Question

First recall that the matrix corresponding to a rotation by an angle θ is given by

Rθ=$\begin{bmatrix} cos(\Theta )&-sin(\Theta)\\ sin(\Theta)& cos(\Theta) \end{bmatrix}$

We build a 2-by-2 matrix by first rotating by θ1=−π/3, then stretching using the diagonal matrix D=$\begin{bmatrix}4 &0\\ 0& 1 \end{bmatrix}$, then rotating again by θ2=π/2, so that A=Rθ2DRθ1.

Now recall that the maximum stretch for a matrix A is computed by

max ||x||=1  ||Ax||

and any vector x of norm 1 such that ||Ax|| attains this max is called the direction of maximum stretch.

Write down the vector of maximum stretch for A in this case:

x=

Answer 1

-sinie7 RIO): ſcosco) I sinie) cosco) RC-7/3) = | Žena R 1 1/2) = 10 thi To LA A = R(01) DRIO,) - TO -11 14 07 12 +13): 7 - Odlo 1] [-J3/2 V2 -17 | 12 + 13/27 I t o Lw5/12] - +13/2 +12 7 i +235 /

| 고 | l 2 25] All | els) - Long7 As - I Bee - 9 1 | [ + [ cose +ssine] || sin( -) | | + Cosld - 3) | Ax) = J sin-lo - (3) † \6 Cos -) = / i + 15Cos ( d - 지) | - / 1 + Is cos -지) | =) o Cos²d-지3) 4 / ㅋ) ) [Ax) < la =) Almax = 4. ㅋ) cos( - 지3) = ) =) Cos -지3) - ) 2) = 4 . | x = | -1/2 7. Il A max = 4 115 !