Exercise 1: Dimension and the Zero Vector in Different Vector Spaces
A
State the zero-vector and the dimension of the following vector spaces.
$\Bbb R^{4}$
$\Bbb C^{4}$
$C^{0}(\left[\,0,\,1\,\right])$
$\Bbb R^{4 \times 2}$
$P_{4}(\Bbb R)$
answer
$(0,0,0,0)$ and $\mathrm{dim(\Bbb R^{4})}=4$.
$(0,0,0,0)$ and $\mathrm{dim(\Bbb C^{4})}=4$.
$f(x)=0$ for all $x\in \left[\,0,\,1\,\right]$ and infinite dimensional.
$\begin{matr}{cc}0&0\newline 0&0\newline 0&0\newline 0&0\end{matr}$ and $\mathrm{dim(\Bbb R^{4\times 2})}=8$.
$P(x)=0+0x+0x^2+0x^3+0x^4=0$ for all $x\in \mathbb R$ and $\mathrm{dim(P_4(\Bbb R))}=5$.
Exercise 2: Linear Dependence or Independence
A
Determine whether the following systems of vectors are linearly dependent or linearly independent. In case the vectors are linearly dependent, write one of the vectors as a linear combination of the other vectors.
1) By hand:
$$(1,2,1,0), (2,7,3,1), (3,12,5,2)\quad \text{ in } \Bbb
R^{4}$$
2) By hand:
$$(1,i), (1+i,-1+i) \quad \text{ in } {\Bbb
C}^{2}$$
Regarding 2) with imaginary vectors: Solve the system of two equations with two complex unknowns that are established from the given vectors as columns in a coefficient matrix. Just like for real vectors.
hint
Regarding 4) with matrices: Rewrite the matrices to column form while paying close attention to moving elements to equivalent positions for all matrices. Now merge the columns into a matrix and solve as usual. When you are done solving the system, the columns can be converted back into matrices when needed.
answer
1) They are linearly dependent, e.g.
$$(3,12,5,2)=-(1,2,1,0)+2(2,7,3,1)\,.$$
2) Linearly dependent. It is readily seen that the second tuple is equal to the first tuple multiplied by a complex number.
By hand: Determine the value of $a$ that must be avoided if the set
$$ \big ( \,(1,2,3),(-1,0,2),(1,6,a)\,\big )$$
is to be a valid basis for $\mathbb R^3\,$.
hint
See Method 11.39 in eNote 11.
hint
Compute the determinant of the coordinate matrix created with the the three vectors as columns with respect to the standard basis.
answer
The number $a=13$ must be avoided.
B
In $\mathbb R^4$ five vectors are given: $\ma_1=(1,-1,2,1),\,\ma_2=(0,1,1,3),\,\ma_3=(1,-2,2,-1)\, \ma_4=(0,1,-1,3)$ and $\mv=(1,-2,2,-3)\,$
Prove that $(\ma_1,\ma_2,\ma_3,\ma_4)$ is a basis for $\mathbb R^4\,$, and compute the coordinate vector $_\mathrm a\mv\,$.
hint
Compute $_\mathrm a\mv\,$ using Method 11.40, point 2 in eNote 11.
answer
The vector set is a basis because their number equals $\mathrm{dim}\mathbb R^4)$, each of them belongs to the vector space $\mathbb R^4$, and they are linearly independent.
Determine the three basis vectors $P_1(x),\,P_2(x)$ and $P_3(x)\,$. (Solve this without referring to a), but you can have a look at a) for a hint.)
hint
See Method 11.40 in eNote 11. In this exercise it is the change-of-basis matrix (with respect to the standard basis) that is unknown and that we are tasked to find.
answer
$$P_1(x)=1+x^2, P_2(x)=-1-x-3x^2\quad\text{ and } \quad P_3(x)=6+x+5x^2$$
Exercise 5: Subspaces (By Hand)
A
Consider the set $\,G3\,$ of geometric vectors in space. Do subspaces of $\,G3\,$ with the dimensions 0, 1, 2, 3 or 4 exist? If they do exist describe them in words.
hint
See Theorem 11.42 i eNote 11.
answer
The zero vector is a 0-dimensional subspace of $G$.
The set of all vectors that have representations on the same straight line through the origin in 3D space is a 1-dimensional subspace of $G$.
The set of all vectors that have representations in the same plane through the origin in 3D space is a 2-dimensional subspace of $G$.
$G$ is a 3-dimensional subspace of itself.
Subspaces of $G$ of dimension greater than 3 do not exist.
B
Is the set $\lbrace\,a\cos(x)+b\sin(x)\,|\,a,b\in\mathbb R\,\rbrace$ a subspace of $C^{0}(\mathbb R)\,$?
hint
Does the set fulfill the two stability requirements in Theorem 11.42 in eNote 11?
answer
Yes! The set is a subspace of $C^{0}(\mathbb R)\,$.
C
Is $\lbrace\,\left (x_1,x_2, x_3, x_4 \right)\, |\, x_1 \cdot x_2 \cdot x_3\cdot x_4=0 \,\rbrace$ a subspace of $\mathbb R^4\,?$
hint
Try to find two vectors in $\mathbb R^4$ whose product of coordinates each gives 0, but where the product of coordinates of their vector sum is different from 0.
answer
The set is not a subspace of $\mathbb R^4\,$.
D
Is the subset of polynomials $\,P_2(\Bbb R)\,$ that have the root 1 a subspace of $\,P_2(\Bbb R)\,?$ If so, find a basis for the subspace.
answer
Set up the general form of an element from the space and then apply the condition so that the element belongs to the subspace. The condition in this case is $P(1)=0$. After applying this condition, your results after rewriting describe the subspace only.
answer
Yes, a 2D subspace. A possible basis is $\,(1-x^2,x-x^2)\,.$
E
Is the subspace of polynomials in $\,P_2(\Bbb R)\,$ that have a double root a subspace of $\,P_2(\Bbb R)\,?$ If so, find a basis for the subspace.
answer
No, it is not a subspace.
Exercise 6: Bases for Subspaces
A
By hand: Explain why the solution set for the system of linear equations
Do $\mb_1$ and $\mb_2$ belong to the spanning of $\ma_1$ and $\ma_2\,$?
answer
The two sets span the same subspace.
Exercise 7: Vectors Within and Outside a Subspace
A
Introductory meditation about identity: A matrix is a vector is a vector is a matrix
(freely after Gertrud Stein, 1913: A rose is a rose is a rose is a rose).
hint
Can a matrix be a vector?
In the vector space $\,\reel^{3\times 3}\,$ four vectors are given:
belong to $\,\mathrm{span}\lbrace P_1(x), P_2(x), P_3(x)\rbrace\,$ and if so state their coordinate vectors with respect to the basis $\,\big (P_1(x), P_2(x)\big)\,$.
answer
$Q_1(x)$ does not belong to the span. $Q_2(x)$ does, and
$$_\mathrm p Q_2(x)=\begin{matr}{c}1\newline 2 \end{matr}\,.$$
C
State the simplest possible basis for $\text{span}\lbrace \,P_1(x), P_2(x), P_3(x),Q_1(x),Q_2(x) \, \rbrace\,.$
