Question

e) the set of polynomials ( p(x) ) with ( int_{-1}^{1} p(x) dx = 0 ).
f) the set of solutions ( y = y(t) ) of ( y + 4y + y = 0 ).
g) the set of solutions ( y = y(t) ) of ( y + 4y + y = 7e^{-t} ).
h) let ( s_f ) be the set of solutions ( u(t) ) of the differential equation ( u - ru = f ) (where ( f ) is a continuous function). is ( s_f ) a linear space? why? note: you are not being asked to actually solve this differential equation.

1. 3 which of the following sets of vectors are bases for ( mathbb{r}^2 )?

a). ( {(0, 1), (1, 1)} ) \t\t\t d). ( {(1, 1), (1, -1)} )
b). ( {(1, 0), (0, 1), (1, 1)} ) \t\t e). ( {(1, 1), (2, 2)} )
c). ( {(1, 0), (-1, 0)} ) \t\t\t f). ( {(1, 2)} )

1. 4 for which real numbers ( c ) are the vectors ( v_1 = (c, 1, 1) ), ( v_2 = (1, c, 1) ), ( v_3 ) (assuming a typo, likely a third vector) linearly independent?

for each of the values of ( c ) that you find, what is the dimension of the subspace that they span?

1. 5 let ( c(mathbb{r}) ) be the linear space of all continuous functions from ( mathbb{r} ) to ( mathbb{r} ) and let ( s ) be the set of differentiable functions ( u(x) ) that satisfy the differential equation

u = 2xu + c
for which value(s) of the real constant ( c ) is this set a linear subspace of ( c(mathbb{r}) )?

1. let ( mathcal{a} ) and ( mathcal{b} ) be linear spaces and ( l : mathcal{a} \to mathcal{b} ) be a linear map. for which vector ( y in mathcal{b} ) is the set

s_y := { x in mathcal{a} mid lx = y }
a linear space?

1. 7 compute the dimension and find bases for the following linear spaces.

a) real anti - symmetric ( 4\times4 ) matrices.
b) cubic polynomials ( p(x)=ax^3 + bx^2 + cx + d ) with ( p(2)=0 ) and ( p(3)=) (likely a typo, maybe ( p(3)=0 ) or another condition).
c) quadratic polynomials ( p(x,y)=ax^2 + bxy + cy^2 + dx + ey + f ) in two variables with the properties: ( p(0,0)=0 ), ( p(1,0)=0 ), and ( p(0,1)=0 ).
d) the space of linear maps ( l : mathbb{r}^5 \to mathbb{r}^3 ) whose kernels contain ( (0, 2, -3, 0, ) (likely a typo, missing a component)).

1. 8a compute the dimension of the intersection of the following two planes

( x + 2y - z = 0 ), ( 3x - 3y + z = 0 ).

Explanation:

Step1: Recall Basis for \(\mathbb{R}^2\)

A basis for \(\mathbb{R}^2\) must satisfy two conditions: (1) The vectors are linearly independent, and (2) They span \(\mathbb{R}^2\) (i.e., the number of vectors is equal to the dimension of \(\mathbb{R}^2\), which is 2).

Step2: Analyze Option a: \(\{(0, 1), (1, 1)\}\)

• Linear Independence: Let \(a(0, 1)+b(1, 1)=(0, 0)\). This gives \(b = 0\) (from the \(x\)-component \(b = 0\)) and \(a + b=0\) (from the \(y\)-component). Since \(b = 0\), then \(a = 0\). So the vectors are linearly independent.
• Spanning: The number of vectors is 2 (equal to \(\dim(\mathbb{R}^2)\)), and they are linearly independent, so they span \(\mathbb{R}^2\). Thus, this set is a basis.

Step3: Analyze Option b: \(\{(1, 0), (0, 1), (1, 1)\}\)

The dimension of \(\mathbb{R}^2\) is 2, so any basis for \(\mathbb{R}^2\) can have at most 2 vectors. This set has 3 vectors, so it cannot be a basis (it is linearly dependent, as \((1, 1)=(1, 0)+(0, 1)\)).

Step4: Analyze Option c: \(\{(1, 0), (-1, 0)\}\)

• Linear Dependence: Let \(a(1, 0)+b(-1, 0)=(0, 0)\). This gives \(a - b = 0\) (from the \(x\)-component) and \(0 = 0\) (from the \(y\)-component). We can choose \(a = 1\), \(b = 1\) (non - trivial solution), so the vectors are linearly dependent. Thus, they cannot be a basis.

Step5: Analyze Option d: \(\{(1, 1), (1, -1)\}\)

• Linear Independence: Let \(a(1, 1)+b(1, -1)=(0, 0)\). This gives \(a + b = 0\) (from the \(x\)-component) and \(a - b = 0\) (from the \(y\)-component). Solving the system \(

$$\begin{cases}a + b=0\\a - b = 0\end{cases}$$

\), we add the two equations: \(2a=0\Rightarrow a = 0\), then \(b = 0\). So the vectors are linearly independent.

• Spanning: There are 2 vectors, and they are linearly independent, so they span \(\mathbb{R}^2\). Thus, this set is a basis.

Step6: Analyze Option e: \(\{(1, 1), (2, 2)\}\)

• Linear Dependence: Notice that \((2, 2)=2(1, 1)\). So there exist non - zero scalars (\(a=-2\), \(b = 1\)) such that \(-2(1, 1)+1(2, 2)=(0, 0)\). The vectors are linearly dependent, so they cannot be a basis.

Step7: Analyze Option f: \(\{(1, 2)\}\)

A single vector cannot span \(\mathbb{R}^2\) (we need at least 2 linearly independent vectors to span \(\mathbb{R}^2\)). Also, it cannot be a basis since the number of vectors is not equal to the dimension of \(\mathbb{R}^2\).

Answer:

a. \(\{(0, 1), (1, 1)\}\)
d. \(\{(1, 1), (1, -1)\}\)