# 307.8.1 Suppose Y_1 and Y_2 form a basis for a 2-dimensional vector space V

• MHB
• karush
In summary, In order to create a basis for a two-dimensional space, vectors must be linearly independent.
karush
Gold Member
MHB
nmh{796}
$\textsf{Suppose$Y_1$and$Y_2$form a basis for a 2-dimensional vector space$V$.}\\$
$\textsf{Show that the vectors$Y_1+Y_2$and$Y_1−Y_2$are also a basis for$V$.}$
$$Y_1=\begin{bmatrix}a\\b\end{bmatrix} \textit{ and }Y_2=\begin{bmatrix}c\\d\end{bmatrix}$$
$\textit{ then }$
$$Y_1+Y_2 = \begin{bmatrix}a\\b\end{bmatrix} -\begin{bmatrix}c\\d\end{bmatrix} =\begin{bmatrix}a+c\\b+d\end{bmatrix}$$
$$Y_1-Y_2 = \begin{bmatrix}a\\b\end{bmatrix} -\begin{bmatrix}c\\d\end{bmatrix} =\begin{bmatrix}a-c\\b-d\end{bmatrix}$$
so far

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It's probably not helpful to write $Y_1$ and $Y_2$ in terms of coordinates. Instead, you could start by trying to show that $Y_1 + Y_2$ and $Y_1 - Y_2$ are linearly independent. Suppose in fact that $\alpha (Y_1 + Y_2) + \beta (Y_1 - Y_2) = 0$. Then $(\alpha + \beta)Y_1 + (\alpha-\beta)Y_2 = 0$. Can you take it from there to show that $\alpha = \beta = 0$? And can you deduce from this that $Y_1 + Y_2$ and $Y_1 - Y_2$ form a basis for $V$?

Opalg said:
It's probably not helpful to write $Y_1$ and $Y_2$ in terms of coordinates. Instead, you could start by trying to show that $Y_1 + Y_2$ and $Y_1 - Y_2$ are linearly independent. Suppose in fact that $\alpha (Y_1 + Y_2) + \beta (Y_1 - Y_2) = 0$. Then $(\alpha + \beta)Y_1 + (\alpha-\beta)Y_2 = 0$. Can you take it from there to show that $\alpha = \beta = 0$? And can you deduce from this that $Y_1 + Y_2$ and $Y_1 - Y_2$ form a basis for $V$?
ok i assume you are referring to vector space properties of:
$$u+v=v+u \text{ for all u and v \in V}$$
$$c(u+v)=cu+cv \text{ for all real numbers c and for all u and v \in V}$$

karush said:
ok i assume you are referring to vector space properties of:
$$u+v=v+u \text{ for all u and v \in V}$$
$$c(u+v)=cu+cv \text{ for all real numbers c and for all u and v \in V}$$
Those properties are certainly needed. But the essential idea that I was hinting at is that the set of vectors in a basis must be linearly independent. Conversely, any two linearly independent vectors in a two-dimensional space form a basis.

The fact that $V$ is a dimension-$2$ vector space is a great help: you just need to prove that $Y_1+Y_2$ and $Y_1-Y_2$ are linearly independent for them to form a basis. If $V$ is of dimension $n>2$, they will still form a basis, provided the field the space $V$ is over does not have characteristic $2$; in this case, $Y_1+Y_2$ and $Y_1-Y_2$ also span $V$.

We can easily prove this – but your first task is to prove that $Y_1+Y_2$ and $Y_1-Y_2$ are linearly independent. (Smile)

Olinguito said:
The fact that $V$ is a dimension-$2$ vector space is a great help: you just need to prove that $Y_1+Y_2$ and $Y_1-Y_2$ are linearly independent for them to form a basis. If $V$ is of dimension $n>2$, they will still form a basis, provided the field the space $V$ is over does not have characteristic $2$; in this case, $Y_1+Y_2$ and $Y_1-Y_2$ also span $V$.

We can easily prove this – but your first task is to prove that $Y_1+Y_2$ and $Y_1-Y_2$ are linearly independent. (Smile)
View attachment 8442

got this from the book
what I see is if $Y_1+Y_2=0$ then $Y_1-Y_2\ne 0$ unless their scalars are 0

so $Y_1+Y_2=0$ in Linearly Independent and $Y_1-Y_2\ne 0$ in not

Then you do not understand what "linearly independent" means!

You are saying that one vector, $Y_1+ Y_2$, is linearly independent while another, $Y_1- Y_2$, is not. That makes NO SENSE. In particular, it makes no sense to talk about a single vector being "independent" or not. Sets of vectors may or may not be linearly independent, not individual vectors. (And a "singleton set", {v}, with v a non-zero vector, is always linearly independent.)

To determine whether the two vectors $Y_1+ Y_2$ and $Y_1- Y_2$ (the set of vectors $\{Y_1+ Y_2, Y1- Y_2\}$) are independent, look at the equation $a(Y_1+ Y_2)+ b(Y_1- Y_2)= 0$. If a and b must both b 0 then $Y_1+ Y_2$ and $Y_1- Y_2$ are linearly independent. If either a or b can be non-zero and that equation still true, they dependent.

We can write $a(Y_1+ Y_2)+ b(Y_1- Y_2)= aY_1+ aY_2+ bY_1- bY2= (a+ b)Y_1+ (a- b)Y_2= 0$. We know that $Y_1$ and $Y_2$ are independent so that we must have their coefficients equal to 0: a+ b= 0 and a- b= 0. What can you conclude about a and b?

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Olinguito said:
The fact that $V$ is a dimension-$2$ vector space is a great help: you just need to prove that $Y_1+Y_2$ and $Y_1-Y_2$ are linearly independent for them to form a basis. If $V$ is of dimension $n>2$, they will still form a basis, provided the field the space $V$ is over does not have characteristic $2$; in this case, $Y_1+Y_2$ and $Y_1-Y_2$ also span $V$.
Sorry, that’s not correct. I made a mistake. Ignore what I said. (Worried)

https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy

## 1. What is a basis for a vector space?

A basis for a vector space is a set of vectors that can be used to represent any other vector in that space through linear combinations.

## 2. How many vectors are needed to form a basis for a 2-dimensional vector space?

Two vectors are needed to form a basis for a 2-dimensional vector space.

## 3. Can any two vectors form a basis for a 2-dimensional vector space?

No, the two vectors must be linearly independent in order to form a basis.

## 4. What is the significance of a basis in a vector space?

A basis provides a framework for understanding and working with vectors in a vector space. It allows for the representation of any vector in the space and simplifies calculations and proofs.

## 5. How can we determine if two vectors form a basis for a 2-dimensional vector space?

We can determine if two vectors form a basis by checking if they are linearly independent and span the entire 2-dimensional vector space. This can be done by solving a system of equations using the vectors as coefficients.

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