<Home>Teaching><MATH 115A (Spring 2023)>Worksheet 1

Worksheet 1

Problem 3

Let V=F[x1,…,xn]V = \F\br{x_1, \ldots, x_n} be the space of polynomials in nn-variables. A polynomial f∈Vf \in V is said to be homogeneous of degree kk if the degree (i.e. the sum of the exponents) of each non-zero term of ff is kk. A polynomial f∈Vf \in V is said to be symmetric if exchanging any two variables yields the same polynomial, namely f(x1,…,xi,…xj,…,xn)=f(x1,…,xj,…,xi,…,xn)f\p{x_1, \ldots, x_i, \ldots x_j, \ldots, x_n} = f\p{x_1, \ldots, x_j, \ldots, x_i, \ldots, x_n} for all i,j∈{1,…,n}i, j \in \set{1, \ldots, n}.

  1. Prove that VV is a vector space.
  2. Prove that the space of homogeneous degree kk polynomials form a subspace of VV.
  3. Prove that the space of symmetric polynomials in nn variables is a subspace of VV.
Solution.

First, note that addition and scalar multiplication are defined as follows. Let

f=∑i1,i2…,in=0∞ai1i2⋯inx1i1⋯xnin,g=∑i1,i2…,in=0∞bi1i2⋯inx1i1⋯xnin, f = \sum_{i_1,i_2\ldots,i_n=0}^\infty a_{i_1 i_2 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n}, \quad g = \sum_{i_1,i_2\ldots,i_n=0}^\infty b_{i_1 i_2 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n},

where ai1i2⋯in,bi1i2⋯ina_{i_1 i_2 \cdots i_n}, b_{i_1 i_2 \cdots i_n} are zero except for only finitely many indices (i1,i2,…,in)\p{i_1, i_2, \ldots, i_n}, and c∈Fc \in \F. Then

f+g≔∑i1,i2…,in=0∞(ai1i2⋯in+bi1i2⋯in)x1i1⋯xnincf≔∑i1,i2…,in=0∞(cai1i2⋯in)x1i1⋯xnin.\begin{aligned} f + g &\coloneqq \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{a_{i_1 i_2 \cdots i_n} + b_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} \\ cf &\coloneqq \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{ca_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n}. \end{aligned}
  1. We just need to check the axioms.

(VS 1) Let f,g∈Vf, g \in V be as above. Then

f+g=∑i1,i2…,in=0∞(ai1i2⋯in+bi1i2⋯in)x1i1⋯xnin=∑i1,i2…,in=0∞(bi1i2⋯in+ai1i2⋯in)x1i1⋯xnin((F,+) is commutative)=g+f.\begin{aligned} f + g &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{a_{i_1 i_2 \cdots i_n} + b_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{b_{i_1 i_2 \cdots i_n} + a_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} && \p{\p{\F, +} \text{ is commutative}} \\ &= g + f. \end{aligned}

(VS 2) Let f,g∈Vf, g \in V be as above, and let

h=∑i1,i2…,in=0∞ci1i2⋯inx1i1⋯xnin.h = \sum_{i_1,i_2\ldots,i_n=0}^\infty c_{i_1 i_2 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n}.

Then

(f+g)+h=∑i1,i2…,in=0∞(ai1i2⋯in+bi1i2⋯in)x1i1⋯xnin+∑i1,i2…,in=0∞ci1i2⋯inx1i1⋯xnin=∑i1,i2…,in=0∞((ai1i2⋯in+bi1i2⋯in)+ci1i2⋯in)x1i1⋯xnin=∑i1,i2…,in=0∞(ai1i2⋯in+(bi1i2⋯in+ci1i2⋯in))x1i1⋯xnin((F,+) is associative)=∑i1,i2…,in=0∞ai1i2⋯inx1i1⋯xnin+∑i1,i2…,in=0∞(bi1i2⋯in+ci1i2⋯in)x1i1⋯xnin=f+(g+h).\begin{aligned} \p{f + g} + h &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{a_{i_1 i_2 \cdots i_n} + b_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} + \sum_{i_1,i_2\ldots,i_n=0}^\infty c_{i_1 i_2 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{\p{a_{i_1 i_2 \cdots i_n} + b_{i_1 i_2 \cdots i_n}} + c_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{a_{i_1 i_2 \cdots i_n} + \p{b_{i_1 i_2 \cdots i_n} + c_{i_1 i_2 \cdots i_n}}} x_1^{i_1} \cdots x_n^{i_n} && \p{\p{\F, +} \text{ is associative}} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty a_{i_1 i_2 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n} + \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{b_{i_1 i_2 \cdots i_n} + c_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} \\ &= f + \p{g + h}. \end{aligned}

(VS 3) The zero vector is

0V=∑i1,i2…,in=0∞0Fx1i1⋯xnin.0_V = \sum_{i_1,i_2\ldots,i_n=0}^\infty 0_\F x_1^{i_1} \cdots x_n^{i_n}.

Indeed, given f∈Vf \in V as above,

f+0V=∑i1,i2…,in=0∞(ai1i2⋯in+0F)x1i1⋯xnin=∑i1,i2…,in=0∞ai1i2⋯inx1i1⋯xnin(0F is the additive identity in F)=f.\begin{aligned} f + 0_V &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{a_{i_1 i_2 \cdots i_n} + 0_\F} x_1^{i_1} \cdots x_n^{i_n} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty a_{i_1 i_2 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n} && \p{0_\F \text{ is the additive identity in }\F} \\ &= f. \end{aligned}

(VS 4) Given f∈Vf \in V as above, note that −ai1i2⋯in∈F-a_{i_1 i_2 \cdots i_n} \in \F since F\F is a field. Thus, if we set

g=∑i1,i2…,in=0∞(−ai1i2⋯in)x1i1⋯xnin,g = \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{-a_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n},

then

f+g=∑i1,i2…,in=0∞(ai1i2⋯in+(−ai1i2⋯in))x1i1⋯xnin=∑i1,i2…,in=0∞0Fx1i1⋯xnin=0V.\begin{aligned} f + g &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{a_{i_1 i_2 \cdots i_n} + \p{-a_{i_1 i_2 \cdots i_n} }} x_1^{i_1} \cdots x_n^{i_n} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty 0_\F x_1^{i_1} \cdots x_n^{i_n} \\ &= 0_V. \end{aligned}

(VS 5) Given f∈Vf \in V as above,

1Ff=∑i1,i2…,in=0∞(1Fai1i2⋯in)x1i1⋯xnin=∑i1,i2…,in=0∞ai1i2⋯inx1i1⋯xnin(1F is the multiplicative identity in F)=f.\begin{aligned} 1_\F f &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{1_\F a_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty a_{i_1 i_2 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n} && \p{1_\F \text{ is the multiplicative identity in }\F} \\ &= f. \end{aligned}

(VS 6) Let f∈Vf \in V be as above, and b,c∈Fb, c \in \F. Then

(bc)f=∑i1,i2…,in=0∞((bc)ai1i2⋯in)x1i1⋯xnin=∑i1,i2…,in=0∞(b(cai1i2⋯in))x1i1⋯xnin((F, ⋅ ) is associative)=b∑i1,i2…,in=0∞(cai1i2⋯in)x1i1⋯xnin=b(cf).\begin{aligned} \p{bc}f &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{\p{bc} a_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{b \p{ca_{i_1 i_2 \cdots i_n}}} x_1^{i_1} \cdots x_n^{i_n} && \p{\p{\F, \,\cdot\,} \text{ is associative}} \\ &= b \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{ca_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} \\ &= b\p{cf}. \end{aligned}

(VS 7) Let f,g∈Vf, g \in V be as above and c∈Fc \in \F. Then

c(f+g)=c∑i1,i2…,in=0∞(ai1i2⋯in+bi1i2⋯in)x1i1⋯xnin=∑i1,i2…,in=0∞(c(ai1i2⋯in+bi1i2⋯in))x1i1⋯xnin=∑i1,i2…,in=0∞(cai1i2⋯in+cbi1i2⋯in)x1i1⋯xnin((F,+, ⋅ ) is distributive)=∑i1,i2…,in=0∞(cai1i2⋯in)x1i1⋯xnin+∑i1,i2…,in=0∞(cbi1i2⋯in)x1i1⋯xnin=cf+cg.\begin{aligned} c\p{f + g} &= c\sum_{i_1,i_2\ldots,i_n=0}^\infty \p{a_{i_1 i_2 \cdots i_n} + b_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{c\p{a_{i_1 i_2 \cdots i_n} + b_{i_1 i_2 \cdots i_n}}} x_1^{i_1} \cdots x_n^{i_n} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{ca_{i_1 i_2 \cdots i_n} + cb_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} && \p{\p{\F, +, \,\cdot\,}\text{ is distributive}} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{ca_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} + \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{cb_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} \\ &= cf + cg. \end{aligned}

(VS 8) Let f∈Vf \in V be as above and b,c∈Fb, c \in \F. Then

(b+c)f=∑i1,i2…,in=0∞((b+c)ai1i2⋯in)x1i1⋯xnin=∑i1,i2…,in=0∞(bai1i2⋯in+cai1i2⋯in)x1i1⋯xnin((F,+, ⋅ ) is distributive)=∑i1,i2…,in=0∞(bai1i2⋯in)x1i1⋯xnin+∑i1,i2…,in=0∞(cai1i2⋯in)x1i1⋯xnin=bf+cf.\begin{aligned} \p{b + c}f &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{\p{b + c}a_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{ba_{i_1 i_2 \cdots i_n} + ca_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} && \p{\p{\F, +, \,\cdot\,}\text{ is distributive}} \\ &= \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{ba_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} + \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{ca_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} \\ &= bf + cf. \end{aligned}

  1. There are three things to check. Let f,g∈Vf, g \in V be as above, c∈Fc \in \F, and we further assume that f,gf, g are homogeneous. Then 0V0_V is homogeneous of degree kk vacuously (there are no non-zero terms to check the condition on).

    Next, let's show that f+gf + g are homogeneous of degree kk. Recall that

    f+g=∑i1,i2…,in=0∞(ai1i2⋯in+bi1i2⋯in)x1i1⋯xnin.f + g = \sum_{i_1,i_2\ldots,i_n=0}^\infty \p{a_{i_1 i_2 \cdots i_n} + b_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n}.

    We need to show that if (ai1i2⋯in+bi1i2⋯in)x1i1⋯xnin\p{a_{i_1 i_2 \cdots i_n} + b_{i_1 i_2 \cdots i_n}} x_1^{i_1} \cdots x_n^{i_n} is a non-zero term, then i1+i2+⋯+in=ki_1 + i_2 + \cdots + i_n = k. Assume this term is non-zero, which means that ai1i2⋯in+bi1i2⋯in≠0a_{i_1 i_2 \cdots i_n} + b_{i_1 i_2 \cdots i_n} \neq 0. Thus, at least one of ai1i2⋯ina_{i_1 i_2 \cdots i_n} and bi1i2⋯inb_{i_1 i_2 \cdots i_n} is non-zero. Without loss of generality, we assume that ai1i2⋯in≠0a_{i_1 i_2 \cdots i_n} \neq 0. Otherwise, we can replace ai1i2⋯ina_{i_1 i_2 \cdots i_n} with bi1i2⋯inb_{i_1 i_2 \cdots i_n} in the following proof.

    Since ai1i2⋯in≠0a_{i_1 i_2 \cdots i_n} \neq 0, we know that ai1i2⋯inx1i1⋯xnina_{i_1 i_2 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n} is a non-zero term of ff. But ff is homogeneous of degree kk, so by definition, i1+i2+⋯+in=ki_1 + i_2 + \cdots + i_n = k. This shows that f+gf + g is homogeneous of degree kk.

    Lastly, we need to show that cfcf is homogeneous of degree kk. If c=0c = 0, then cf=0cf = 0, which we already know is homogeneous of degree kk. Otherwise, suppose (cai1i2⋯in)x1i1⋯xnin\p{ca_{i_1 i_2 \cdots i_n}}x_1^{i_1} \cdots x_n^{i_n} is a non-zero term of cfcf. Then because c≠0c \neq 0, we know that ai1i2⋯in≠0a_{i_1 i_2 \cdots i_n} \neq 0. Thus, ai1i2⋯inx1i1⋯xnina_{i_1 i_2 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n} is a non-zero term of ff, so by degree-kk homogeneity of ff, we have i1+i2+⋯+in=ki_1 + i_2 + \cdots + i_n = k, so cfcf is homogeneous of degree kk.

  2. Let f,g∈Vf, g \in V be symmetric and let c∈Fc \in \F. Note that 0F0_\F is symmetric since all its terms are zero, so after exchanging any two variables, all the terms will still remain zero. Next,

    (f+g)(x1,…,xj,…,xi,…,xn)=f(x1,…,xj,…,xi,…,xn)+g(x1,…,xj,…,xi,…,xn)=f(x1,…,xi,…,xj,…,xn)+g(x1,…,xi,…,xj,…,xn)(f,g are symmetric)=(f+g)(x1,…,xi,…,xj,…,xn)\begin{aligned} \p{f + g}\p{x_1, \ldots, x_j, \ldots, x_i, \ldots, x_n} &= f\p{x_1, \ldots, x_j, \ldots, x_i, \ldots, x_n} + g\p{x_1, \ldots, x_j, \ldots, x_i, \ldots, x_n} \\ &= f\p{x_1, \ldots, x_i, \ldots, x_j, \ldots, x_n} + g\p{x_1, \ldots, x_i, \ldots, x_j, \ldots, x_n} && \p{f, g\text{ are symmetric}} \\ &= \p{f + g}\p{x_1, \ldots, x_i, \ldots, x_j, \ldots, x_n} \end{aligned}

    Similarly,

    (cf)(x1,…,xj,…,xi,…,xn)=cf(x1,…,xj,…,xi,…,xn)=cf(x1,…,xi,…,xj,…,xn)(f is symmetric)=(cf)(x1,…,xi,…,xj,…,xn)\begin{aligned} \p{cf}\p{x_1, \ldots, x_j, \ldots, x_i, \ldots, x_n} &= cf\p{x_1, \ldots, x_j, \ldots, x_i, \ldots, x_n} \\ &= cf\p{x_1, \ldots, x_i, \ldots, x_j, \ldots, x_n} && \p{f\text{ is symmetric}} \\ &= \p{cf}\p{x_1, \ldots, x_i, \ldots, x_j, \ldots, x_n} \end{aligned}

Problem 5

Let V={(a1,a2)∣a1,a2∈R}V = \set{\p{a_1, a_2} \mid a_1, a_2 \in \R}. Define addition of elements of VV coordinate-wise. For (a1,a2)∈V\p{a_1, a_2} \in V and c∈Rc \in \R, define

c(a1,a2)={(0,0)if c=0,(ca1,a2c)if c≠0.c\p{a_1, a_2} = \begin{cases} \p{0, 0} & \text{if } c = 0, \\ \p{ca_1, \frac{a_2}{c}} & \text{if } c \neq 0. \end{cases}

Is VV a vector space over R\R with these operations? Justify your answer.

Solution.

No. For example, (VS 8) fails:

(1+1)⋅(1,1)=2⋅(1,1)=(2,12)\p{1 + 1} \cdot \p{1, 1} = 2 \cdot \p{1, 1} = \p{2, \frac{1}{2}}

and

1⋅(1,1)+1⋅(1,1)=(1,1)+(1,1)=(2,2),1 \cdot \p{1, 1} + 1 \cdot \p{1, 1} = \p{1, 1} + \p{1, 1} = \p{2, 2},

but these are different since 2≠122 \neq \frac{1}{2}.