Linear algebra: subspaces, linear independence, dimension

[morsel]

Homework Statement

1. Consider three linearly independent vectors v1, v2, v3 in Rn. Are the vectors v1, v1+v2, v1+v2+v3 linearly independent as well?

2. Consider a subspace V of Rn. Is the orthogonal complement of V a subspace of Rn as well?

3. Consider the line L spanned by
[1
2
3]. Find a basis of the orthogonal complement of L.

4. Consider a nonzero vector v in Rn. What is the dimension of the space of all vectors in Rn that are perpendicular to v?

Homework Equations

The Attempt at a Solution

1. I think since the three vectors are linearly independent, adding them together doesn't create redundancies. But this seems like an inadequate explanation..

2. The orthogonal complement of V is the kernel of V. Since the kernel is a subspace, the orthogonal complement is a subspace as well.

3. I was thinking of row reducing this matrix to get the orthogonal complement..
[1 0
2 0
3 0]

4. I don't even know where to start with this one..

Thanks in advance!

 

[Mark44]

Homework Statement

1. Consider three linearly independent vectors v1, v2, v3 in Rn. Are the vectors v1, v1+v2, v1+v2+v3 linearly independent as well?

2. Consider a subspace V of Rn. Is the orthogonal complement of V a subspace of Rn as well?

3. Consider the line L spanned by
[1
2
3]. Find a basis of the orthogonal complement of L.

4. Consider a nonzero vector v in Rn. What is the dimension of the space of all vectors in Rn that are perpendicular to v?

Homework Equations

The Attempt at a Solution

1. I think since the three vectors are linearly independent, adding them together doesn't create redundancies. But this seems like an inadequate explanation..

Yes, I agree about your explanation. A better explanation would be to show that v1, v1 + v2, and v1 + v2 + v3 are linearly independent.

2. The orthogonal complement of V is the kernel of V. Since the kernel is a subspace, the orthogonal complement is a subspace as well.

No the orthogonal complement of V is not the kernel of V. It's the set of all vectors in Rⁿ that are perpendicular to each vector in V.

3. I was thinking of row reducing this matrix to get the orthogonal complement..
[1 0
2 0
3 0]

The vector you are given is the basis of a one-dimensional subspace of R³. That means that the orthogonal complement of L is a two-dimensional subspace of R³. If you take an arbitrary vector in the orthogonal complement of L and dot it with your given vector, what should you get?

4. I don't even know where to start with this one..

I'm not sure you understand what orthogonal complement means. Look up the definition. This is related to what is being asked in #4.

 

[morsel]

Thanks for your help. I understand how to approach the other problems but I'm still unsure about #4.

Is the dimension n-1 because there's a free variable? In other words, n-1 number of leading 1's in the rref form?

 

[Mark44]

Thanks for your help. I understand how to approach the other problems but I'm still unsure about #4.

Is the dimension n-1 because there's a free variable? In other words, n-1 number of leading 1's in the rref form?

Yes, the dimension of the subspace of vectors in Rⁿ that are perpendicular to the given vector v is n - 1. No, it's not because there's a free variable - it's because there are n - 1 free variables.

What matrix in RREF form are you talking about? There is only one equation, and it has n variables. Do you know where this equation comes from?