3

Homework 3 1.pdf

(15 pts) Are the following subsets \(S\) of the vector space \(V\) subspaces?

(a) \(S = \{ax^3 + bx^2 + cx + d \in V : a + b + c + d = 0\}\) where \(V = \{p \in \mathbb{R}[x] : \deg(p) \leq 3\}\).

Let's check two things
1. Take \(p(x)=ax^3+bx^2+cx+d:a+b+c+d=0\) and \(h(x)=ex^3+fx^2+gx+h:e+f+g+h=0\)
  
  Then \(p(x)+h(x)=\left(a+e\right)x^3+\left(b+f\right)x^2+\left(c+g\right)x+\left(d+h\right):a+b+c+d+e+f+g+h=0\)
  
  Then \((a+e)+(b+f)+(c+g)+(d+h)=0\)
  
  Thus \(p(x)+h(x)\in V\) 2. Take \(\lambda\in\mathbb{R},\lambda p(x)=a\lambda x^3+b\lambda x^2+c\lambda x+d\lambda\)
  
  Since \(a+b+c+d=0\), then \(\lambda\left(a+b+c+d\right)=0\Rightarrow\lambda a+\lambda b+\lambda c+\lambda d=0\)
  
  Thus \(\lambda p(x)\in V\)
Thus \(S\) is a subspace

(b) \(S = \{A \in M_2(\mathbb{R}) : \det(A) = 0\}\) where \(V = M_2(\mathbb{R})\).

Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\). Since \(\det(A) = 0\), then \(ad=bc\Rightarrow d=\frac{bc}{a}\) when \(a\neq 0\)

Thus \(A = \begin{pmatrix} a & b \\ c & \frac{bc}{a} \end{pmatrix}\)

Let's check two things:
1. Take \(A = \begin{pmatrix} a & b \\ c & \frac{bc}{a} \end{pmatrix}\), \(B=\begin{pmatrix}e & f\\ g & \frac{fg}{e}\end{pmatrix}\)
Then \(A+B=\begin{pmatrix}a+e & b+f\\ c+g & \frac{bc}{a}+\frac{fg}{e}\end{pmatrix}\)

Then \(\left(a+e\right)\left(\frac{bc}{a}+\frac{fg}{e}\right)-\left(b+f\right)\left(c+g\right)=bc+\frac{afg}{e}+\frac{bce}{a}+gf-bc-bg-fg=\frac{afg}{e}+\frac{bce}{a}-bg-fc\neq0.\)

Counterexample: \(A=\begin{pmatrix}1 & 2\\ 2 & 4\end{pmatrix},B=\begin{pmatrix}1 & \sqrt2\\ \sqrt2 & 2\end{pmatrix}\)

\(A+B=\begin{pmatrix}2 & 2+\sqrt2\\ 2+\sqrt2 & 6\end{pmatrix},\quad\det(A+B)=12-\left(4+2+4\sqrt2\right)\neq0.\)

Thus \(S\) is not a vector subspace

(c) \(S = \{A \in M_n(\mathbb{R}) : A^T = A\}\) where \(V = M_n(\mathbb{R})\).

Since \(A^T = A\), the matrix is a square matrix. Then we have \(a_{ij} = a_{ji}, \ \forall i, j \leq n\).

Let’s take \(A, B\). Check two things:
1. For \(A + B\), \((A + B)^T = (a_{ij} + b_{ij})^T = a_{ji} + b_{ji} = A + B\). Thus, \((A + B)^T = A + B\).
2. Let \(\lambda \in \mathbb{R}\), \(\lambda A = \lambda a_{ij}\). Then \((\lambda A)^{T}\left.=(\lambda a_{ij}\right)^{T}=\lambda(a_{ij})^{T}=\lambda a_{ji}=\lambda a_{ij}\)
  
  Thus \((\lambda A)^{T}=\lambda A\)
Thus \(S\) is a vector subspace.
(20 pts) Prove or give a counterexample: if \(U_1\), \(U_2\) and \(W\) are subspaces of \(V\) such that \(V = U_1 \oplus W\) and \(V = U_2 \oplus W\), then \(U_1 = U_2\).

No! Let \(V = \mathbb{R}^3\), \(W=\{(0,0,z):z\in\mathbb{R}\}\)

\(U_1 = \{(x, y, 0) : x, y \in \mathbb{R}\}\) \(U_2=\{(x,y,-x-y):y\in\mathbb{R}\}\)

Since \(U_1+W=\{(x,y,z):x,y,z\in\mathbb{R}\}=V\) and \(U_2+W=\{(x,y,-x-y+z):x,y,z\in\mathbb{R}\}=V\)

And \(\left(0,0,0\right)\in U_1+W\) can be expressed uniquely only when \(x=y=z=0\) and \((0,0,0)\in U_2+W\) can be expressed uniquely only when \(x=y=0,z=0\)

Thus \(V = U_1 \oplus W\) and \(V = U_2 \oplus W\)

But \(U_1\neq U_2\)
(30 pts) Let \(S_1\) and \(S_2\) be subsets of a vector space \(V\) and denote the subspaces spanned by \(S_1\) and \(S_2\) as \(\operatorname{Span}(S_1)\) and \(\operatorname{Span}(S_2)\) respectively. Prove that \(\operatorname{Span}(S_1 \cup S_2) = \operatorname{Span}(S_1) + \operatorname{Span}(S_2)\).

Proof

We need to prove it in two directions

\(\subseteq\)) We need to prove \(\operatorname{Span}(S_1\cup S_2)\subseteq\operatorname{Span}(S_1)+\operatorname{Span}(S_2)\)

Let's take a vector \(p=a_1v_1+\cdots+a_{m}v_{m}:a_1,\ldots,a_{m}\in F\) in \(Span\left(S_1\cup S_2\right)\), then \(v_1,v_2,...,v_{m}\in S_1\cup S_2\)

Then we can split the list into two part: \(v_{1a},v_{1b},\ldots\in S_1,v_{2a},v_{2b},\ldots\in S_2\) such that \(v_1,v_2,...,v_{m}=v_{1a},v_{1b},\ldots,v_{2a},v_{2b},\ldots\)

Then let \(m=a_{1a}v_{1a}+a_{1b}v_{1b}\cdots\in Span\left(S_1\right)\) and \(n=a_{2a}v_{2a}+a_{2b}v_{2b}\cdots\in Span(S_2)\)

Then \(m+n=a_{1a}v_{1a}+a_{1b}v_{1b}\cdots+a_{2a}v_{2a}+a_{2b}v_{2b}\cdots=a_1v_1+\cdots+a_{m}v_{m}=p\in Span\left(S_1\right)+Span\left(S_2\right)\)

\(\supseteq\)) We need to prove \(\operatorname{Span}(S_1)+\operatorname{Span}(S_2)\subseteq\operatorname{Span}(S_1\cup S_2)\)

Let's take \(p\in \operatorname{Span}(S_1)+\operatorname{Span}(S_2)\). Since we are in the vectorspace, \(v=m+n\) where \(m\in \operatorname{Span}(S_1)\) and \(n\in\operatorname{Span}(S_2)\)

Then let \(m=a_{1a}v_{1a}+a_{1b}v_{1b}\cdots\in Span\left(S_1\right)\) and \(n=a_{2a}v_{2a}+a_{2b}v_{2b}\cdots\in Span(S_2)\) where \(v_{1a},v_{1b},\ldots\in S_1,v_{2a},v_{2b},\ldots\in S_2\)

Then \(v_{1a},v_{1b},\ldots v_{2a},v_{2b},\ldots\in S_1\cup S_2\) and \(m+n=a_{1a}v_{1a}+a_{1b}v_{1b}\cdots+a_{2a}v_{2a}+a_{2b}v_{2b}\cdots=a_1v_1+\cdots+a_{m}v_{m}\operatorname{\in Span}(S_1\cup S_2)\)

Thus \(\operatorname{Span}(S_1 \cup S_2) = \operatorname{Span}(S_1) + \operatorname{Span}(S_2)\).
(35 pts) Homogeneous system of linear equations.

(a) Let \(W_A\) be the set of all solutions \(x = (x_1, x_2, x_3) \in \mathbb{R}^3\) of the homogeneous system of linear equations \(Ax = 0\) where \(A \in M_3(\mathbb{R})\) is given by
\(A = \begin{pmatrix} -1 & 1 & 1 \\ 3 & -1 & 0 \\ 2 & -4 & -5 \end{pmatrix}\)
Find a finite set of vectors in \(\mathbb{R}^3\) which spans \(W_A\).

\(\begin{cases}-x_1+x_2+x_3=0\\ 3x_1-x_2=0\\ 2x_1-4x_2-5x_3=0\end{cases}\Rightarrow\begin{cases}x_1=x_2+x_3\\ 3x_1=x_2\\ 2x_1=4x_2+5x_3\end{cases}\Rightarrow\begin{cases}2x_1+x_3=0\\ 3x_1=x_2\\ 2x_1+x_3=0\end{cases}\Rightarrow\begin{cases}2x_1=-x_3\\ 3x_1=x_2\end{cases}\)

Thus, the solutions are \((x_1, 3x_1, -2x_1)\)\(\Rightarrow W_A = \langle (1, 3, -2) \rangle\)

(b) Let \(W_B\) be the set of all solutions \(x = (x_1, x_2, x_3) \in \mathbb{R}^3\) of the homogeneous system of linear equations \(Bx = 0\) where \(B \in M_3(\mathbb{R})\) is given by
\(B = \begin{pmatrix} 1 & -3 & 1 \\ 2 & -6 & 2 \\ 3 & -9 & 3 \end{pmatrix}\)
Find a finite set of vectors in \(\mathbb{R}^3\) which spans \(W_B\).

\(\begin{cases} x_1 - 3x_2 + x_3 = 0 \\ 2x_1 - 6x_2 + 2x_3 = 0 \\ 3x_1 - 9x_2 + 3x_3 = 0 \end{cases} \Rightarrow \begin{cases} x_1 = 3x_2 - x_3 \\ x_1 = 3x_2 - x_3 \\ x_1 = 3x_2 - x_3 \end{cases}\)

Thus, the solutions are \((3x_2 - x_3, x_2, x_3)\) \(\Rightarrow W_{B}=\langle(3,1,0),(-1,0,1)\rangle\)

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