); B := mat((-1,2),(0,3),(1,-2)); A^2; A*B; B*A; B^2;
Suppose that \(B\) is a square matrix and \(B^2 = O\). Prove that
\(I+B\) is invertible, and find its inverse. Illustrate with a
\(2\times2\) example.
% (I + B)(I - B) = I^2 - B^2 = I, hence (I + B)^-1 = (I - B) B := mat((0,1),(0,0)); B^2; Id := mat((1,0),(0,1)); A := Id + B; A^-1 = Id - B;
For each of the following statements about \(n\times n\) matrices
\(A\) and \(B\), state whether it is true or false, and illustrate
with general \(2\times2\) matrices.
- \(\det(A + B) = \det A + \det B\);
- \(\det(A + B^T) = \det(B + A^T)\);
- \(\det(BA) = \det(AB)\);
- \(\det(\lambda A) = \lambda \det A\), where \(\lambda\in\mathbb{R}\).
matrix A, B; operator aa, bb; for i := 1:2 do for j := 1:2 do <<A(i,j) := aa(i,j); B(i,j) := bb(i,j)>>; A; B; % (a) false det(A + B) = det A + det B; if det(A + B) = det A + det B then true else false; % (b) true det(A + tp B)= det(tp A + B); if det(A + tp B)= det(tp A + B) then true else false; % (c) true det(B*A) = det(A*B); if det(B*A) = det(A*B) then true else false; % (d) false (unless n=1) det(lam*A) = lam*det(A); if det(lam*A) = lam*det(A) then true else false;
Let \(V = \mathbb{R}^4\), and let
\[
S = \{(0,2,1,-3)^T, (1,1,1,2)^T, (3,0,-1,-1)^T, (-1,4,4,2)^T\}.
\]
Either determine whether or not \(S\) is a spanning set for \(V\),
or determine whether or not \(S\) is a linearly dependent set.
% Using lists to represent vectors: S := {{0,2,1,-3}, {1,1,1,2}, {3,0,-1,-1}, {-1,4,4,2}}; rank S; % or equivalently S := mat((0,2,1,-3), (1,1,1,2), (3,0,-1,-1), (-1,4,4,2)); rank S; % S is not a spanning set because it contains only 3 linearly independent vectors.
Let \(P_2\) denote the set of polynomials of degree at most \(2\), that is
\[
P_2 = \{\mathbf{p} \mid \mathbf{p}(x) = a_2x^2 + a_1x + a_0 \mbox{ for some }
a_0, a_1, a_2 \in \mathbb{R}\}.
\]
Let \(D : P_2 \to P_2\) be the mapping given by \(D(\mathbf{p}) =
\mathbf{q}\), where \(\mathbf{p}'\) denotes the derivative of
\(\mathbf{p}\), and
\[
\mathbf{q}(x) = x^2\mathbf{p}(0) + \mathbf{p}'(x).
\]
- Show that \(D\) is a linear transformation.
- Is \(D\) surjective? Justify your answer.
- Is \(D\) injective? Justify your answer.
operator D; let D(~p) => x^2*sub(x=0,p) + df(p,x); p1 := a2*x^2 + a1*x + a0; p2 := b2*x^2 + b1*x + b0; % (a) D(alpha*p1 + beta*p2) = alpha*D(p1) + beta*D(p2); if D(alpha*p1 + beta*p2) = alpha*D(p1) + beta*D(p2) then true else false; % (b) Yes if we can find p1 that maps to an arbitrary p2: D(p1) = p2; coeff(lhs ws - rhs ws, x); solve(ws, {a2,a1,a0}); % (c) Yes, since precisely one p1 maps to each p2.
Let \(H\) be a subspace of \(\mathbb{R}^n\) and let \(H^\perp\) denote
the orthogonal complement of \(H\). Suppose \(n = 4\) and
\[
H = \{(x,y,z,t)^T \mid x + 2y - 3z = 0, 3y + 2z - t = 0\}.
\]
Compute \(\dim H\), \(\dim H^\perp\), and a basis for \(H^\perp\).
% First, ensure that arbitrary constants have the expected indices: % (See MacCallum and Wright, footnote 31 on page 35.) symbolic(!!arbint := 0); solve({x+2y-3z=0, 3y+2z-t=0}, {x,y,z,t}); h := for each el in first ws collect rhs el; % 2 parameters, so dim H = 2, with basis vectors h1 := sub(arbcomplex(1)=1, arbcomplex(2)=0, h); h2 := sub(arbcomplex(1)=0, arbcomplex(2)=1, h); % Hence a vector {x,y,z,t} in Hperp must satisfy solve({-2x+1y+0z+3t=0, 3x+0y+1z+2t=0}, {x,y,z,t}); hp := for each el in first ws collect rhs el; % 2 parameters, so dim Hperp = 2, with basis vectors hp1 := sub(arbcomplex(3)=1, arbcomplex(4)=0, hp); hp2 := sub(arbcomplex(3)=0, arbcomplex(4)=1, hp);
Let
\[
A = \left(\begin{array}{ccc}
-1 & 0 & 1 \\
2 & 3 & 2 \\
2 & 4 & 0
\end{array}\right), \quad
\mathbf{v}_1 = \left(\begin{array}{c}
-2 \\
1 \\
0
\end{array}\right), \quad
\mathbf{v}_2 = \left(\begin{array}{c}
1 \\
0 \\
-1
\end{array}\right).
\]
- Show that \(\mathbf{v}_1\) and \(\mathbf{v}_2\) are eigenvectors of \(A\) and find the corresponding eigenvalues.
- Find the characteristic polynomial of \(A\).
- Determine all eigenvalues of \(A\) and find bases for the corresponding eigenspaces.
- Find an invertible matrix \(P\) and a diagonal matrix \(D\) such that \(P^{-1}AP = D\).
A := mat((-1,0,1),(2,3,2),(2,4,0)); v1 := mat((-2),(1),(0)); v2 := mat((1),(0),(-1)); % (a) A*v1; A*v2; % Hence the eigenvalues are respectively -1 and -2. % (b) det(lam*mat((1,0,0),(0,1,0),(0,0,1))-A); % (c) solve ws; ev := mateigen(A, lam); for each el in ev collect third el; bases := (ws where arbcomplex(~j) => 1); % (d) % Construct D from the eigenvalues: for each el in ev collect rhs first solve first el; clear D; matrix D(3,3); for i := 1:3 do D(i,i) := part(ws, i); D; % Construct P from the eigenspace bases: matrix P(3,3); for i := 1:3 do for j := 1:3 do P(i,j) := <<v:=part(bases, j); v(i,1)>>; % Check P^(-1)*A*P = D;