Linear Dependence

**Roam** · September 29th 2009, 09:03 PM

Set $\alpha = \sqrt{2}$ and $\beta= \sqrt{3}$ . Note that $\alpha^2 = 2$ and $\beta^2 = 3$ . $Q(\alpha, \beta)$ is the 4 dimensional vector space of R having basis $\{1, \alpha, \beta, \alpha \beta \}$ .

(a) Show that real numbers $(\alpha, \beta)^2$ and $(\alpha, \beta)^4$ are both in space $Q(\alpha, \beta)$ by expanding the powers and expressing them as a linear combination of the basis vectors.

(b) Show that the vectors $\{(\alpha + \beta)^4), (\alpha+ \beta)^2,1\}$ are linearly dependent.

My Attempt:

(a) $(\alpha, \beta)^2 = \alpha^2 + 2 \alpha \beta + \beta^2 = 2+ 2 \alpha \beta +3$ , which can be written as:

$2.1 + 2 \alpha \beta +3 . 1$ (since $1 \in Q(\alpha, \beta)$ and $2,3 \in R$ )

Now, $(\alpha, \beta)^4 = \alpha^4 + 4 \alpha^3 \beta + 6 \alpha^2 \beta^2 + 4 \alpha \beta^3 + b^4$

$(2.2).1+(4.2) \alpha \beta + (6.2.3).1+(3.4)\alpha \beta + (4).1$ $= (4).1+(8) \alpha \beta + (46).1+(12) \alpha \beta + (4).1$

Is this right?

(b) This is the HARD part! I know that the vectors $\{(\alpha + \beta)^4), (\alpha+ \beta)^2,1\}$ are linearly dependent if there are $\lambda_1, \lambda_2, \lambda_3$ in reals, NOt all 0, such that

$\lambda_1 (\alpha + \beta)^4), \lambda_2 (\alpha+ \beta)^2,\lambda_3 1 = 0$

But I cant' figure out how to prove/represent this. Can anyone show me?

**HallsofIvy** · September 30th 2009, 08:47 AM

Originally Posted by Roam

Set $\alpha = \sqrt{2}$ and $\beta= \sqrt{3}$ . Note that $\alpha^2 = 2$ and $\beta^2 = 3$ . $Q(\alpha, \beta)$ is the 4 dimensional vector space of R having basis $\{1, \alpha, \beta, \alpha \beta \}$ .

(a) Show that real numbers $(\alpha, \beta)^2$ and $(\alpha, \beta)^4$ are both in space $Q(\alpha, \beta)$ by expanding the powers and expressing them as a linear combination of the basis vectors.

(b) Show that the vectors $\{(\alpha + \beta)^4), (\alpha+ \beta)^2,1\}$ are linearly dependent.

My Attempt:

(a) $(\alpha, \beta)^2 = \alpha^2 + 2 \alpha \beta + \beta^2 = 2+ 2 \alpha \beta +3$ , which can be written as:

$2.1 + 2 \alpha \beta +3 . 1$ (since $1 \in Q(\alpha, \beta)$ and $2,3 \in R$ )

Now, $(\alpha, \beta)^4 = \alpha^4 + 4 \alpha^3 \beta + 6 \alpha^2 \beta^2 + 4 \alpha \beta^3 + b^4$

$(2.2).1+(4.2) \alpha \beta + (6.2.3).1+(3.4)\alpha \beta + (4).1$ $= (4).1+(8) \alpha \beta + (46).1+(12) \alpha \beta + (4).1$

Is this right?

(b) This is the HARD part! I know that the vectors $\{(\alpha + \beta)^4), (\alpha+ \beta)^2,1\}$ are linearly dependent if there are $\lambda_1, \lambda_2, \lambda_3$ in reals, NOt all 0, such that

$\lambda_1 (\alpha + \beta)^4), \lambda_2 (\alpha+ \beta)^2,\lambda_3 1 = 0$

But I cant' figure out how to prove/represent this. Can anyone show me?

You just said that $(\alpha+ \beta)^2= 5+ \alpha\beta$ and that $(\alpha+ \beta)^4= 20+ 20\alpha\beta$ . (I didn't check to see if you were right!)

So you need to determine if there exist $\lambda_1$ , $\lambda_2$ , and $\lambda_3$ such that $\lambda_1(5+ \alpha\beta)+ \lambda_2(20+ 20\alpha\beta)+ \lambda_3= (5\lambda_1+ 20\lambda_2+ \lambda_3)+ (\lambda_1+ 20\lambda_2)\alpha\beta= 0$ . That gives you two equations for $\lambda_1$ , $\lambda_2$ , and $\lambda_3$ .

That gives you two equations

**Roam** · September 30th 2009, 11:52 AM

Originally Posted by HallsofIvy

You just said that $(\alpha+ \beta)^2= 5+ \alpha\beta$ and that $(\alpha+ \beta)^4= 20+ 20\alpha\beta$ . (I didn't check to see if you were right!)

No. I didn't say that in my last post.

So you need to determine if there exist $\lambda_1$ , $\lambda_2$ , and $\lambda_3$ such that $\lambda_1(5+ \alpha\beta)+ \lambda_2(20+ 20\alpha\beta)+ \lambda_3= (5\lambda_1+ 20\lambda_2+ \lambda_3)+ (\lambda_1+ 20\lambda_2)\alpha\beta= 0$ . That gives you two equations for $\lambda_1$ , $\lambda_2$ , and $\lambda_3$ .

That gives you two equations

I can't see where to get the two equations from (and those equations probably wouldn't be linear)?

**Roam** · October 1st 2009, 02:15 AM

What I meant was that $(\alpha+ \beta)^2$ simplifies to $5+2 \alpha\beta$ and also $(\alpha, \beta)^4 = \alpha^4 + 4 \alpha^3 \beta + 6 \alpha^2 \beta^2 + 4 \alpha \beta^3 + \beta^4$ simplifies to $4+8 \alpha \beta + 36 +12 \alpha \beta +9$ $= 49+20 \alpha \beta$ .

Now I want to show that for $\lambda_1, ... \lambda_n \in R$ not all zeros,

$<br /> \lambda_1 (5+2 \alpha\beta) + \lambda_2 (49+20 \alpha \beta) = 0<br />$

I don't know how this can be proved.

Thread: Linear Dependence

LinkBack

Thread Tools

Display