LADR Exercises 2B

Problem 1

Find all vector spaces that have exactly one basis

The only one is $\{0\}$. If there is some nonzero vector $v$ in the basis, we can get a new basis by changing $v$ to $2v$.

Problem 2

Verify all of the assertions in 2.27.

Omitted – easy but tedious.

Problem 3

• (a) Find the basis of $U$, where $U$ is the subspace of ${\mathbb{R}}^5$ defined by

$$U=\{(x_1,x_2,x_3,x_4,x_5)\in {\mathbb{R}}^5:x_1=3x_2\text{ and }{x}_3=7x_4\}$$

• (b) Extend the basis in (a) to a basis of ${\mathbb{R}}^5$.
• (c) Find a subspace $W$ of ${\mathbb{R}}^5$ such that ${\mathbb{R}}^5=U\oplus W$.

(a) We can parametrize the set as:

$$(3x_2,x_2,7x_4,x_4,x_5)$$

which leads to choosing

$$(3,1,0,0,0),(0,0,7,1,0),(0,0,0,0,1)$$

as a a basis; it is linearly independent and spans the space.

(b) We can extend by adding basis vectors for that place nonzero entries that are currently “missed” by our basis vectors.

$$(1,0,0,0,0),(0,0,1,0,0)$$

(c) $W=\text{span}((1,0,0,0,0),(0,0,1,0,0))$ by (b). We can show that $U\cap W=\{0\}$ by considering $u\in U\cap W$ such that

$$\begin{array}{r}{a}_1{v}_1+a_2{v}_2+a_3{v}_3=b_1{v}_4+b_2{v}_5\\ a_1{v}_1+a_2{v}_2+a_3{v}_3-b_1{v}_4-b_2{v}_5=0\end{array}$$

Since $v_1,\dots ,v_5$ are linearly independent, the only solution is where $a$‘s and $b$‘s are all zero, and so we have $U\cap W=\{0\}$.

Problem 4

• (a) Find the basis of $U$, where $U$ is the subspace of ${\mathbb{C}}^5$ defined by

$$U=\{(z_1,z_2,z_3,z_4,z_5)\in {\mathbb{C}}^5:6z_1=z_2\text{ and }{z}_3+2z_4+3z_5=0\}$$

• (b) Extend the basis in (a) to a basis of ${\mathbb{C}}^5$ .
• (c) Extend the basis $W$ of ${\mathbb{C}}^5$ such that ${\mathbb{C}}^5=U\oplus W$

(a) We can parametrize by writing $z_1=a$, which gives $z_2=6a$. We can also make $z_4=b$ and $z_5=c$ so $z_3=-2b-3c$. This gives:

$$\begin{array}{rl}(z_1,z_2,z_3,z_4,z_5)& =(a,6a,-2b-3c,b,c)\\ & =a(1,6,0,0,0)+b(0,0,-2,1,0)+c(0,0,-3,0,1)\end{array}$$

Thus the vectors corresponding forming our basis are

$$v_1=(1,6,0,0,0),v_2=(0,0,-2,1,0),v_3=(0,0,-3,0,1)$$

(b) We need vectors that are independent from the current basis (not in span of $U$). We can choose a vector where $z_2\ne 6z_1$ and a vector where $z_3\ne -2z_4-3z_5$. Thus, we can extend to ${\mathbb{C}}^5$ by adding

$$(0,1,0,0,0),(0,0,1,0,0)$$

(c) $W=\text{span}((0,1,0,0,0),(0,0,1,0,0))$.

Problem 5

Suppose $V$ is finite-dimensional and $U,W$ are subspaces of $V$ such that $V=U+W$. Prove that there exists a basis of $V$ consisting of vectors in $U\cup W$.