Linear transformations

1. \(x_2\) and \(y_2\) are defined in terms of \(x_1\) and \(y_1\) by the equation $$\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$$ \(G_1\) is the graph with equation $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ and \(G_2\) is the graph with equation $$\frac{\left(\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{9} + \frac{\left(-\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{4} = 1$$ Show that, if \((x_1, y_1)\) is a point on \(G_1\), then \((x_2, y_2)\) is a point on \(G_2\). Show that \(G_2\) is an anti-clockwise rotation of \(G_1\) through \(45^\circ\) about the origin.
2. 1. The matrix $$\begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix}$$ represents a reflection. Find the line of invariant points of this matrix.
   2. Sketch, on the same axes, the graphs with equations $$y = 2^x \text{ and } 0.8x + 0.6y = 2^{-0.6x+0.8y}$$
3. Sketch, on the same axes, for \(0 \leq x \leq 2\pi\), the graphs with equations $$y = \sin x \text{ and } y = \sin(x - 2y)$$ You should determine the exact co-ordinates of the points on the graph with equation \(y = \sin(x - 2y)\) where the tangent is horizontal and those where it is vertical.

1. The matrix \(\mathbf{R}\) represents an anticlockwise rotation through angle \(\varphi\) (\(0^\circ \leqslant \varphi < 360^\circ\)) in two dimensions, and the matrix \(\mathbf{R} + \mathbf{I}\) also represents a rotation in two dimensions. Determine the possible values of \(\varphi\) and deduce that \(\mathbf{R}^3 = \mathbf{I}\).
2. Let \(\mathbf{S}\) be a real matrix with \(\mathbf{S}^3 = \mathbf{I}\), but \(\mathbf{S} \neq \mathbf{I}\). Show that \(\det(\mathbf{S}) = 1\). Given that \[ \mathbf{S} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \] show that \(\mathbf{S}^2 = (a+d)\mathbf{S} - \mathbf{I}\). Hence prove that \(a + d = -1\).
3. Let \(\mathbf{S}\) be a real \(2 \times 2\) matrix. Show that if \(\mathbf{S}^3 = \mathbf{I}\) and \(\mathbf{S} + \mathbf{I}\) represents a rotation, then \(\mathbf{S}\) also represents a rotation. What are the possible angles of the rotation represented by \(\mathbf{S}\)?

The matrix A is given by $$\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.$$

1. You are given that the transformation represented by A has a line \(L_1\) of invariant points (so that each point on \(L_1\) is transformed to itself). Let \((x, y)\) be a point on \(L_1\). Show that \(((a - 1)(d - 1) - bc)xy = 0\). Show further that \((a - 1)(d - 1) = bc\). What can be said about A if \(L_1\) does not pass through the origin?
2. By considering the cases \(b \neq 0\) and \(b = 0\) separately, show that if \((a - 1)(d - 1) = bc\) then the transformation represented by A has a line of invariant points. You should identify the line in the different cases that arise.
3. You are given instead that the transformation represented by A has an invariant line \(L_2\) (so that each point on \(L_2\) is transformed to a point on \(L_2\)) and that \(L_2\) does not pass through the origin. If \(L_2\) has the form \(y = mx + k\), show that \((a - 1)(d - 1) = bc\).

Let \(R_{\alpha}\) be the \(2\times2\) matrix that represents a rotation through the angle \(\alpha\) and let $$A=\begin{pmatrix}a&b\\b&c\end{pmatrix}.$$

1. Find in terms of \(a\), \(b\) and \(c\) an angle \(\alpha\) such that \(R_{-\alpha}AR_{\alpha}\) is a diagonal matrix (i.e. has the value zero in top-right and bottom-left positions).
2. Find values of \(a\), \(b\) and \(c\) such that the equation of the ellipse \[x^2+(y+2x\cot2\theta)^2=1\qquad(0 < \theta < \tfrac{1}{4}\pi)\] can be expressed in the form \[\begin{pmatrix}x&y\end{pmatrix}A\begin{pmatrix}x\\y\end{pmatrix}=1.\] Show that, for this \(A\), \(R_{-\alpha}AR_{\alpha}\) is diagonal if \(\alpha=\theta\). Express the non--zero elements of this matrix in terms of \(\theta\).
3. Deduce, or show otherwise, that the minimum and maximum distances from the centre to the circumference of this ellipse are \(\tan\theta\) and \(\cot\theta\).

The transformation \(T\) of the point \(P\) in the \(x\),\(y\) plane to the point \(P'\) is constructed as follows: \hfil\break Lines are drawn through \(P\) parallel to the lines \(y=mx\) and \(y=-mx\) to cut the line \(y=kx\) at \(Q\) and \(R\) respectively, \(m\) and \(k\) being given constants. \(P'\) is the fourth vertex of the parallelogram \(PQP'R\). Show that if \(P\) is \((x_1,y_1)\) then \(Q\) is $$ \left( {mx_1-y_1 \over m-k}, {k(mx_1-y_1)\over m-k}\right). $$ Obtain the coordinates of \(P'\) in terms of \(x_1\), \(y_1\), \(m\) and \(k\), and express \(T\) as a matrix transformation. Show that areas are transformed under \(T\) into areas of the same magnitude.

The transformation \(T\) from \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} X \\ Y \end{pmatrix}\) is given by \[ \begin{pmatrix}X\\ Y \end{pmatrix}=\frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix}. \] Show that \(T\) leaves the vector \(\begin{pmatrix} 1\\ 2 \end{pmatrix}\) unchanged in direction but multiplied by a scalar, and that \(\begin{pmatrix} 2\\ -1 \end{pmatrix}\) is similarly transformed. The circle \(C\) whose equation is \(x^{2}+y^{2}=1\) transforms under \(T\) to a curve \(E\). Show that \(E\) has equation \[ 8X^{2}+12XY+17Y^{2}=80, \] and state the area of the region bounded by \(E\). Show also that the greatest value of \(X\) on \(E\) is \(2\sqrt{17/5}.\) Find the equation of the tangent to \(E\) at the point which corresponds to the point \(\frac{1}{5}(3,4)\) on \(C\).

The linear transformation \(\mathrm{T}\) is a shear which transforms a point \(P\) to the point \(P'\) defined by

1. \(\overrightarrow{PP'}\) makes an acute angle \(\alpha\) (anticlockwise) with the \(x\)-axis,
2. \(\angle POP'\) is clockwise (i.e. the rotation from \(OP\) to \(OP'\) clockwise is less than \(\pi),\)
3. \(PP'=k\times PN,\) where \(PN\) is the perpendicular onto the line \(y=x\tan\alpha,\) where \(k\) is a given non-zero constant.

Show Solution

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We can see that \(\mathbf{M}\) sends \(\begin{pmatrix} 1 \\ \tan \alpha \end{pmatrix}\) to itself, and \(\begin{pmatrix} -\tan \alpha \\ 1 \end{pmatrix}\) to \(\begin{pmatrix} -\tan \alpha \\ 1 \end{pmatrix} + k \begin{pmatrix} 1 \\ \tan \alpha \end{pmatrix}\) Therefore, we have: \begin{align*} && \mathbf{M} \begin{pmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{pmatrix} &= \begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix} \\ && \sec \alpha \mathbf{M} \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} &= \begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix} \\ \Rightarrow && \mathbf{M} &= \cos \alpha\begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix}\begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \\ &&&= \cos\alpha \begin{pmatrix}\cos \alpha -k\sin\alpha + \frac{\sin^2 \alpha}{\cos \alpha} & \sin \alpha + k \cos \alpha - \sin \alpha \\ \sin \alpha - \sin \alpha - k\frac{\sin^2 \alpha}{\cos \alpha} & \frac{\sin^2 \alpha}{\cos \alpha} + \cos\alpha + k \sin \alpha \end{pmatrix} \\ &&&= \begin{pmatrix}1-k\sin\alpha\cos\alpha & k\cos^{2}\alpha\\ -k\sin^{2}\alpha & 1+k\sin\alpha\cos\alpha \end{pmatrix} \end{align*} Suppose $\begin{pmatrix}p & q\\ r & t \end{pmatrix} \mathbf{M} = \mathbf{M} \begin{pmatrix}p & q\\ r & t \end{pmatrix}$ then, \begin{align*} && \begin{pmatrix}p & q\\ r & t \end{pmatrix} \mathbf{M} &= \mathbf{M} \begin{pmatrix}p & q\\ r & t \end{pmatrix} \\ \Leftrightarrow && \small \begin{pmatrix} p(1-k\sin\alpha\cos\alpha) + q(-k\sin^{2}\alpha) & pk\cos^{2}\alpha + q(1+k\sin\alpha\cos\alpha)\\ r(1-k\sin\alpha\cos\alpha) + t(-k\sin^{2}\alpha) & rk\cos^{2}\alpha + t(1+k\sin\alpha\cos\alpha)\end{pmatrix} &= \\ && \qquad \small \begin{pmatrix} p(1-k\sin\alpha\cos\alpha) + rk\cos^{2}\alpha & q(1-k\sin\alpha\cos\alpha) + tk\cos^{2}\alpha \\ -pk\sin^{2}\alpha + r(1+k\sin\alpha\cos\alpha) & -qk\sin^{2}\alpha+t (1+k\sin\alpha\cos\alpha) \end{pmatrix} \\ \Leftrightarrow && \begin{cases} p(1-k\sin\alpha\cos\alpha) + q(-k\sin^{2}\alpha) &= p(1-k\sin\alpha\cos\alpha) + rk\cos^{2}\alpha \\ pk\cos^{2}\alpha + q(1+k\sin\alpha\cos\alpha) &=q(1-k\sin\alpha\cos\alpha) + tk\cos^{2}\alpha \\ r(1-k\sin\alpha\cos\alpha) + t(-k\sin^{2}\alpha) &=-pk\sin^{2}\alpha + r(1+k\sin\alpha\cos\alpha) \\ rk\cos^{2}\alpha + t(1+k\sin\alpha\cos\alpha) &= -qk\sin^{2}\alpha+t (1+k\sin\alpha\cos\alpha) \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ p\cos^{2}\alpha + q\sin\alpha\cos\alpha &=-q\sin\alpha\cos\alpha + t\cos^{2}\alpha \\ -r\sin\alpha\cos\alpha + -t\sin^{2}\alpha &=-p\sin^{2}\alpha + r\sin\alpha\cos\alpha \\ r &= -q\tan^{2}\alpha \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ 2q\sin\alpha\cos\alpha &=(t-p)\cos^{2}\alpha \\ (p-t)\sin^{2}\alpha &=2r\sin\alpha\cos\alpha \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ 2q\tan \alpha &=(t-p) \end{cases} \\ \end{align*} as required

Show that, if the lengths of the diagonals of a parallelogram are specified, then the parallogram has maximum area when the diagonals are perpendicular. Show also that the area of a parallelogram is less than or equal to half the square of the length of its longer diagonal. The set \(A\) of points \((x,y)\) is given by \begin{alignat*}{1} \left|a_{1}x+b_{1}y-c_{1}\right| & \leqslant\delta,\\ \left|a_{2}x+b_{2}y-c_{2}\right| & \leqslant\delta, \end{alignat*} with \(a_{1}b_{2}\neq a_{2}b_{1}.\) Sketch this set and show that it is possible to find \((x_{1},y_{1}),(x_{2},y_{2})\in A\) with \[ (x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}\geqslant\frac{8\delta^{2}}{\left|a_{1}b_{2}-a_{2}b_{1}\right|}. \]

Show Solution

In a parallelogram the diagonals meet at their mid points. Fixing one diagonal, we can look at the two triangles formed by the other diagonal. Suppose the angle between them is \(\theta\). Then the area of the triangles will be \(\frac12 \frac{l_1}{2} \frac{l_2}2 \sin \theta+\frac12 \frac{l_1}{2} \frac{l_2}2 \sin (\pi -\theta) = \frac{l_1l_2}{4} \sin \theta\). This will be true on both sides. Therefore we can maximise this area by setting \(\theta = \frac{\pi}{2}\).

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Consider the (darker) shaded area. This is our set \(A\). The area of the set is indifferent to a parallel shift in the lines, so without loss of generality, we can consider \(c_1 = 0, c_2 = 0\), so our lines meet at the origin. Now also consider the linear transformation \(\begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}^{-1}\) which takes the coordinate axes to these lines. This will take the square \([-\delta, \delta] \times [-\delta, \delta]\) which has area \(4\delta^2\) to our square, which will have area \(\frac{4 \delta^2}{ |a_1b_2 - a_2b_1|}\). If we consider the length of the two diagonals of this area, \(l_1, l_2\) we know that \(\frac{l_1l_2}2 \sin \theta = \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\), if we consider the larger of \(l_1\) and \(l_2\) (wlog \(l_1\)) we must have that \(\frac{l_1^2}{2} \geq \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\) and so points on opposite ends of the diagonal will satisfy the inequality in the question.