LADR Exercises 2C

Problem 1

Show that the subspaces of ${\mathbb{R}}^2$ are precisely $\{0\}$, all lines in ${\mathbb{R}}^2$ containing the origin, and ${\mathbb{R}}^2$.

A subspace of ${\mathbb{R}}^3$ can have a basis of length 0, 1, 2.

• 0: The only basis of length 0 is $\{0\}$.
• 1: Any basis of length 1 contains a single $x\in {\mathbb{R}}^n$. Note that $\text{span}(x)=\{ax\in {\mathbb{R}}^n:a\in \mathbb{R}\}$, and hence bases of length $1$ are lines through the origin.
• 2: Any basis of length 2 is simply a basis of ${\mathbb{R}}^2$ and hence generates all ${\mathbb{R}}^2$ with its span.

Problem 2

Show that the subspaces of ${\mathbb{R}}^3$ are precisely $\{0\}$, all lines in ${\mathbb{R}}^3$ containing the origin, all planes in ${\mathbb{R}}^3$ containing the origin, and ${\mathbb{R}}^3$.

A subspace of ${\mathbb{R}}^3$ can have a basis of length 0, 1, 2, 3.

• 0: The only basis of length 0 is $\{0\}$.
• 1: Any basis of length 1 contains a single $x\in {\mathbb{R}}^n$. Note that $\text{span}(x)=\{ax\in {\mathbb{R}}^n:a\in \mathbb{R}\}$, and hence bases of length $1$ are lines through the origin.
• 2: Any basis of length 2 contains two linearly independent $x,y\in {\mathbb{R}}^n$. Notice $\text{span}(x,y)=\{ax+by\in {\mathbb{R}}^n:x,y\in \mathbb{R}\}$, hence bases of length 2 generate planes through the origin.
• 3: Any basis of length 3 is simply a basis of ${\mathbb{R}}^3$ and hence generates all ${\mathbb{R}}^3$ with its span.

Problem 3

• (a) Let $U=\{p\in {\mathcal{P}}_4(\mathbb{F}):p(6)=0\}$. Find a basis of $U$.
• (b) Extend the basis in (a) to a basis of ${\mathcal{P}}_4(\mathbb{F})$.
• (c) Find the subspace $W$ of ${\mathcal{P}}_4(\mathbb{F})$ such that ${\mathcal{P}}_4(\mathbb{F})=U\oplus W$.

(a) The set $U$ consists of polynomials in ${\mathcal{P}}_4$ such that $p(6)=0$. This gives a single constraint, so our basis needs 4 elements. A basis would be

$$\{x-6,(x-6)x,(x-6)x^2,(x-6)x^3\}$$

which are linearly independent. Any linear combination of these would still be a member of $U$. Note that $\dim U=4$.

(b) We can simply add one element $q\notin U$, since we just need $\dim U=5$ (linearly independent list of the right length is a basis). We can add $1$, so that we have

$$\{1,x-6,(x-6)x,(x-6)x^2,(x-6)x^3\}$$

(c) We need $P_4(\mathbb{F})=U+W$ and $U\cap W=\{0\}$. Based on (b) above, we can let $W=\{c:c\in \mathbb{F}\}$, then ${\mathcal{P}}_4(\mathbb{F})=U\oplus W$ by the previous part.

Problem 4

• (a) Let $U=\{p\in {\mathcal{P}}_4(\mathbb{F}):p^{\prime \prime }(6)=0\}$. Find a basis of $U$.
• (b) Extend the basis in (a) to a basis of ${\mathcal{P}}_4(\mathbb{F})$.
• (c) Find the subspace $W$ of ${\mathcal{P}}_4(\mathbb{F})$ such that ${\mathcal{P}}_4(\mathbb{F})=U\oplus W$.

(a) The set $U$ consists of degree-4 polynomials such that $p^{\prime \prime }(6)=0$.

$$\begin{array}{rl}p(x)& =a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4\\ p^{\prime }(x)& =a_1+2a_{2}x+3x^2{a}_3+4x^3{a}_4\\ p^{\prime \prime }(x)& =2a_2+6a_{3}x+12a_4{x}^2\\ p^{\prime \prime }(6)& =2a_2+36a_3+432a_4\end{array}$$

Since $p^{\prime \prime }(6)=0$, we can write

$$\begin{array}{r}2a_2+36a_3+432a_4=0\\ a_2=-18a_3-216a_4\end{array}$$

This means that any $p\in U$ has the form

$$\begin{array}{rl}p(x)& =a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4\\ p(x)& =a_0+a_{1}x+(-18a_3-216a_4)x^2+a_3{x}^3+a_4{x}^4\end{array}$$

which gives us a basis of

$$1,x,-18x^2+x^3,-216x^2+x^4$$

(b) We just need to find an element of ${\mathcal{P}}_4(\mathbb{F})$ to $U$ that is not in $U$ already, because we already have $\dim U=4$. We can simply add $x^2$, since $\frac{d^2}{dx}(x^2)=2\ne 6$. This gives us

$$\{x^2,1,x,-18x^2+x^3,-216x^2+x^4\}$$

(c) We can use $W=\text{span}(x^2)$.

Problem 5

• (a) Let $U=\{p\in {\mathcal{P}}_4(\mathbb{F}):p(2)=p(5)\}$. Find a basis of $U$.
• (b) Extend the basis in (a) to a basis of ${\mathcal{P}}_4(\mathbb{F})$.
• (c) Find the subspace $W$ of ${\mathcal{P}}_4(\mathbb{F})$ such that ${\mathcal{P}}_4(\mathbb{F})=U\oplus W$.

(a) We can write $p(2)$ and $p(5)$ and equate them to each other:

$$\begin{array}{rl}{a}_0+2a_1+4a_2+8a_3+16a_4& =a_0+5a_1+25a_2+125a_3+625a^4\\ 0& =3a_1+21a_2+117a_3+609a_4\end{array}$$

This gives a constraint on the coefficients. We can solve the constraint explicitly for one variable:

$$a_1=-7a_2-39a_3-203a_4$$

Then we can re-write the original polynomial as:

$$\begin{array}{rl}p(x)& =a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4\\ p(x)& =a_0+(-7a_2-39a_3-203a_4)x+a_2{x}^2+a_3{x}^3+a_4{x}^4\end{array}$$

which means that any $p\in U$ has the form

$$p(x)=a_0+a_2(x^2-7x)+a_3(x^3-39x)+a_4(x^4-203x)$$

This results in a basis of:

$$\{1,x^2-7x,x^3-39x,x^4-203x\}$$

(b) We can add $x$

$$\{1,x,x^2-7x,x^3-39x,x^4-203x\}$$

(c) We can have $W=\text{span}(x)$.

Problem 6

• (a) Let $U=\{p\in {\mathcal{P}}_4(\mathbb{F}):p(2)=p(5)=p(6)=0\}$. Find a basis of $U$.
• (b) Extend the basis in (a) to a basis of ${\mathcal{P}}_4(\mathbb{F})$.
• (c) Find the subspace $W$ of ${\mathcal{P}}_4(\mathbb{F})$ such that ${\mathcal{P}}_4(\mathbb{F})=U\oplus W$.

(a) The condition $U=\{p\in P_4(\mathbb{F}):p(2)=p(5)=p(6)=0\}$ means that any polynomial $p(x)\in U$ must have the form

$$p(x)=(x-2)(x-5)(x-6)q(x)$$

where $q(x)$ is some polynomial of at most degree 1, since $p(x)\in P_4(\mathbb{F})$ and the degree of $(x-2)(x-5)(x-6)$ is already 3. This means that:

$$p(x)=a_3(x-2)(x-5)(x-6)+a_{4}x(x-2)(x-5)(x-6)$$

where $a_3,a_4\in \mathbb{F}$. The basis is thus

$$\{(x-2)(x-5)(x-6),x(x-2)(x-5)(x-6)\}$$

(b) The dimension of $P_4(\mathbb{F})$ is 5, and the basis of $U$ in part (a) only had 2 elements. Thus, we need three additional linearly independent polynomials. A simple choice is

$$\{1,x,x^2\}$$

Thus, the extended basis is

$$\{(x-2)(x-5)(x-6),x(x-2)(x-5)(x-6),1,x,x^2\}$$

(c) $W=\text{span}(1,x,x^2)$.

Problem 7

• (a) Let $U=\left\{p\in {\mathcal{P}}_4(\mathbb{F}):{\int }_{-1}^{1}p=0\right\}$. Find a basis of $U$.
• (b) Extend the basis in (a) to a basis of ${\mathcal{P}}_4(\mathbb{F})$.
• (c) Find the subspace $W$ of ${\mathcal{P}}_4(\mathbb{F})$ such that ${\mathcal{P}}_4(\mathbb{F})=U\oplus W$.

(a) We can write $p(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4$. The integral is then

$$\begin{array}{rl}{\int }_{-1}^{1}p(x)dx& ={\int }_{-1}^1{a}_{0}dx+{\int }_{-1}^1{a}_{1}xdx+{\int }_{-1}^1{a}_2{x}^{2}dx+{\int }_{-1}^1{a}_3{x}^{3}dx+{\int }_{-1}^1{a}_4{x}^{4}dx\\ & =2a_0+\frac{2}{3}{a}_2+\frac{2}{5}{a}_4\end{array}$$

so for $p(x)=0$, we have

$$\begin{array}{r}2a_0+\frac{2}{3}{a}_2+\frac{2}{5}{a}_4=0\\ a_0+\frac{1}{3}{a}_2+\frac{1}{5}{a}_4=0\end{array}$$

which gives

$$p(x)=\left(-\frac{1}{3}{a}_2-\frac{1}{5}{a}_4\right)+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4$$

Thus, the basis for $U$ is

$$\left\{x,x^2-\frac{1}{3},x^3,x^4-\frac{1}{5}\right\}$$

(b) We have $\dim U=4$ and $\dim P_4=5$, so we need to find one more linearly independent vector.

$$\left\{1,x,x^2-\frac{1}{3},x^3,x^4-\frac{1}{5}\right\}$$

(c) We have $W=\text{span}(1)$.

Problem 8

Suppose $v_1,\dots ,v_m$ is linearly independent in $V$ and $w\in V$. Prove that

$$\dim \text{span}(v_1+w,\dots ,v_m+w)\ge m-1$$

The span is

$$\text{span}(v_1+w,v_2+w,\dots ,v_m+w)=\text{span}(w,v_1,v_2,\dots ,v_m)$$

Since $v_1,\dots ,v_m$ are linearly independent, their span has dimension $m$. Adding $w$ changes the span, depending on whether $w$ is in the span of $v_1,\dots ,v_m$.

If $w\in \text{span}(v_1,\dots ,v_m)$, then $\text{span}(v_1,\dots ,v_m)=\text{span}(w,v_1,\dots ,v_m)$. In this case, $\dim \text{span}(w,v_1,\dots ,v_m)=m$.

If $w\notin \text{span}(v_1,\dots ,v_m)$, then $w$ adds an additional dimension, and the dimension becomes $m+1$.

Thus, in both cases, the minimum dimension is $m-1$.

Alternatively, we can note that

$$v_2-v_1=(v_2+w)-(v_1+w)$$

it follows that $v_2-v_1\in \text{span}(v_1+w,\dots ,v_m+w)$. Similarly, $v_i-v_1\in \text{span}(v_1+w,\dots ,v_m+w)$ for all $2\le i\le m$.

The list $v_2-v_1,\dots ,v_m-v_1$ is linearly independent since $v_1,\dots ,v_m$ is linearly independent in $V$. Thus, we have $\dim \text{span}(v_1+w,\dots ,v_m+w)\ge m-1$.

Problem 9

Suppose $m$ is a positive integer and $p_0,p_1,\dots ,p_m\in \mathcal{P}(\mathbb{F})$ are such that each $p_k$ has degree $k$. Prove that $p_0,p_1,\dots ,p_m$ is a basis of ${\mathcal{P}}_m(\mathbb{F})$.

To prove that $p_0,p_1,\dots ,p_m$ is a basis, they must span $P_m(\mathbb{F})$ and they must be linearly independent.

Any polynomial $q(x)\in P_m(\mathbb{F})$ can be written as a linear combination of polynomials of degree $0,1,\dots ,m$. Since the list $p_0,p_1,\dots ,p_m$ contains a polynomial from each of these degrees, their span must cover $P_m(\mathbb{F})$.

By constructions, the list is linearly independent since each of them is a polynomial with a different degree. We can also think of this recursively (the highest degree polynomial $p_m$ has a term $x^m$ that doesn’t exist in any other $p$‘s), so for $c_0{p}_0+\cdots +c_m{p}_m=0$, $c_m$ must be zero, and then we can work backward to see that all coefficients must be zero.

Problem 10

Suppose $m$ is a positive integer. For $0\le k\le m$, let

$$p_k(x)=x^k(1-x{)}^{m-k}$$

Show that $p_0,\dots ,p_m$ is a basis of ${\mathcal{P}}_m(\mathbb{F})$.

We need to show that they span $P_m(\mathbb{F})$ and they are linearly independent.

For linear independence, we need

$$\sum _{k=0}^m{c}_k{p}_k(x)=0\text{for all }x\in \mathbb{F}$$

where $c_k\in \mathbb{F}$ are constants. Substituting $p_k(x)=x^k(1-x{)}^{m-k}$, this becomes

$$\sum _{k=0}^m{c}_k{x}^k(1-x{)}^{m-k}=0$$

Each term $x^k(1-x{)}^{m-k}$ is a polynomial of degree $m$. These polynomials are linearly independent because they involve distinct contributions of powers of $x$ and $(1-x)$, which cannot cancel each other out unless all coefficients $c_k=0$.

Since the list contains $m+1$ linearly independent polynomials and $\dim (P_m(\mathbb{F}))=m+1$, it suffices to show linear independence. Thus, the list is a basis because linearly independent list of the right length is a basis.

Problem 11

Suppose $U$ and $W$ are both four-dimensional subspaces of ${\mathbb{C}}^6$. Prove that there exist two vectors in $U\cap W$ such that neither of these vectors is a scalar multiple of the other.

We treat ${\mathbb{C}}^6$ as a vector space over $\mathbb{C}$. We have

$$\dim (U+W)=\dim U+\dim W-\dim (U\cap W)$$

and thus $\dim U=\dim W=4$. Then we have

$$\dim (U+W)=8-\dim (U\cap W)$$

Since $U+W$ is a subspace of ${\mathbb{C}}^6$ and $\dim {\mathbb{C}}^6=6$, we must have $\dim (U+W)\le 6$. Since $\dim (U+W)\ge \max (\dim U,\dim W)=4$, we have

$$4\le \dim (U+W)\le 6$$

and hence

$$2\le \dim (U\cap W)\le 4$$

Thus, $U\cap W$ has a basis of length at least two. Since vectors in a basis are linearly independent from each other, there exist two vectors in $U\cap W$ such that neither is a scalar multiple of the other.

Problem 12

Suppose $U$ and $W$ are both five-dimensional subspaces of ${\mathbb{R}}^9$, such that $\dim U=3$, $\dim W=5$, and $U+W={\mathbb{R}}^8$. Prove that ${\mathbb{R}}^8=U\oplus W$.

We have

$$\dim (U+W)=\dim U+\dim W-\dim (U\cap W)$$

and since $U+W={\mathbb{R}}^8$ with $\dim U=5$ and $\dim W=3$, we have

$$8=3+5-\dim (U\cap W)$$

and hence we must have $\dim (U\cap W)=0$. Therefore, we must have $U\cap W=\{0\}$. By the condition for direct sum of two subspaces, we have ${\mathbb{R}}^8=U\oplus W$.

Problem 13

Suppose $U$ and $W$ are both five-dimensional subspaces of ${\mathbb{R}}^9$. Prove that $U\cap W\ne \{0\}$.

Since $U$ and $W$ are subspaces of ${\mathbb{R}}^9$, their sum is also a subspace (sums and scalar multiples of elements from $U+W$ remain in $U+W$). Thus, we must have:

$$\begin{array}{rl}\dim (U+W)& \le \dim ({\mathbb{R}}^9)\\ \dim (U+W)& \le 9\end{array}$$

We can write $\dim (U+W)$ as

$$\dim (U+W)=\dim U+\dim W-\dim (U\cap W)$$

which gives us

$$\begin{array}{rl}\dim U+\dim W-\dim (U\cap W)& \le 9\\ 5+5-\dim (U\cap W)& \le 9\end{array}$$

which means $\dim (U\cap W)\ge 1$. This means that $U\cap W\ne 0$, as this would give $\dim (U\cap W)=0$.

Problem 14

Suppose $V$ is a ten-dimensional vector space and $V_1,V_2,V_3$ are subspaces of $V$ with $\dim V_1=\dim V_2=\dim V_3=7$. Prove that $V_1\cap V_2\cap V_3\ne \{0\}$.

We must have

$$\dim (V_1\cap V_2)=\dim V_1+\dim V_2-\dim (V_1+V_2)$$

Since $V_1,V_2\subseteq V$, we must have $\dim (V_1+V_2)\le \dim V=10$. This implies that

$$\dim (V_1\cap V_2)\ge 7+7-10=4$$

Now we can consider the intersection $V_1\cap V_2\cap V_3$. Let $W=V_1\cap V_2$, and observe that

$$V_1\cap V_2\cap V_3=W\cap V_3$$

Since $\dim W\ge 4$ and $\dim (W+V_3)\le \dim V=10$ we now have

$$\begin{array}{rl}\dim (W\cap V_3)& =\dim W+\dim V_3-\dim (W+V_3)\\ & \ge 4+7-10\\ & \ge 1\end{array}$$

Because $\dim (V_1\cap V_2\cap V_3)\ge 1$, the intersection is non-trivial, so $V_1\cap V_2\cap V_3\ne \{0\}$.

Problem 15

Suppose $V$ is finite-dimensional and $V_1,V_2,V_3$ are subspaces of $V$ with $\dim V_1+\dim V_2+\dim V_3>2\dim V$. Prove that $V_1\cap V_2\cap V_3\ne \{0\}$.

Let $\dim V=n,\dim V_1=a,\dim V_2=b,\dim V_3=c$. We have $a+b+c>2n$.

We have

$$\dim (V_1+V_2)=a+b-\dim (V_1\cap V_2)$$

Then, we have

$$\begin{array}{rl}\dim (V_1+V_2+V_3)& =\dim (V_1+V_2)+c-\dim ((V_1+V_2)\cap V_3)\\ & \le a+b-\dim (V_1\cap V_2)-\dim ((V_1+V_2)\cap V_3)\end{array}$$

Since $V_1+V_2+V_3\subseteq V$, we have $\dim (V_1+V_2+V_3)\le n$. Thus:

$$n\ge a+b+c-\dim ((V_1+V_2)\cap V_3)$$

Re-arranging:

$$\dim ((V_1+V_2)\cap V_3)\ge a+b+c-n$$

but

$$\dim ((V_1+V_2)\cap V_3)=\dim V_3+\dim (V_1+V_2)-\dim (V_1+V_2+V_3)$$

because $\dim (A\cap B)=\dim A+\dim B-\dim (A+B)$).

$$\dim (V_1\cap V_2\cap V_3)\ge a+b+c-2n$$

From the assumption that $a+b+c>2n$, this implies

$$\dim (V_1\cap V_2\cap V_3)$$

Problem 16

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$ with $U\ne V$. Let $n=\dim V$ and $m=\dim U$. Prove that there exist $n-m$ subspaces of $V$, each of dimension $n-1$, whose intersection equals $U$.

We aim to construct $n-m$ subspaces $W_1,W_2,\dots ,W_{n-m}$, each of dimension $n-1$, such that

$$W_1\cap W_2\cap \cdots \cap W_{n-m}=U$$

Since $U$ is a subspace of $V$, we can extend a basis of $U$ to a basis of $V$. That is, the basis

$$\{u_1,u_2,\dots ,u_m\}$$

can be extended to a basis of $V$ by adding $n-m$ additional vectors:

$$\{u_1,u_2,\dots ,u_m,v_1,v_2\dots ,v_m\}$$

The additional $v_1,v_2,\dots ,v_{n-m}$ generate the complement of $U$ in $V$.

For each $v_i$ in $i=1,2,\dots ,n-m$, we can define a subspace $W_i$ of $V$ as follows:

$$W_i=\{x\in V:x\cdot v_i=0\}$$

Intuitively, $W_i$ is the hyperplane of $V$ orthogonal to $v_i$.

Each $W_i$ has dimension $n-1$, since imposing a single linear constraint reduced the dimension of $V$. We also have $U\subseteq W_i$, because elements of $U$ are linear combinations of $u_1,\dots ,u_m$, which are independent of $v_i$, so $U\subseteq W_i$. Thus, $x\in U⟹x\cdot v_i=0$.

Now, consider the intersection $W_1\cap W_2\cap \cdots \cap W_{n-m}$. Each $W_i$ imposes a constraint orthogonal to one of the $v_i$, ensuring that any $x\in W_1\cap W_2\cap \cdots \cap W_{n-m}$ satisfies $x\cdot v_i=0$ for all $i$.

The subspace of vectors orthogonal to all $v_1,\dots ,v_{n-m}$ is precisely $U$, because $\{v_1,\dots ,v_{n-m}\}$ spans the complement of $U$. Enforcing orthogonality to the complement leaves us with exactly $U$. Thus:

$$W_1\cap W_2\cap \cdots \cap W_{n-m}=U$$

With this, we have constructed $n-m$ subspaces $W_1,\dots ,W_{n-m}$, each of dimension $n-1$, such that their intersection is $U$.

Problem 17

Suppose $V_1,\dots ,V_m$ are finite-dimensional subspaces of $V$. Prove that $V_1+\cdots +V_m$ is finite-dimensional and

$$\dim (V_1+\cdots +V_m)\le \dim V_1+\cdots +\dim V_m$$

For each $i=1,\dots ,m$, choose a basis for $V_i$. Combine these bases to form a single list of vectors in $V$. This list clearly spans $V_1+\cdots +V_m$ by construction.

Hence, $V_1+\cdots +V_m$ is finite-dimensional, and has a dimension less than or equal to the number of vectors in the list we constructed. Notice that the number of vectors in the list is equal to $\dim V_1+\cdots +\dim V_m$, since the list is just combined bases for each $V_i$. That is,

$$\dim (V_1+\cdots +V_m)\le \dim V_1+\cdots +\dim V_m$$

as desired.

Problem 18

Suppose $V$ is finite-dimensional with $\dim V=n\ge 1$. Prove that there exist one-dimensional subspaces $V_1,\dots ,V_n$ of $V$ such that

$$V=V_1\oplus \cdots \oplus V_n$$

Since $\dim V=n$, there exists a basis $v_1,\dots ,v_n$ of $V$. Let $U_i=\text{span}(v_i)$ for $i=1,\dots ,n$, so that each $U_k$ has dimension $1$. Clearly,

$$V=U_1+\cdots +U_n$$

so we need to show that this sum is direct.

For some $u\in U_1+\cdots +U_n$, there exist $a_1,\dots ,a_n\in \mathbb{F}$, such that

$$u=a_1{v}_1+\cdots +a_n{v}_n$$

But since $v_1,\dots ,v_n$ is a basis, this representation of $u$ as a linear combination of $v_1,\dots ,v_n$ is unique, and thus the sum is direct.

Problem 19

Explain why you might guess, motivated by some analogy with the formula for the number of elements in the union of three finite sets, that if $V_1,V_2,V_3$ are subspaces of finite-dimensional vector space, then

\begin{align} \dim (V_{1}+ & V_{2}+V_{3}) \\ & = \dim V_{1}+\dim V_{2}+\dim V_{3}\\ & \quad - \dim (V_{1}\cap V_{2})-\dim (V_{1}\cap V_{3})-\dim (V_{2}\cap V_{3}) \\ & \quad +\dim (V_{1}\cap V_{2}\cap V_{3})

\end{align}

The statement is false. Consider

$$\begin{array}{rl}{V}_1& =\{(x,0)|x\in \mathbb{R}\}\\ V_2& =\{(x,x)\in {\mathbb{R}}^2|x\in \mathbb{R}\}\\ V_3& =\{(0,y)|y\in \mathbb{R}\}\end{array}$$

We have

$$\begin{array}{rl}& \dim (V_1+V_2+V_3)=\dim {\mathbb{R}}^2=2\\ & \dim V_1=\dim V_2=\dim V_3=1\\ & \dim (V_1\cap V_2)=\dim (V_2\cap V_3)=1\\ & \dim (V_1\cap V_3)=\dim (V_1\cap V_2\cap V_3)=0\end{array}$$

and therefore

$$\begin{array}{rl}\dim (V_1+V_2+V_3)& \ne \dim V_1+\dim V_2+\dim V_3\\ & -\dim (V_1\cap V_2)-\dim (V_1\cap V_3)-\dim (V_2\cap V_3)\\ & +\dim (V_1\cap V_2\cap V_3)\\ 2& \ne 1\end{array}$$

Problem 20

Prove that if $V_1$, $V_2$ and $V_3$ are subspaces of finite-dimensional vector space, then

$$\begin{array}{rl}\dim (& V_1+V_2+V_3)\\ & =\dim V_1+\dim V_2+\dim V_3\\ & -\frac{\dim (V_1\cap V_2)+\dim (V_1\cap V_3)+\dim (V_2\cap V_3)}{3}\\ & -\frac{\dim ((V_1+V_2)\cap V_3)+\dim ((V_1+V_3)\cap V_2)+\dim ((V_2+V_3)\cap V_1)}{3}\end{array}$$

This can be shown by successively applying the dimension formula for sums of subspaces, $\dim (A+B)=\dim A+\dim B-\dim (A\cap B)$.

We can treat $V_1+V_2$ as a single subspace, and add $V_3$. We have

$$\begin{array}{rl}\dim ((V_1+V_2)+V_3)& =\dim (V_1+V_2)+\dim V_3-\dim ((V_1+V_2)\cap V_3)\\ & =(\dim V_1+V_2-\dim (V_1\cap V_2))+\dim V_3-\dim ((V_1+V_2)\cap V_3)\\ & =\dim V_1+\dim V_2+\dim V_3-\dim (V_1\cap V_2)-\dim ((V_1+V_2)\cap V_3)\end{array}$$

We can apply this symmetrically for the other two ways to group and sum the subspaces

$$\dim ((V_1+V_3)+V_2)=\dim V_1+\dim V_3+\dim V_2-\dim (V_1\cap V_3)-\dim ((V_1+V_3)\cap V_2)$$

and

$$\dim ((V_2+V_3)+V_1)=\dim V_2+\dim V_3+\dim V_1-\dim (V_2\cap V_3)-\dim ((V_2+V_3)\cap V_1)$$

All three expressions are valid and symmetric. We take the average of these three expressions:

$$\dim (V_1+V_2+V_3)=\frac{1}{3}[(\text{Expression 1})+(\text{Expression 2})+(\text{Expression 3})]$$

Substituting gives us:

$$\begin{array}{rl}\dim (& V_1+V_2+V_3)\\ & =\dim V_1+\dim V_2+\dim V_3\\ & -\frac{\dim (V_1\cap V_2)+\dim (V_1\cap V_3)+\dim (V_2\cap V_3)}{3}\\ & -\frac{\dim ((V_1+V_2)\cap V_3)+\dim ((V_1+V_3)\cap V_2)+\dim ((V_2+V_3)\cap V_1)}{3}\end{array}$$