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

[karush]

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

 

[Opalg]

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 $V$?

 

[karush]

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:
$$u+v=v+u \text{ for all $u$ and $v \in V$}$$
$$c(u+v)=cu+cv \text{ for all real numbers $c$ and
for all $u$ and $v \in V$} $$

 

[Opalg]

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.

 

[Olinguito]

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)

 

[karush]

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

 

[HOI]

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?

 

[Olinguito]

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)

 

[karush]

https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy