direct sum question

**jayshizwiz** · July 27th 2010, 11:51 PM

Given two sets
$U = \{p(x) \in R_4[x] { | } p(2) = p(-2) = 0\}$
$W = \{p(x) \in R_4[x] { | } p(1) = p(-1) = p(0) \}$

Prove $R_4[x] = U \oplus W$

Can someone please give some direction on how to solve this.
Thanks!

**tonio** · July 28th 2010, 02:56 AM

Originally Posted by jayshizwiz

Given two sets
$U = \{p(x) \in R_4[x] { | } p(2) = p(-2) = 0\}$
$W = \{p(x) \in R_4[x] { | } p(1) = p(-1) = p(0) \}$

Prove $R_4[x] = U \oplus W$

Can someone please give some direction on how to solve this.
Thanks!

1) Prove first that $\dim U=3\,,\,\,\dim W=2$

2) Now prove that $U\cap W=\{0\}$

Tonio

**jayshizwiz** · July 28th 2010, 03:13 AM

Thanks. I'll try to prove that. But I'm blanking out...

Why is dimR_4[x]=5?

I'm assuming you meant in your reply that dimU + dimW = 5 = dimR_4[x]

**HallsofIvy** · July 28th 2010, 03:17 AM

Originally Posted by jayshizwiz

Thanks. I'll try to prove that. But I'm blanking out...

Why is dimR_4[x]=5?

I'm assuming you meant in your reply that dimU + dimW = 5 = dimR_4[x]

Because $R_4[x]$ is the space of fourth degree polynomials with variable x: any member is of the form $ax^4+ bx^3+ cx^2+ dx+ e$ which has the 5 polynomials, $x^4$ , $x^3$ , $x^2$ , $x$ , and $1$ , as basis.

**tonio** · July 28th 2010, 03:56 AM

Originally Posted by jayshizwiz

Thanks. I'll try to prove that. But I'm blanking out...

Why is dimR_4[x]=5?

I'm assuming you meant in your reply that dimU + dimW = 5 = dimR_4[x]

If you didn't know that then perhaps it's be a good idea to first learn a little more about polynomials spaces:
Linear Algebra: More Vector Spaces; Isomorphism - CliffsNotes

Tonio

**jayshizwiz** · July 28th 2010, 04:36 AM

Thanks, that's a great website.

I'm just having one of those days where I seem to forget everything I've learned in Linear Algebra...no biggie

**jayshizwiz** · July 29th 2010, 04:36 AM

Originally Posted by tonio

1) Prove first that $\dim U=3\,,\,\,\dim W=2$

2) Now prove that $U\cap W=\{0\}$

Tonio

But first don't I have to prove that $R_4[x] = U + W$ ??? That's the hard part and I have no idea how to do that.

**jayshizwiz** · October 5th 2010, 05:22 AM

Originally Posted by HallsofIvy

Because $R_4[x]$ is the space of fourth degree polynomials with variable x: any member is of the form $ax^4+ bx^3+ cx^2+ dx+ e$ which has the 5 polynomials, $x^4$ , $x^3$ , $x^2$ , $x$ , and $1$ , as basis.

Actually, the way I am learning it is that $R_4[x]$ = $\alpha_0+\alpha_1x+\alpha_2x^2+\alpha_3x^3$ . It doesn't go to the 4th power. So dim $R_4[x]$ =4

Long story short, I proved $R_4[x]$ = U+W.

U=(4s, 4t, -s, -t)
W=(i, j, 0, -j)

How do I find dimU and dimV? Or how do I find a basis for U and for W?

(and I'm pretty sure dimU=dimV=2 but I don't know how to prove it)

**I think I got it...

U=(4s,4t,-s,-t) = (4s, 0, -s, 0) + (0, 4t, 0, -t) $\rightarrow$ dimU=2
W=(i, j, 0, -j) = (i, 0, 0, 0) + (0, j, 0, -j) $\rightarrow$ dimW = 2

dimU + dimW = 4 = dim $R_4[x]$

Thread: direct sum question

LinkBack

Thread Tools

Display

direct sum question

Similar Math Help Forum Discussions

Direct Sum Question

Direct sum question

Direct Sum of Subspaces - Question

subgroup question with direct product

Direct Proof Question

Search Tags