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Goodmorning,

I have a question related to Linear Algebra:

The line L through the origin and (1,2) is given. Furthermore we know that for any vector v the vector Av is given by the projection of v onto the line L. Can you determine the eigenvalues and eigenvectors of A (without determining the matrix A explicitly)? If not, write down the answer 'no'.

What I did is the following:

1) First determine the equation of the line. The line passes through points (0,0) and (1,2) so the directional coefficient is 2-1 =2. The line passes through the origin, which gives equation y=2x for the line.

2) Determine the matrix by the projection: (([x y] * [1 2])/ ([1 2] . [1 2]))[1 2] (note that . indicates a dot product and I wrote all values in between brackets in column form, but I don't know how to write the values in brackets below each other).

3) This gives ((x+2y)/5)[1 2]. Using e1=[1 0] and e2=[0 1] with x,y we get a matrix: [1/5 2/5] [2/5 4/5]

4) Next I used eigenvalues -1 and +1 without any good reasoning (don't really know which eigenvalues to use). The matrix becomes: [(1/5)-lambda (2/5)] [(2/5) (4/5)-lambda]

5) Substituting the value of -1 for lambda first, we can solve the matrix [(1/5)+1 (2/5) 0] [(2/5) (4/5)+1 0]

If we have a free variable in the solution we can determine the corresponding eigenvector. The same for the other eigenvalue. I have the question whether the method I've used above is correct? I expect that I'm doing something wrong. Which eigenvalues should I use? The question explicitly states:"without determining matrix A", but with my method I do determine matrix A. How can I best solve this problem? Thank you in advance for the replies. I apologies for the above notation of the formulas, but unfortunately I don't know how programs such as LaTeX work.

Rik

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  • $\begingroup$ Welcome to MSE. What do you mean by “the projection of $v$ onto the line $L$”? Do you mean orthogonal projection? $\endgroup$ – José Carlos Santos Jan 3 at 9:46
  • $\begingroup$ Thank you for your reply. The above question was directly copy pasted so unfortunately doesn't specify this information. I am following the Linear Algebra 1 course and so far we've only discussed orthogonal projections, so I assume they mean orthogonal projection. $\endgroup$ – Misterrik Jan 3 at 9:52
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In the spirit of this question, you're not supposed to set up the projection matrix $A$. The question seems to be designed to check if you understand, from a geometrical point of view, what eigenvectors and eigenvalues are.

Recall that an eigenvector is a vector which is mapped to a scalar multiple of itself with the corresponding eigenvalue being precisely that scalar multiple. For an orthogonal projection (in the plane) onto a line: which vector(s) will be mapped to a scalar multiple of itself?

If you have no idea: draw a few projections onto this line to get a feeling.

Hoover over for a hint:

Look at two special case: what happens to vectors parallel to the line (i.e. along the line) and what happens to vectors perpendicular to the line; how are they mapped?

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  • $\begingroup$ Unfortunately I still have no idea how to solve the above problem using this comment. I haven't really developed a geometrical understanding of eigenvalues and eigenvectors yet. Referring to your question, algebraically I know that if the direction vector dot product the normal vector equals zero, the line and a plane are parallel. If a scalar exists such that the direction vector equals the normal vector, then the line is perpendicular to the plane. But I'm not sure how this can be of any help answering the question stated in my initial post $\endgroup$ – Misterrik Jan 5 at 15:13

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