Assignment 2

Jet Hughes 9474308

1.

Let V = P_2(\mathbb{R}) with the ususal vector addition and scalar multiplication. For each of the following subsets of V, either prove that it is a basis of V or explain why it is not a basis of V . You may use any result from class.

(a) \{759, 20+2x+43x^2\}
(b) \{1-x, 2x^{2},3+x^2\}
(c) \{2x, 4+2x-x^{2}, -4-6x+x^2\}
(d) \{-1+3x, 1+x^{2,}x-3x^{2,}4+4x-11x^2\}

Since P_2(\mathbb{R}) has dimension 3, by default all bases of P_2(\mathbb{R}) have three elements. Hence (a) and (d) cannot possibly be bases of P_2(\mathbb{R})

(b) Since this set has 3 vectors and P_2(\mathbb{R}) has dimension 3, it is enough to check either that is is linearly independent or that it spans P_2(\mathbb{R}). To show linear independence, if a(1-x)+b(2x^2)+c(3+x^2)=(0x^2 + 0x + 0), we have 2b+c=0, -a=0 and a+3c = 0. So a=0 which implies c=0 which then implies b=0. So the only linear combination equal to the zero vector is the one where a=b=c=0, hence this set in linearly independent. Since it is linearly independent and has three vectors, its span is a subspace of P_2(\mathbb{R}) of dimension 3, i.e., all of P_2(\mathbb{R})

(c) Since this set has 3 vectors and P_2(\mathbb{R}) has dimension 3, it is enough to check either that is is linearly independent or that it spans P_2(\mathbb{R}). To show linear independence, if a(0 +2x+0x^2)+b(4+2x-x^2)+c(-4-6x+x^2)=(0+0x+0x^2), we have 4b-4c=0, 2a+2b-6c=0 and -b+c = 0. So b=c which implies -c+c=0 and 2a+2c-6c=0 so a=2c=2b. But this does no force a,b,c t obe zero; we could have e.g., a=2 and b=c=1. Therefore we have a linear combination of the vectors that gives the zero vector when the coefficients are not all zero. So it is linearly dependent, and therefore can't be a basis.

2.

Let V be a vector space such that dim(V)= 10. Let U and W denote subspaces of V.

(a). What is the max value of dim(U\cap W), assuming dim(U)=7 and dim(W)=4 ?

The largest possible dimension of (U\cap W) is 4 which occurs when W is entirely contained in U.

(b). What is the min value of dim(U+W)

dim(U\cap W) \leq 4 so from the formula dim(U+W) = dim(U) + dim(W) - dim(U \cap W) we have dim(U+W) \geq 7 + 4 - 4= 7.

(c). Max value of dim(U+W)

We have dim(U+W) \leq dim(U) + dim(W) = 11. But since U+W is a subspace of V which has dimension 10 , we must have dim(U+W) \leq 10. To show that this can be achieved, we could let \{v_1,...,v_{10}\} be a basis for V, and choose U = span\{v_1, v_2, v_3, v_4, v_5, v_5, v_7)\} and W=span\{v_7,v_8,v_9,v_{10}\}. Then U+W will contain the basis for V and hence U+W=V. So the maximum possible value of dim(U+W) is 10.

(d). Since dim(U+W) \leq 10 and dim(U+W) = 11 - dim(U\cap W), we have dim(U\cap W) \geq 11-10= 1

(c). No, because if U+W is direct then dim(U+W) = 0, but we showed that dim(U\cap W) \geq 1

3.

Let U = \{p\in P_2(\mathbb{R}): p(x) is divisible by x-3\}. Then U is a subspace of P2 (a) Find a basis of U.

Suppose a polynomial of degree 2 is divisible by x-3 . It can be written as (x-3)(ax+b) where a,b \in \mathbb{R}. Then we have a basis \{(x-3), x(x-3)\}

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Assignment 2

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