{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

M341F09HW2

# M341F09HW2 - M341 F10 Homework 2 With the exception of 1.4...

This preview shows pages 1–2. Sign up to view the full content.

M341- F10 - Homework 2. With the exception of 1.4, problems 1 and 3 which are purely computational, expla- nations in English are given as they should. Mostly in section 1.3 even when there are a computations (in Math there are almost always some), reasoning is more important than computing. See the blend of mathematical statements and English language, see also how you should always try and explain until what you have in mind seems (or should seem) obvious to the reader (to you in fact). Section 1.3. 12. This problem is a bit diﬃcult. Do not hesitate to read the solution several times if necessary Let us assume that three mutually orthogonal non zero vectors exist in R 2 , say x = [ x 1 , x 2 ] , y = [ y 1 , y 2 ] , z = [ z 1 , z 2 ]. Since [ x 1 , x 2 ] ̸ = [0 , 0], one of its coordinates is not 0. We may assume, changing the numbering of the coordinates if necessary that x 1 ̸ = 0. If y 1 = z 1 = 0, [ y 1 , y 2 ] . [ z 1 , z 2 ] = y 1 z 1 + y 2 z 2 = y 2 z 2 = 0, would bring that either y 2 = 0 or z 2 = 0 so that either y or z would be the zero vector, which is not true. So at least one of y 1 and z 1 is not zero. Changing the names of y and z if necessary, we may assume that y 1 ̸ = 0. If both x 1 and y 1 are not zero, we may consider a = x 2 /x 1 and b = y 2 /y 1 (observe that a or b might be 0). [ x 1 , x 2 ] = x 1 [1 , a ] and [ y 1 , y 2 ] = y 2 [1 , b ], so that the orthogonality conditions being preserved and [1 , a ] , [1 , b ], and [ z 1 , z 2 ] are still mutually orthogonal: [1 , a ] . [ z 1 , z 2 ] = z 1 + az 2 = 0, and also [1 , b ] . [ z 1 , z 2 ] = z 1 + bz 2 = 0. From z 1 + az 2 = 0 and z 1 + bz 2 = 0, we get ( a b ) z 2 = 0, so that either a = b which is not possible since it would mean that x and y are parallel instead of being orthogonal, or z 2 = 0 which is the only possibility. But z 2 = 0 and z 1 + az 2 = z 1 + bz 2 = 0, would mean that z 1 = z 2 = 0 which is not possible since z ̸ = 0.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

M341F09HW2 - M341 F10 Homework 2 With the exception of 1.4...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online