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The Question

If ${x_1,x_2}$ is a basis for a subspace $X$show that ${x_1 +x_2,x_1 −x_2}$ is also a basis for X .


So what I thought was to show that $x_1+x_2$ , $x_1-x_2$ is a basis. How I did it:

Since $x_1$, $x_2$ is a valid subspace of X, then it must obey the three conditions: contain zero, closed under addition, closed under scalar multiplication.

  1. Since $x_1$, $x_2$ contain the zero vector, by $x_1=0$, $x_2=0$ This means that $x_1+x_2=0+0=0$ and $x_1-x_2=0-0=0$. Therefore it is contains zero.

  1. Since $x_1$, $x_2$ are closed under addition this means that we know that $x_1+x_2$ should belong to X. And since $x_1,x_2$ are closed under scalar multiplication this means that $(-1)x_2$ belongs to the subspace. Hence, $x_1-x_2$ belongs to the subspace since $x_1,-x_2$ are closed under addition.

  1. I am uncertain about how to do show this is closed under scalar multiplication.

Is my work correct, and isn't there a simpler less confusing way to prove this than my method? And how do I do the last part?

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  • $\begingroup$ This is a bit sloppy, $x_1,x_2$ are fixed vectors, they can't be set equal to $0$. $\endgroup$
    – fretty
    Commented Aug 9, 2015 at 11:35
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    $\begingroup$ ...also this is not true over a field of characteristic $2$! $\endgroup$
    – fretty
    Commented Aug 9, 2015 at 11:36
  • $\begingroup$ Also, you don't need to show about anything that it is closed under scalar multiplication (or addtion, or ...) $\endgroup$ Commented Aug 9, 2015 at 11:36
  • $\begingroup$ @HagenvonEitzen then? $\endgroup$
    – M.S.E
    Commented Aug 9, 2015 at 11:37

3 Answers 3

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$x_{1}$ and $x_{2}$ is a basis for $X$ means that any $v \in X$ can be written as a linear combination of $x_{1}$ and $x_{2}$, and that $x_{1}$ and $x_{2}$ are linearly independant. We need to show that $x_{1}+x_{2}$ and $x_{1}-x_{2}$ are linearly independent. Assume there are real numbers (I am assuming $X$ is a real vector space) $a$ and $b$ such that $a(x_{1}+x_{2})+b(x_{1}-x_{2})=\vec{0}$, then $(a+b)x_{1}+(a-b)x_{2}=\vec{0}$. Then since $x_{1}$ and $x_{2}$ are linearly independent, $a+b=0$ and $a-b=0$ so $2a=0$ (adding both) so $a=0$, so $b=0$. Now, $x_{2}=\frac{1}{2}[(x_{1}+x_{2})-(x_{1}-x_{2})]$ and $x_{1}=\frac{1}{2}((x_{1}+x_{2})+(x_{1}-x_{2})$. So any $v \in X$ can be written in terms of the new vectors!

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  • $\begingroup$ (+1) "$x_{2}=\frac{1}{2}[(x_{1}+x_{2})-(x_{1}-x_{2})]$ and $x_{1}=\frac{1}{2}((x_{1}+x_{2})+(x_{1}-x_{2})$" What does this mean? Like I can't understand why you wrote $x_1$ and $x_2$ in this form? Also, Thanks. $\endgroup$
    – M.S.E
    Commented Aug 9, 2015 at 11:45
  • $\begingroup$ Oh I got it, THANK YOU! $\endgroup$
    – M.S.E
    Commented Aug 9, 2015 at 11:48
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What is a basis? It is a set of vectors that span a vector space and are linearly independent.

So we must show that if $x_1,x_2$ have these properties then $x_1+x_2, x_1-x_2$ also have them.

Linear Independence:

Suppose $\alpha(x_1+x_2) + \beta(x_1-x_2) = 0$ for some $\alpha,\beta$ in the underlying field.

Then $(\alpha+\beta)x_1 + (\alpha-\beta)x_2 = 0$.

But AHAAA, we know that $x_1,x_2$ are linearly independent, so we are forced to have $\alpha+\beta=\alpha-\beta=0$. Solving gives $\alpha=\beta=0$.

Span:

We must show that every $x\in V$ can be written as $\gamma(x_1+x_2) + \delta(x_1-x_2)$ for some $\gamma,\delta$ in the underlying field.

Now we know for sure that we can find $\gamma',\delta'$ such that $\gamma'x_1 + \delta'x_2 = x$. Then notice that:

$\frac{\gamma'+\delta'}{2}(x_1+x_2) + \frac{\gamma'-\delta'}{2}(x_1-x_2) = x$ and so we are done.

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  • $\begingroup$ (+1) Thank you. However, I dont understand why you wrote in terns of the first derivative (the last part). $\endgroup$
    – M.S.E
    Commented Aug 9, 2015 at 11:49
  • $\begingroup$ I didn't...it was just notation. $\endgroup$
    – fretty
    Commented Aug 9, 2015 at 11:51
  • $\begingroup$ oh now i see. thank you very much :) $\endgroup$
    – M.S.E
    Commented Aug 9, 2015 at 12:04
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The elements of the second basis can clearly be written as linear combinations of your first basis. Since the subspace has dimension 2, all you need to show is that the elements in the second basis are linearly independent. Hint: you know that ${x_{1},x_{2}}$ are linearly independent.

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