MHB Solving Linear Transformations in R2: Step by Step Guide

AI Thread Summary
To solve the linear transformation problem in R2, the transformation phi is defined by its action on the basis vectors, with phi((1,0)) = (4,5) and phi((0,1)) = (9,11). The image of any vector (x,y) under phi can be expressed as phi(x,y) = (4x + 9y, 5x + 11y). To find the matrix representation of phi with respect to the basis B formed by vectors w1 = (3,5) and w2 = (10,17), one must compute phi(w1) and phi(w2) and express these results as linear combinations of the basis vectors. This leads to a system of equations to determine the coefficients α1, α2, β1, and β2 for the matrix M_B^B(φ). The final matrix representation can then be constructed using these coefficients.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
I quote a question from Yahoo! Answers

Could anyone give me a step by step method on how to solve these type of questions please? Very confused and need to know for my exams.

Let V denote the real vector space R2 and phi: V -> V be a real linear transformation such that phi((1,0)) = (4,5) and phi((0,1)) = (9,11). Express the imagine phi((x,y)) of (x,y) in terms of x and y.

Assume that w1 = (3,5) and w2 = (10,17) form an ordered basis B for V. Working from the definition determine the matrix M(subscript and superscript B) (phi) of phi with respect to the basis B.

I have given a link to the topic there so the OP can see my response.
 
Mathematics news on Phys.org
Using that $\phi$ is a linear map:
$$\phi(x,y)=\phi[x(1,0)+y(0,1)]=x\phi (1,0)+y\phi(0,1)\\=x(4,5)+y(9,11)=(4x+9y,5x+11y)$$
On the other hand:
$$\phi(w_1)=\phi (3,5)=(4\cdot 3+9\cdot 5,\;5\cdot 3+11\cdot 5)=(57,70)$$
Now, write $\phi(w_1)=(57,70)=\alpha_1w_1+\alpha_2 w_2$ and you'll easily find $\alpha_1$ and $\alpha_2$ by means of a simple system. In a similar way, you'll get $\phi(w_2)=\beta_1w_1+\beta_2 w_2$. Then,
$$M_B^B= \begin{bmatrix}{\alpha_1}&{\beta_1}\\{\alpha_2}&{ \beta_2}\end{bmatrix}=\ldots$$
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Replies
1
Views
1K
Replies
12
Views
2K
Replies
3
Views
2K
Replies
1
Views
3K
Replies
16
Views
4K
Replies
6
Views
3K
Replies
1
Views
1K
Back
Top