# Linear algebra - finding bases

## Homework Statement

I have three vectors F, G and H and I want to find the basis for R^5 that contains the three vectors F, G and H.

## The Attempt at a Solution

I write the matrix A <F G H I>, where I is the 5x5 identitymatrix. I bring this matrix to rref, but which columns must I choose afterwards? And why?

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Oh yeah, I have to choose the columns before the rref that have leading 1's after rref - these columns are linearly independant and therefor span R^5.

HallsofIvy
Homework Helper
Hold on a moment- you were specifically asked for a basis of R5 that contains F, G, H. Assuming F, G, H did not all already have leading 1s then "the columns that have leading 1s" will not include them. Assuming F, G, H are independent, then you just need to find two vectors that are also independent of F, G, H and each other.

Hmm, F, G and H are:

F = (1,0,1,0,0)^T
G = (0,-1,-1,0,0)^T and
H = (0,0,0,-1,1)^T.

I have to find a basis B for R^5 which containts these three vectors. Is my approach in #2 wrong?

Perhaps I should add that when I do what I did in #2, I get leading 1s in the columns where F, G and H are - so they are part of the basis.

Here is one way to do it.

In $$\mathbb{R}^n$$ you an always define the scalar product

$$\vec{a}\cdot\vec{b}=\sum_{i=1}^n a_i\,b_i$$

Every basis element must be orthogonal to each over, if not then you can write one of them as linear combination of the others which is not possible because the basis vectors are independent.

Your vectors are written with respect to the normal basis $$\vec{e}_i$$ as

$$\vec{F}=\vec{e}_1+\vec{e}_3,\, \vec{G}=-\vec{e}_2-\vec{e}_3,\, \vec{H}=-\vec{e}_4+\vec{e}_5$$

Let now an arbitrary vector on $$\mathbb{R}^n$$

$$\vec{A}=\sum_{i=1}^n a_i \,\vec{e}_i$$

and consider the scalar products of $$\vec{A}$$ with the basis elements. We will end up with a simple system on $$a_i$$ from where you an read the desired vectors.

HallsofIvy
Homework Helper
Rainbow child's explanation, while correct, might be just a little misleading. It is always possible, given any basis for a vector space, to define an inner product in which the basis vectors are orthonormal, but "beginners" might mistake that for the "usual" dot product on Rn.

It does happen here, and I didn't notice it, that the F, G, and H given here are all orthogonal (though not of length 1) in the usual dot product. If you can find two vectors that are orthogonal (have zero dot product) to each of F, G, and H, and with each other, then you have a basis.

(0, 0, 0, 1, 1) leaps to mind immediately. Now can you find a 5th vector that has 0 dot product with (1,0,1,0,0), (0,-1,-1,0,0), (0,0,0,-1,1), and (0, 0, 0, 1, 1)?

Hmm, I am a little confused.

The five column-vectors that span R^5 must all be perpendicular to one another - so the dot-product must be zero for all the vectors (or all the different combinations)?

And (1,0,1,0,0) does have zero dot-prodcut with (0,-1,-1,0,0)?

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Rainbow child's explanation, while correct, might be just a little misleading. It is always possible, given any basis for a vector space, to define an inner product in which the basis vectors are orthonormal, but "beginners" might mistake that for the "usual" dot product on Rn. ...
That's why I began my post with

In $$\mathbb{R}^n$$ you an always define the scalar product... Hmm, I am a little confused.

The five column-vectors that span R^5 must all be perpendicular to one another - so the dot-product must be zero for all the vectors (or all the different combinations)?
In a general vector space there is no such thing as dot product although you can define on it. But from the postulates of a vector space you don't have a way to "multiply" vectors.

And (1,0,1,0,0) does have zero dot-prodcut with (0,-1,-1,0,0)?
HallsofIvy did a mistake with his calculations, it happens to everyone The vectors (1,0,1,0,0), (0,-1,-1,0,0) have dot-product -1 not zero.

If you carry the calculations within my previous post you will find the desired vectors

$$\vec{a}_1=\vec{e}_1+\vec{e}_2-\vec{e}_3, \quad \vec{a}_1=\vec{e}_4+\vec{e}_5$$

which of course is not the only choice.

HallsofIvy