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 $U\cup W$.

Suppose we have a set $\{v_1,v_2,\dots ,v_m\}$ as a basis for $U\cup W$. We can extend to to a basis of $U$ by adding vectors $\{u_1,u_2,\dots ,u_{p-m}\}$ from $U$ such that $U=\text{span}(v_1,\dots ,v_m,u_1,\dots ,u_{p-m})$. Similarly, we can extend to basis for $W$ by adding vectors from $W$ such that $W=\text{span}(v_1,\dots ,v_m,w_1,\dots ,w_{q-m})$.

• Note that $u_i$ are vectors exclusive to $U$ such that $u_i\in U\setminus U\cap W$
• Note that $w_i$ are vectors exclusive to $V$ such that $w_i\in W\setminus U\cap W$

Let’s construct a set $S$ that contains all of these.

$$S=\{u_1,\dots ,u_{p-m},v_1,\dots v_m,w_1,\dots ,w_{q-m}\}$$

Note that all elements of $S$ are in $U\cup W$.

Let’s show that $S$ is linearly independent. Suppose there exist scalars $c_i$ and $d_j$ such that

$$\sum _{i=1}^{p-m}{c}_i{u}_i+\sum _{k=1}^m{c}_{p-m+k}{v}_k+\sum _{j=1}^{q-m}{d}_j{w}_j=0$$

Re-arranging:

$$\left(\sum _{i=1}^{p-m}{c}_i{u}_i+\sum _{k=1}^m{c}_{p-m+k}{v}_k\right)+\left(\sum _{j=1}^{q-m}{d}_j{w}_j\right)=0$$

The first bracket is in $U$ (since $v_k\in U\cap V$) and the second bracket is in $W$. Note that we could do this proof grouping $v$ with $w$ as well.

The sum being zero implies that their negatives are equal. Furthermore, since the left side is in $U$ and the right side is in $W$, they must both be in $U\cap W$:

$$\left(\sum _{i=1}^{p-m}{c}_i{u}_i+\sum _{k=1}^m{c}_{p-m+k}{v}_k\right)=-\left(\sum _{j=1}^{q-m}{d}_j{w}_j\right)\in U\cap W$$

Since $\{u_i\}$ are in $U\setminus (U\cap W)$, their coefficients must be zero, such that $c_i=0$ for $i=1,\dots ,p-m$. Similarly, $\{w_j\}$ are in $W\setminus (U\cap W)$, their coefficients must be zero, such that $d_j=0$ for $j=1,\dots ,q-m$. Thus, we have

$$\sum _{k=1}^m{c}_{p-m+k}{v}_k=0$$

Since $\{v_k\}$ is a basis, all $c_{p-m+k}=0$. Therefore, $S$ is linearly independent.

Now we can show that $S$ spans $V$ – every vector in $V$ can be expressed as a sum of vectors from $U$ and $W$, because $V=U+W$. Since $S$ contains bases for $U$ and $W$, it spans $V$.

We have constructed a basis $S$ of $V$ consisting entirely of vectors from $U\cup W$; therefore, such a basis exists.

Problem 6

Prove or give a counterexample: If $p_0,p_1,p_2,p_3$ is a list in ${\mathcal{P}}_3(\mathbb{F})$ such that none of the polynomials $p_0,p_1,p_2,p_3$ has degree 2, then $p_0,p_1,p_2,p_3$ is not a basis of ${\mathcal{P}}_3(\mathbb{F})$.

Consider the list

$$p_0,p_1,p_2,p_3=1,x,x^2+x^3,x^3$$

which contains no polynomial of degree 2.

We can prove that $\text{span}(1,x,x^2+x^3,x^3)={\mathcal{P}}_3(\mathbb{F})$.

Let $q\in {\mathcal{P}}_3(\mathbb{F})$. Then, there exist

$$a_0,\dots ,a_3\in \mathbb{F}$$

such that $q=a_0+a_{1}x+a_2{x}^2+a_3{x}^3$. But notice that we can write

$$\begin{array}{rl}{a}_0{p}_0+a_1{p}_1+a_2{p}_2+(a_3-a_2)p_3& =a_0+a_{1}x+a_2(x^2+x^3)+(a_3-a_2)x^3\\ & =a_0+a_{1}x+a_2{x}^2+a_3{x}^3\\ & =q\end{array}$$

To see the list is linearly independent, suppose $b_0,\dots ,b_3\in \mathbb{F}$ such that

$$b_0{p}_0+b_1{p}_1+b_2{p}_2+b_3{p}_3=0$$

It follows that

$$b_0+b_{1}x+b_2{x}^2+(b_2+b_3)x^3=0$$

which is true if and only if all coefficients are zero. Thus, this is a basis as claimed. $■$

Problem 7

Suppose that $v_1,v_2,v_3,v_4$ is a basis of $V$. Prove that

$$v_1+v_2,v_2+v_3,v_3+v_4,v_4$$

is also a basis of $V$.

First, we need to show that the list is linearly independent. Suppose that

$$\begin{array}{rl}a(v_1+v_2)+b(v_2+v_3)+c(v_3+v_4)+d{v}_4& =0\\ a{v}_2+(a+b)v_2+(b+c)v_4+(c+d)v_4& =0\end{array}$$

If $a=0$, then we must have $b=0$, which in turn also gives us $c=0,d=0$. Thus, we have shown that the list is linearly independent.

Now note that we can write:

$$\begin{array}{rl}{v}_3& =(v_3+v_4)-v_4\\ v_2& =(v_2+v_3)-v_3=(v_2+v_3)-(v_3+v_4)+v_4\\ v_1& =(v_1+v_2)-v_2=(v_1+v_2)-(v_2+v_3)+(v_3+v_4)-v_4\end{array}$$

and of course, $v_4=v_4$.

Since we know that $v_1,v_2,v_3,v_4$ is a basis of $V$, and we can write these basis vectors as linear combinations of elements from our list, then our list must span $V$ as well.

Therefore, our list is a span of $V$.

Problem 8

Prove or give a counterexample: If $v_1,v_2,v_3,v_4$ is a basis of $V$ and $U$ is a subspace of $V$ such that $v_1,v_2\in U$ and $v_3\notin U$ and $v_4\notin U$, then $v_1,v_2$ is a basis of $U$.

The statement is false. Let $V={\mathbb{R}}^4$ and let

$$\begin{array}{r}{v}_1=(1,0,0,0)\\ v_2=(0,1,0,0)\\ v_3=(0,0,1,0)\\ v_4=(0,0,0,1)\end{array}$$

Clearly $v_1,v_2,v_3,v_4$ is a basis of $V$.

Define

$$U=\{(x_1,x_2,x_3,x_4)\in {\mathbb{R}}^4:x_3=x_4\}$$

Then $v_1,v_2\in U$ but $v_3,v_4\notin U$.

However $v_1,v_2$ does not span $U$, since no combination of them can yield $(0,0,1,1)$.

Problem 9

Suppose $v_1,\dots ,v_m$ is a list of vectors in $V$. For $k\in \{1,\dots ,m\}$, let

$$w_k=v_1+\cdots +v_k$$

Show that $v_1,\dots ,v_m$ is a basis of $V$ if and only if $w_1,\dots ,w_m$ is a basis of $V$.

First, we want to show that if $v_1,\dots ,v_m$ is a basis of $V$, then $w_1,\dots ,w_m$ is a basis of $V$.

We can first show that $w_1,\dots ,w_m$ spans $V$ if $v_1,\dots ,v_m$ is a basis. This can be done by considering that any element $v_k$ can be written as:

$$\begin{array}{rl}{v}_k& =w_k-w_{k-1}\\ & =(v_1+\cdots +v_k)-(v_1+\cdots +v_{k-1})\end{array}$$

if we define $v_0=0$. Since we can write all any element $v_k$ as a linear combination of $w_k$‘s, and $v_1,\dots ,v_m$ is a basis, then $w_1,\dots ,w_m$ span $V$ as well.

We can then show that $w_1,\dots ,w_m$ is linearly independent by considering

$$\begin{array}{rl}{a}_1{w}_1+a_2{w}_2+\cdots +a_m{v}_m& =0\\ a_1{v}_1+a_2(v_1+v_2)\cdots +a_m(v_1+\cdots +v_m)& =0\\ (a_1+\cdots +a_m)v_1+(a_2+\cdots +a_m)v_2+\cdots +a_m{v}_m& =0\end{array}$$

Since $v_1,\dots ,v_m$ are linearly independent, the coefficients of each $v_k$ needs to be zero. We can start from $a_m=0$ and work backward to get all $a_1=\cdots =a_m=0$.

Then, we want to show the opposite direction; if $w_1,\dots ,w_m$ is a basis of $V$, then $v_1,\dots ,v_m$ is a basis of $V$.

We can first show that $v_1,\dots ,v_m$ spans $V$. Since $w_1,\dots ,w_m$ is basis in $V$, every vector in $W$ can be written as a linear combination of $w_1,\dots ,w_m$. Since each $w_k$ can be written in terms of $v_1,\dots ,v_k$, it follows that any vector in $V$ can also be written as a linear combination of $v_1,\dots ,v_m$.

We can then show that $v_1,\dots ,v_m$ are linearly independent similar to above:

$$\begin{array}{rl}{a}_1{v}_1+a_2{v}_2+\cdots +a_m{v}_m& =0\\ a_1{w}_1+a_2(w_2-w_1)+\cdots +a_m(w_m-w_{m-1})& =0\\ (a_1-a_2)w_1+(a_2-a_3)w_2+\cdots +(a_{m-1}-a_m)w_{m-1}+a_m{w}_m& =0\end{array}$$

Since $w_1,\dots ,w_m$ are a basis of $V$, they are linearly independent, so each of the coefficients must be zero. We can work backward from $a_m=0$ back to get $a_1=\cdots =a_m=0$.

Problem 10

Suppose $U$ and $W$ are subspaces of $V$ such that $V=U\oplus W$. Suppose also that $u_1,\dots ,u_m$ is a basis of $U$ and $w_1,\dots ,w_n$ is a basis of $W$. Prove that

$$u_1,\dots ,u_m,w_1,\dots ,w_n$$

is a basis of $V$.

First, we show that $u_1,\dots ,u_m,w_1,\dots ,w_n$ is linearly independent. If there exist $a_1,\dots ,a_m\in \mathbb{F}$ and $b_1,\dots ,b_n\in \mathbb{F}$ such that

$$a_1{u}_1+\cdots +a_m{u}_m+b_1{w}_1+\cdots +b_n{w}_n=0$$

Then

$$a_1{u}_1+\dots a_m{u}_m=-(b_1{w}_1+\cdots +b_n{w}_n)\in U\cap W$$

since $V=U\oplus W$, we have $U\cap W=\{0\}$.

However, note that $u_1,\dots ,u_m$ is a basis of $U$ and $w_1,\dots ,w_n$ is a basis of $W$. It follows that $a_1=\cdots =a_m=0$ and $b_1=\cdots =b_n=0$. Hence, $u_1,\dots ,u_m,w_1,\dots ,w_n$ is linearly independent.

Second, we need to show that $u_1,\dots ,u_m,w_1,\dots ,w_n$ spans $V$. For any $v\in V$, there exist $u\in U$ and $w\in W$ such that $v=u+w$ since $V=U\oplus W$. Note that $u_1,\dots ,u_m$ is a basis of $U$ and $w_1,\dots ,w_n$ is a basis of $W$. It follows that for any $u$ and $w$, there exist $a_1,\dots ,a_m\in \mathbb{F}$ and $b_1,\dots ,b_n\in \mathbb{F}$ such that

$$\begin{array}{r}u=a_1{u}_1+\cdots +a_m{u}_m\\ w=b_1{w}_1+\cdots +b_n{w}_n\end{array}$$

Hence,

$$v=u+w=a_1{u}_1+\cdots +a_m{v}_m+b_1{w}_1+\cdots +b_n{w}_n$$

which means that $u_1,\dots ,u_m,w_1,\dots ,w_n$ spans $V$.

This shows that

$$u_1,\dots ,u_m,w_1,\dots ,w_n$$

is a basis of $V$.

Problem 11

Suppose $V$ is a real vector space. Show that if $v_1,\dots ,v_n$ is a basis of $V$ (as a real vector space), then $v_1,\dots ,v_n$ is also a basis of the complexification $V_{\mathbb{C}}$ (as a complex vector space).

• See LADR Exercises 1B for the definition of complexification.

Recall that the complexification $V_{\mathbb{C}}$ of $V$ is defined as:

$$V_{\mathbb{C}}=V\times V$$

and we write elements of $V_{\mathbb{C}}$ as $u+iv$, where $u,v\in V$. Scalar multiplication by a complex scalar $a+bi$ is defined as $(a+bi)(u+iv)=(au-bv)+i(av+bu)$.

Suppose there exist complex scalars $c_1,\dots ,c_n\in \mathbb{C}$ such that

$$c_1{v}_1+c_2{v}_2+\cdots +c_n{v}_n=0$$

We can write each $c_j$ as $c_j=a_j+i{b}_j$ and express the sum

$$c_1{v}_1+c_2{v}_2+\cdots +c_n{v}_n=(a_1{v}_1+a_2{v}_2+\cdots +a_n{v}_n)+i(b_1{v}_1+b_2{v}_2+\cdots +b_n{v}_n)$$

For this to be zero, we need the real and imaginary parts to be equal in $V$. Thus:

$$\begin{array}{r}{a}_1{v}_1+a_2{v}_2+\cdots +a_n{v}_n=0\\ b_1{v}_1+b_2{v}_2+\cdots +b_n{v}_n=0\end{array}$$

Since $v_1,\dots ,v_n$ is a basis of $V$ as a real vector space, they are linearly independent over $\mathbb{R}$. Therefore

$$\begin{array}{r}{a}_1=a_2=\cdots =a_n=0\\ b_1=b_2=\cdots =b_n=0\end{array}$$

This shows linear independence.

Next, to show that $v_1,\dots ,v_n$ spans $V_{\mathbb{C}}$, we can write any arbitrary element of $V_{\mathbb{C}}$ in the form $u+iv$, where $u,v\in V$. Since $v_1,\dots v_n$ is a basis of $V$, we can write

$$u=\sum _{j=1}^n{a}_j{v}_j,v=\sum _{j=1}^n{b}_j{v}_j$$

for some $a_j,b_j\in \mathbb{R}$. Substituting these into $u+iv$:

$$u+iv=\sum _{j=1}^n{a}_j{v}_j+i\sum _{j=1}^n{b}_j{v}_j=\sum _{j=1}^n(a_j+i{b}_j)v_j$$

The coefficients $a_j+i{b}_j$ are complex scalars, so $u+iv$ is a complex linear combination of $v_1,\dots ,v_n$. Thus, $v_1,\dots v_n$ span $V_{\mathbb{C}}$ over $\mathbb{C}$.

Since $v_1,\dots ,v_n$ are linearly independent and span $V_{\mathbb{C}}$, they form a basis of $V_{\mathbb{C}}$ as a complex vector space.