Latest revision as of 07:20, 22 February 2010

HW February 19

Discuss HW Problems due Feb 19 here. :) Josh Hunsberger

Does someone know how to solve for number 31? --Give an example of a matrix A such that im(A) is the plane with normal vector [1;3;2] in R^3. Kelsey Underwood

Sorry this is too late to actually help you on the homework. There are actually many ways of solving this, but I think one of the most straightforward ways is to make A the projection of a vector onto a plane, specifically the plane with normal vector [1;3;2]. Using a projection guarantees your output will be on the plane since thats what a projection does. You can find the projection of a vector onto a plane by subtracting the perpendicular component of the vector from the vector. And you find the perpendicular component by taking the projection of the vector onto the normal vector. From that you should be able to find a matrix.

Since we're not allowed to actually do solutions, I'll do an example for im(A) is in the plane with normal vector [2;5;5].

First find the projection of the x vector onto the normal:

$ \text{proj}_{\vec n} \vec x = \frac{\vec x \cdot \vec n}{\vec n \cdot \vec n}\vec n=\frac{2x_1+5x_2+5x_3}{54} \begin{bmatrix} 2 \\ 5 \\ 5 \end{bmatrix}=\frac{1}{54}\begin{bmatrix}4x_1+10x_2+10x_3 \\ 10x_1+25x_2+25x_3 \\ 10x_1+25x_2+25x_3\end{bmatrix} $

Now just subtract this from original vector to get projection onto plane.

$ \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} -\frac{1}{54} \begin{bmatrix}4x_1+10x_2+10x_3 \\ 10x_1+25x_2+25x_3 \\ 10x_1+25x_2+25x_3\end{bmatrix}=\frac{1}{54} \begin{bmatrix}50x_1+44x_2+44x_3 \\ 44x_1+29x_2+29x_3 \\ 10x_1+29x_2+29x_3\end{bmatrix} $

Now A is the matrix we take times x vector, so it is just the coefficients of the elements of x

$ \mathbf{A}=\frac{1}{54} \begin{bmatrix}50 & 44 & 44 \\ 44 & 29 & 29 \\ 10 & 29 & 29 \end{bmatrix} $

I hope this helps. Next time I hope answers will be given before homework is due. Josh Hunsberger

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