# Given basis spanning the vector space

I am learning Linear algebra nowadays. I had a small doubt and I know it's an easy one. But still I am not able to get it. Recently I came across a statement saying "((1,2),(3,5)) is a basis of $F^2$ ".

According to the statement a linear combination of the vectors in the list,i.e., $a(1,2)+b(3,5)$ (where a and b belong to F of course) must span the vector space $F^2$. I wanted to know how we can get all the vectors in the vector space using linear combination of these two given vectors here (if the two vectors were ((1,0),(0,1)) which is the standard basis, it would have been fine).

But how can we get all the points in the vector space $F^2$ using ((1,2),(3,5))??? suppose I want to get the vector (1,1). I can't think of getting it using linear combination of the given two vectors.

Kindly help me with this. I know its an easy one but still could not help and had to ask you guys. Thanks.

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## 4 Answers

A basis does not need to have orthogonal vectors in order to allow other vectors in the space to be expressed as linear combinations (they just can't be linearly dependent). To construct a vector in your $F^2$ , you would need to solve the equation $\ \langle x , y \rangle \ = \ a \langle 1 , 2 \rangle + b \langle 3 , 5 \rangle \ ,$ which, for your example, is equivalent to solving the set of linear equations

$$a \cdot 1 \ + \ b \cdot 3 \ = \ 1 \ \ \text{and} \ \ a \cdot 2 \ + \ b \cdot 5 \ = \ 1 \ .$$

This gives us $\ a = -2 , b = 1 ,$ and thus $\ \langle 1,1 \rangle \ = \ (-2) \cdot \langle 1,2 \rangle \ + \ \langle 3,5 \rangle \ .$

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yes...right...got it now..thanks :)) – under-root Jun 6 '13 at 22:13

So we want to represent, say $(1,1)$ as a linear combination of the given vectors, that is we are to find $a,b \in \mathbb F$ satisfying \begin{align*} 1a + 3b &= 1\\ 2a + 5b &= 1 \end{align*} which are just the two components of $a(1,2) + b(3,5) = (1,1)$. So we must solve this linear system, subtracting equation (1) two times from the second, we get \begin{align*} 1a + 3b &= 1\\ -b &= -1 \end{align*} This gives $b = 1$ and $a = 1 -3b = -2$. That is $(1,1) = -2\cdot (1,2) + (3,5)$. For a general vector $(x,y) \in \mathbb F^2$ you can do it along the same lines, I'm sure.

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Note that if you can get to $(1,0)$ and $(0,1)$ you can get anywhere. So, can you find linear combinations $a(1,2) + b(3,5)$ that get you to these two vectors? Note that you can expand the equation, say, $a(1,2) + b(3,5) = (1,0)$ into two equations with two unknowns by looking at each coordinate separately.

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$(1,1)=a(1,2)+b(3,5)\implies a+3b=1, 2a+5b=1\implies a=-2, b=1$. Since $(1,2)$ and $(3,5)$ are linearly independent, they form a basis, ie. any vector in $F^2$ can be formed by linear combinations of them.

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thank u for your feedback... i should have thought about it more...its so easy and cleared everything.. thanks... – under-root Jun 6 '13 at 22:13