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anonymous

  • one year ago

What matrix 2x2 P1 projects the vector (x, y) onto the x axis to produce (x, 0)? What matrix P2 projecs onto the y axis to produce (0, y)?

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  1. beginnersmind
    • one year ago
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    |dw:1442100623341:dw| Which gives a = 1, b=c=d=0. You can do P_2 the same way, noting that P_2*(x,y) = (0,y)

  2. beginnersmind
    • one year ago
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    |dw:1442101038643:dw|

  3. anonymous
    • one year ago
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    Thank you. Your answer is very straight foward. Later I found the answer with a time consumin and painful reasoning, thinking about the equation of the line that gives the x-axis as the answer, instead of a single point. From there, I figured that it looked like the identity matrix (2 by 2) times the vector (x, y) gone wrong, instead of giving x+y, would give just x, after playing a little, I end up with this: \[\left[\begin{matrix}1 &0 \\ 0 &0\end{matrix}\right]\left[\begin{matrix}x \\ y\end{matrix}\right]\] Too much pain for a very simple question. The way you solved it is much better.

  4. beginnersmind
    • one year ago
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    BTW, there's a useful fact about multiplying a matrix with a column vector. The result is a linear combination of the columns of the matrix, with the coordinates of the vector as coefficients. Written out: |dw:1442280304311:dw| So the first coordinate of the vector multiplies the first column, the second coordinate the second column, etc. (It work both in 2 dimensions or 3. Or any other number actually). A consequence of this is that there's a very easy way to calculate the product of a standard unit vector with a matrix. A*(1,0) is just the first column of the matrix. A*(0,1) is the second column of the matrix. You can use this idea backward as well. If you are given the image of the standard unit vectors under a (unknown) matrix and are told to find the entries of the matrix you are in luck: The image of (1,0) is the first column of the matrix and the image of (0,1) is the second column. Or in 3 dimensions, the image of (1,0,0) is the first column, etc.

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