Extending a Set of 3 Orthogonal Vectors to a Basis in R^5

[MathewsMD]

I've attached the question to this post. The answer is true, but I'm trying to figure out why.

Using Gram-Schmidt, I can only necessarily find 3 orthogonal vectors given 3 linearly independent vectors from $R^5$. How then is it possible to extend this set of 3 vectors that are linearly independent to form a basis of $R^5$? Unless I'm missing something here, the question is asking if this set can be extended to span ALL of $R^5$ (i.e. a basis of this space) and not just a subspace, correct? How exactly can 3 vectors do this? I believe recalling an analog to the cross product in $R^5$ but we surely did not cover this in our class. Any help would be greatly appreciated!

 

[mathman]

The point is that it is possible to find two vectors which added to the given three form a set of five linearly independent vectors. Example: (1,0,0,0,0) and (0,1,0,0,0) should work.

 

[statdad]

Have you checked to see whether these three vectors are independent? If they are, what would that tell you? (Hint: you should know a very simple basis for $\mathbb{R}^5$. If these are independent throw these vectors in with that basis and use your knowledge about how a basis works). Note that nothing in the question says you need to end up with an orthonormal basis at the end.

If you haven't checked to see if these are independent, do so first: Try starting
$$\begin{align*} x = \begin{bmatrix} 1 & 1 & 1 & 1 & 5 \end{bmatrix} & \\ y = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \end{bmatrix} & \\ z = \begin{bmatrix} 1 & 4 & 9 & 16 & 25 \end{bmatrix} \end{align*}$$

Then see whether the equation $ay + bz = x$ has any non-trivial solutions.

 

[MathewsMD]

The point is that it is possible to find two vectors which added to the given three form a set of five linearly independent vectors. Example: (1,0,0,0,0) and (0,1,0,0,0) should work.

Ahh, ok. Thank you. I misinterpreted what they were asking.

 

[blue_raver22]

I can provide some insights into why this is possible. First, let's define what a basis is. A basis for a vector space is a set of linearly independent vectors that span the entire space. In other words, every vector in the space can be written as a linear combination of the basis vectors.

In this case, we are dealing with $R^5$, which is a 5-dimensional vector space. This means that any vector in this space can be written as a linear combination of 5 basis vectors. So, even though you can only find 3 orthogonal vectors using Gram-Schmidt, it is possible to extend this set to span the entire space using the remaining 2 vectors.

You are correct in thinking about the cross product in $R^5$. The cross product is a way to find a vector that is orthogonal to two other vectors in $R^3$. In a similar way, we can use a generalized cross product to find vectors that are orthogonal to a set of vectors in higher dimensions, including $R^5$. This can help us extend our set of 3 orthogonal vectors to span the entire space.

In summary, while it may seem counterintuitive that 3 vectors can form a basis for a 5-dimensional space, it is possible to extend this set using techniques such as the generalized cross product. This is a fundamental concept in linear algebra and is essential for understanding higher-dimensional vector spaces. I hope this helps clarify the situation for you.