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Class 25 : Question 1

Class 25 : Question 1. Which of the following vectors is orthogonal to the row space of A ?. Class 25 : Answer 1: A.

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Class 25 : Question 1

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  1. Class 25: Question 1 Which of the following vectors is orthogonal to the row space of A?

  2. Class 25: Answer 1: A The row space is a subspace of Rn, as is the nullspace. The two are orthogonal complements of each other which means any vector in one is orthogonal to vectors in the other. Thus one way to solve the problem is to find the nullspace of A, which ends up being span{(-1,-1,1)}. Thus the only vector in the 4 given which fits is (1,1,-1)

  3. Class 25: Question 2 Which of the following vectors is orthogonal to the column space of A?

  4. Class 25: Answer 2: C The column space is a subspace of Rm, as is the left nullspace. The two are orthogonal complements of each other which means any vector in one is orthogonal to vectors in the other. Since it’s easier to deal with the column space, let’s do that. The column space of A ends up being span{(1,0,0),(0,2,1)}. Thus we need to see which of the vectors is orthogonal to these two. One way would be to take the cross product, but it’s easier just to check the vectors we have Thus the only vector in the 4 given which fits is (0,1,-2)

  5. Class 25: Question 3 Which of the following vectors is orthogonal to the nullspace of A?

  6. Class 25: Answer 3: D This question is asking about vectors which are orthogonal to the nullspace, thus they must lie in the row space. The row space of A is clearly span{(1,0,1),(0,1,1)}. Thus we need to see which of the vectors is in this vector space. Thus the only vector in the 4 given which fits is (2,0,2)

  7. Class 25: Question 4 TRUE or FALSE: Any set of non-zero orthogonal vectors must be linearly independent. 1. TRUE.2. FALSE.

  8. Class 25: Answer 4: A In order to check whether a set of vectors is linearly independent we need to solve the homogeneous linear system Ax=0where the columns of A are the given vectors and x and 0are vectors. In other words: If the set of vectors is orthogonal to each other then if we take the dot product of both sides Thus c1=0 since these are nonzero vectors. The process can be repeated for all n vectors to show that all the constants must be zero and thus the set of orthogonal vectors is linearly independent.

  9. Class 25: Question 5 1. TRUE.2. FALSE.

  10. Class 25: Answer 5 (B) There are multiple bases for R2 so there must be multiple orthogonal bases for R2. All you need to do is pick two vectors that are not scalar multiples of each other, are orthogonal and then normalize them to produce multiple orthonormal bases in R2. Here are some more

  11. Class 25: Question 6 Let Q be a square matrix with orthonormal columns. TRUE or FALSE: Q-1=QT. 1. TRUE.2. FALSE.

  12. Class 25: Answer 6: A If Q is a square matrix with orthonormal columns then QTQ must equal the identity matrix. This is clear because if you multiply a matrix by it’s transpose then each element of the product consists of dot products of the rows of the matrix with it’s own columns. Since the matrix has orthonormal columns those dot products will produce either 1 or 0, which will result in the identity matrix. The product has 1 along the diagonal because that is when a particular column vector is being dotted with itself. Thus since QTQ=I that must mean that Q-1=QT.

  13. Class 25: Question 7

  14. Class 25: Answer 7 (A)

  15. Class 25: Question 8

  16. Class 25: Answer 8: A The key thing to understand about this question is that it is the same question as the one before! In other words, to make an orthogonal projection onto a subspace one needs to have a basis for that subspace. What is a basis for the subspace corresponding to the line y=x/2? The line corresponds to span{(2,1)}. Thus (2,1) is a basis for the subspace. Thus the orthogonal projection of (-3,1) onto (2,1) is (2,1).

  17. Class 25: Question 9

  18. Class 25: Answer 9: D The line named lis y=3x. Thisis the subspace span{(1,3)}. If zis the projection of b onto this subspace it will be some scalar multiple of (1,3), i.e. (c,3c). However, the more important part of the problem is the interpretation of b-z which equals b-projl(b). Recall that this new vector b-z will be orthogonal to lby definition. There’s no way b-z could be a point on land orthogonal to it. Thus only two of the statements are true.

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