# Linear Algebra/Combining Subspaces/Solutions
## Solutions

    This exercise is recommended for all readers.

### Problem 1

Decide if $\mathbb {R} ^{2}$ is the direct sum of each $W_{1}$ and $W_{2}$.
* $W_{1}=\{{\begin{pmatrix}x\\0\end{pmatrix}}\,{\big |}\,x\in \mathbb {R} \}$, $W_{2}=\{{\begin{pmatrix}x\\x\end{pmatrix}}\,{\big |}\,x\in \mathbb {R} \}$
* $W_{1}=\{{\begin{pmatrix}s\\s\end{pmatrix}}\,{\big |}\,s\in \mathbb {R} \}$, $W_{2}=\{{\begin{pmatrix}s\\1.1s\end{pmatrix}}\,{\big |}\,s\in \mathbb {R} \}$
* $W_{1}=\mathbb {R} ^{2}$, $W_{2}=\{{\vec {0}}\}$
* $W_{1}=W_{2}=\{{\begin{pmatrix}t\\t\end{pmatrix}}\,{\big |}\,t\in \mathbb {R} \}$
* $W_{1}=\{{\begin{pmatrix}1\\0\end{pmatrix}}+{\begin{pmatrix}x\\0\end{pmatrix}}\,{\big |}\,x\in \mathbb {R} \}$, $W_{2}=\{{\begin{pmatrix}-1\\0\end{pmatrix}}+{\begin{pmatrix}0\\y\end{pmatrix}}\,{\big |}\,y\in \mathbb {R} \}$

### Answer

With each of these we can apply Lemma 4.15.
* Yes. The plane is the sum of this $W_{1}$ and $W_{2}$ because for any scalars $a$ and $b$ shows that the general vector is a sum of vectors from the two parts. And, these two subspaces are (different) lines through the origin, and so have a trivial intersection.
    ${\begin{pmatrix}a\\b\end{pmatrix}}={\begin{pmatrix}a-b\\0\end{pmatrix}}+{\begin{pmatrix}b\\b\end{pmatrix}}$
* Yes. To see that any vector in the plane is a combination of vectors from these parts, consider this relationship. We could now simply note that the set is a basis for the space (because it is clearly linearly independent, and has size two in $\mathbb {R} ^{2}$), and thus ther is one and only one solution to the above equation, implying that all decompositions are unique. Alternatively, we can solve to get that $c_{2}=10(-a+b)$ and $c_{1}=11a-10b$, and so we have as required. As with the prior answer, each of the two subspaces is a line through the origin, and their intersection is trivial.
    ${\begin{pmatrix}a\\b\end{pmatrix}}=c_{1}{\begin{pmatrix}1\\1\end{pmatrix}}+c_{2}{\begin{pmatrix}1\\1.1\end{pmatrix}}$
    $\{{\begin{pmatrix}1\\1\end{pmatrix}},{\begin{pmatrix}1\\1.1\end{pmatrix}}\}$
    ${\begin{array}{*{2}{rc}r}c_{1}&+&c_{2}&=&a\\c_{1}&+&1.1c_{2}&=&b\end{array}}\;{\xrightarrow[]{-\rho _{1}+\rho _{2}}}\;{\begin{array}{*{2}{rc}r}c_{1}&+&c_{2}&=&a\\&&0.1c_{2}&=&-a+b\end{array}}$
    ${\begin{pmatrix}a\\b\end{pmatrix}}={\begin{pmatrix}11a-10b\\11a-10b\end{pmatrix}}+{\begin{pmatrix}-10a+10b\\1.1\cdot (-10a+10b)\end{pmatrix}}$
* Yes. Each vector in the plane is a sum in this way and the intersection of the two subspaces is trivial.
    ${\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}x\\y\end{pmatrix}}+{\begin{pmatrix}0\\0\end{pmatrix}}$
* No. The intersection is not trivial.
* No. These are not subspaces.
    This exercise is recommended for all readers.

### Problem 2

Show that $\mathbb {R} ^{3}$ is the direct sum of the $xy$-plane with each of these.
* the $z$-axis
* the line
    $\{{\begin{pmatrix}z\\z\\z\end{pmatrix}}\,{\big |}\,z\in \mathbb {R} \}$

### Answer

With each of these we can use Lemma 4.15.
* Any vector in $\mathbb {R} ^{3}$ can be decomposed as this sum. And, the intersection of the $xy$-plane and the $z$-axis is the trivial subspace.
    ${\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}x\\y\\0\end{pmatrix}}+{\begin{pmatrix}0\\0\\z\end{pmatrix}}$
* Any vector in $\mathbb {R} ^{3}$ can be decomposed as and the intersection of the two spaces is trivial.
    ${\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}x-z\\y-z\\0\end{pmatrix}}+{\begin{pmatrix}z\\z\\z\end{pmatrix}}$

### Problem 3

Is ${\mathcal {P}}_{2}$ the direct sum of $\{a+bx^{2}\,{\big |}\,a,b\in \mathbb {R} \}$ and $\{cx\,{\big |}\,c\in \mathbb {R} \}$?

### Answer

It is. Showing that these two are subspaces is routine. To see that the space is the direct sum of these two, just note that each member of ${\mathcal {P}}_{2}$ has the unique decomposition $m+nx+px^{2}=(m+px^{2})+(nx)$.
    This exercise is recommended for all readers.

### Problem 4

In ${\mathcal {P}}_{n}$, the even polynomials are the members of this set
    ${\mathcal {E}}=\{p\in {\mathcal {P}}_{n}\,{\big |}\,p(-x)=p(x){\text{ for all }}x\}$
and the odd polynomials are the members of this set.
    ${\mathcal {O}}=\{p\in {\mathcal {P}}_{n}\,{\big |}\,p(-x)=-p(x){\text{ for all }}x\}$
Show that these are complementary subspaces.

### Answer

To show that they are subspaces is routine. We will argue they are complements with Lemma 4.15. The intersection ${\mathcal {E}}\cap {\mathcal {O}}$ is trivial because the only polynomial satisfying both conditions $p(-x)=p(x)$ and $p(-x)=-p(x)$ is the zero polynomial. To see that the entire space is the sum of the subspaces ${\mathcal {E}}+{\mathcal {O}}={\mathcal {P}}_{n}$, note that the polynomials $p_{0}(x)=1$, $p_{2}(x)=x^{2}$, $p_{4}(x)=x^{4}$, etc., are in ${\mathcal {E}}$ and also note that the polynomials $p_{1}(x)=x$, $p_{3}(x)=x^{3}$, etc., are in ${\mathcal {O}}$. Hence any member of ${\mathcal {P}}_{n}$ is a combination of members of ${\mathcal {E}}$ and ${\mathcal {O}}$.

### Problem 5

Which of these subspaces of $\mathbb {R} ^{3}$
$W_{1}$: the $x$-axis, $W_{2}$:the $y$-axis, $W_{3}$:the $z$-axis,  $W_{4}$:the plane $x+y+z=0$, $W_{5}$:the $yz$-plane
can be combined to
* sum to $\mathbb {R} ^{3}$?
* direct sum to $\mathbb {R} ^{3}$?

### Answer

Each of these is $\mathbb {R} ^{3}$.
* These are broken into lines for legibility. $W_{1}+W_{2}+W_{3}$, $W_{1}+W_{2}+W_{3}+W_{4}$, $W_{1}+W_{2}+W_{3}+W_{5}$, $W_{1}+W_{2}+W_{3}+W_{4}+W_{5}$,  $W_{1}+W_{2}+W_{4}$, $W_{1}+W_{2}+W_{4}+W_{5}$, $W_{1}+W_{2}+W_{5}$, $W_{1}+W_{3}+W_{4}$, $W_{1}+W_{3}+W_{5}$, $W_{1}+W_{3}+W_{4}+W_{5}$, $W_{1}+W_{4}$, $W_{1}+W_{4}+W_{5}$, $W_{1}+W_{5}$, $W_{2}+W_{3}+W_{4}$, $W_{2}+W_{3}+W_{4}+W_{5}$, $W_{2}+W_{4}$, $W_{2}+W_{4}+W_{5}$, $W_{3}+W_{4}$, $W_{3}+W_{4}+W_{5}$, $W_{4}+W_{5}$
* $W_{1}\oplus W_{2}\oplus W_{3}$, $W_{1}\oplus W_{4}$, $W_{1}\oplus W_{5}$, $W_{2}\oplus W_{4}$, $W_{3}\oplus W_{4}$
    This exercise is recommended for all readers.

### Problem 6

Show that ${\mathcal {P}}_{n}=\{a_{0}\,{\big |}\,a_{0}\in \mathbb {R} \}\oplus \dots \oplus \{a_{n}x^{n}\,{\big |}\,a_{n}\in \mathbb {R} \}$.

### Answer

Clearly each is a subspace. The bases $B_{i}=\langle x^{i}\rangle$ for the subspaces, when concatenated, form a basis for the whole space.

### Problem 7

What is $W_{1}+W_{2}$ if $W_{1}\subseteq W_{2}$?

### Answer

It is $W_{2}$.

### Problem 8

Does Example 4.5 generalize? That is, is this true or false:if a vector space $V$ has a basis $\langle {\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{n}\rangle$ then it is the direct sum of the spans of the one-dimensional subspaces $V=[\{{\vec {\beta }}_{1}\}]\oplus \dots \oplus [\{{\vec {\beta }}_{n}\}]$?

### Answer

True by Lemma 4.8.

### Problem 9

Can $\mathbb {R} ^{4}$ be decomposed as a direct sum in two different ways? Can $\mathbb {R} ^{1}$?

### Answer

Two distinct direct sum decompositions of $\mathbb {R} ^{4}$ are easy to find. Two such are $W_{1}=[\{{\vec {e}}_{1},{\vec {e}}_{2}\}]$ and $W_{2}=[\{{\vec {e}}_{3},{\vec {e}}_{4}\}]$, and also $U_{1}=[\{{\vec {e}}_{1}\}]$ and $U_{2}=[\{{\vec {e}}_{2},{\vec {e}}_{3},{\vec {e}}_{4}\}]$. (Many more are possible, for example $\mathbb {R} ^{4}$ and its trivial subspace.)
In contrast, any partition of $\mathbb {R} ^{1}$'s single-vector basis will give one basis with no elements and another with a single element. Thus any decomposition involves $\mathbb {R} ^{1}$ and its trivial subspace.

### Problem 10

This exercise makes the notation of writing "$+$" between sets more natural. Prove that, where $W_{1},\dots ,W_{k}$ are subspaces of a vector space,
    $W_{1}+\dots +W_{k}=\{{\vec {w}}_{1}+{\vec {w}}_{2}+\dots +{\vec {w}}_{k}\,{\big |}\,{\vec {w}}_{1}\in W_{1},\dots ,{\vec {w}}_{k}\in W_{k}\},$
and so the sum of subspaces is the subspace of all sums.

### Answer

Set inclusion one way is easy: $\{{\vec {w}}_{1}+\dots +{\vec {w}}_{k}\,{\big |}\,{\vec {w}}_{i}\in W_{i}\}$ is a subset of $[W_{1}\cup \dots \cup W_{k}]$ because each ${\vec {w}}_{1}+\dots +{\vec {w}}_{k}$ is a sum of vectors from the union.
For the other inclusion, to any linear combination of vectors from the union apply commutativity of vector addition to put vectors from $W_{1}$ first, followed by vectors from $W_{2}$, etc. Add the vectors from $W_{1}$ to get a ${\vec {w}}_{1}\in W_{1}$, add the vectors from $W_{2}$ to get a ${\vec {w}}_{2}\in W_{2}$, etc. The result has the desired form.

### Problem 11

(Refer to Example 4.19. This exercise shows that the requirement that pariwise intersections be trivial is genuinely stronger than the requirement only that the intersection of all of the subspaces be trivial.) Give a vector space and three subspaces $W_{1}$, $W_{2}$, and $W_{3}$ such that the space is the sum of the subspaces, the intersection of all three subspaces $W_{1}\cap W_{2}\cap W_{3}$ is trivial, but the pairwise intersections $W_{1}\cap W_{2}$, $W_{1}\cap W_{3}$, and $W_{2}\cap W_{3}$ are nontrivial.

### Answer

One example is to take the space to be $\mathbb {R} ^{3}$, and to take the subspaces to be the $xy$-plane, the $xz$-plane, and the $yz$-plane.
    This exercise is recommended for all readers.

### Problem 12

Prove that if $V=W_{1}\oplus \dots \oplus W_{k}$ then $W_{i}\cap W_{j}$ is trivial whenever $i\neq j$. This shows that the first half of the proof of Lemma 4.15 extends to the case of more than two subspaces. (Example 4.19 shows that this implication does not reverse; the other half does not extend.)

### Answer

Of course, the zero vector is in all of the subspaces, so the intersection contains at least that one vector. By the definition of direct sum the set $\{W_{1},\dots ,W_{k}\}$ is independent and so no nonzero vector of $W_{i}$ is a multiple of a member of $W_{j}$, when $i\neq j$. In particular, no nonzero vector from $W_{i}$ equals a member of $W_{j}$.

### Problem 13

Recall that no linearly independent set contains the zero vector. Can an independent set of subspaces contain the trivial subspace?

### Answer

It can contain a trivial subspace; this set of subspaces of $\mathbb {R} ^{3}$ is independent: $\{\{{\vec {0}}\},x{\text{-axis}}\}$. No nonzero vector from the trivial space $\{{\vec {0}}\}$ is a multiple of a from the $x$-axis, simply because the trivial space has no nonzero vectors to be candidates for such a multiple (and also no nonzero vector from the $x$-axis is a multiple of the zero vector from the trivial subspace).
    This exercise is recommended for all readers.

### Problem 14

Does every subspace have a complement?

### Answer

Yes. For any subspace of a vector space we can take any basis $\langle {\vec {\omega }}_{1},\dots ,{\vec {\omega }}_{k}\rangle$ for that subspace and extend it to a basis $\langle {\vec {\omega }}_{1},\dots ,{\vec {\omega }}_{k},{\vec {\beta }}_{k+1},\dots ,{\vec {\beta }}_{n}\rangle$ for the whole space. Then the complement of the original subspace has this for a basis: $\langle {\vec {\beta }}_{k+1},\dots ,{\vec {\beta }}_{n}\rangle$.
    This exercise is recommended for all readers.

### Problem 15

Let $W_{1},W_{2}$ be subspaces of a vector space.
* Assume that the set $S_{1}$ spans $W_{1}$, and that the set $S_{2}$ spans $W_{2}$. Can $S_{1}\cup S_{2}$ span $W_{1}+W_{2}$? Must it?
* Assume that $S_{1}$ is a linearly independent subset of $W_{1}$ and that $S_{2}$ is a linearly independent subset of $W_{2}$. Can $S_{1}\cup S_{2}$ be a linearly independent subset of $W_{1}+W_{2}$? Must it?

### Answer

* It must. Any member of $W_{1}+W_{2}$ can be written ${\vec {w}}_{1}+{\vec {w}}_{2}$ where ${\vec {w}}_{1}\in W_{1}$ and ${\vec {w}}_{2}\in W_{2}$. As $S_{1}$ spans $W_{1}$, the vector ${\vec {w}}_{1}$ is a combination of members of $S_{1}$. Similarly ${\vec {w}}_{2}$ is a combination of members of $S_{2}$.
* An easy way to see that it can be linearly independent is to take each to be the empty set. On the other hand, in the space $\mathbb {R} ^{1}$, if $W_{1}=\mathbb {R} ^{1}$ and $W_{2}=\mathbb {R} ^{1}$ and $S_{1}=\{1\}$ and $S_{2}=\{2\}$, then their union $S_{1}\cup S_{2}$ is not independent.

### Problem 16

When a vector space is decomposed as a direct sum, the dimensions of the subspaces add to the dimension of the space. The situation with a space that is given as the sum of its subspaces is not as simple. This exercise considers the two-subspace special case.
* For these subspaces of ${\mathcal {M}}_{2\!\times \!2}$ find $W_{1}\cap W_{2}$, $\dim(W_{1}\cap W_{2})$, $W_{1}+W_{2}$, and $\dim(W_{1}+W_{2})$.
    $W_{1}=\{{\begin{pmatrix}0&0\\c&d\end{pmatrix}}\,{\big |}\,c,d\in \mathbb {R} \}\qquad W_{2}=\{{\begin{pmatrix}0&b\\c&0\end{pmatrix}}\,{\big |}\,b,c\in \mathbb {R} \}$
* Suppose that $U$ and $W$ are subspaces of a vector space. Suppose that the sequence $\langle {\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{k}\rangle$ is a basis for $U\cap W$. Finally, suppose that the prior sequence has been expanded to give a sequence $\langle {\vec {\mu }}_{1},\dots ,{\vec {\mu }}_{j},{\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{k}\rangle$ that is a basis for $U$, and a sequence $\langle {\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{k},{\vec {\omega }}_{1},\dots ,{\vec {\omega }}_{p}\rangle$ that is a basis for $W$. Prove that this sequence is a basis for for the sum $U+W$.
    $\langle {\vec {\mu }}_{1},\dots ,{\vec {\mu }}_{j},{\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{k},{\vec {\omega }}_{1},\dots ,{\vec {\omega }}_{p}\rangle$
* Conclude that $\dim(U+W)=\dim(U)+\dim(W)-\dim(U\cap W)$.
* Let $W_{1}$ and $W_{2}$ be eight-dimensional subspaces of a ten-dimensional space. List all values possible for $\dim(W_{1}\cap W_{2})$.

### Answer

* The intersection and sum are which have dimensions one and three.
    $\{{\begin{pmatrix}0&0\\c&0\end{pmatrix}}\,{\big |}\,c\in \mathbb {R} \}\qquad \{{\begin{pmatrix}0&b\\c&d\end{pmatrix}}\,{\big |}\,b,c,d\in \mathbb {R} \}$
* We write $B_{U\cap W}$ for the basis for $U\cap W$, we write $B_{U}$ for the basis for $U$, we write $B_{W}$ for the basis for $W$, and we write $B_{U+W}$ for the basis under consideration. To see that that $B_{U+W}$ spans $U+W$, observe that any vector $c{\vec {u}}+d{\vec {w}}$ from $U+W$ can be written as a linear combination of the vectors in $B_{U+W}$, simply by expressing ${\vec {u}}$ in terms of $B_{U}$ and expressing ${\vec {w}}$ in terms of $B_{W}$. We finish by showing that $B_{U+W}$ is linearly independent. Consider can be rewritten in this way. Note that the left side sums to a vector in $U$ while right side sums to a vector in $W$, and thus both sides sum to a member of $U\cap W$. Since the left side is a member of $U\cap W$, it is expressible in terms of the members of $B_{U\cap W}$, which gives the combination of ${\vec {\mu }}$'s from the left side above as equal to a combination of ${\vec {\beta }}$'s. But, the fact that the basis $B_{U}$ is linearly independent shows that any such combination is trivial, and in particular, the coefficients $c_{1}$, ..., $c_{j}$ from the left side above are all zero. Similarly, the coefficients of the ${\vec {\omega }}$'s are all zero. This leaves the above equation as a linear relationship among the ${\vec {\beta }}$'s, but $B_{U\cap W}$ is linearly independent, and therefore all of the coefficients of the ${\vec {\beta }}$'s are also zero.
    $c_{1}{\vec {\mu }}_{1}+\dots +c_{j+1}{\vec {\beta }}_{1}+\dots +c_{j+k+p}{\vec {\omega }}_{p}={\vec {0}}$
    $c_{1}{\vec {\mu }}_{1}+\dots +c_{j}{\vec {\mu }}_{j}=-c_{j+1}{\vec {\beta }}_{1}-\dots -c_{j+k+p}{\vec {\omega }}_{p}$
* Just count the basis vectors in the prior item:$\dim(U+W)=j+k+p$, and $\dim(U)=j+k$, and $\dim(W)=k+p$, and $\dim(U\cap W)=k$.
* We know that $\dim(W_{1}+W_{2})=\dim(W_{1})+\dim(W_{2})-\dim(W_{1}\cap W_{2})$. Because $W_{1}\subseteq W_{1}+W_{2}$, we know that $W_{1}+W_{2}$ must have dimension greater than that of $W_{1}$, that is, must have dimension eight, nine, or ten. Substituting gives us three possibilities $8=8+8-\dim(W_{1}\cap W_{2})$ or $9=8+8-\dim(W_{1}\cap W_{2})$ or $10=8+8-\dim(W_{1}\cap W_{2})$. Thus $\dim(W_{1}\cap W_{2})$ must be either eight, seven, or six. (Giving examples to show that each of these three cases is possible is easy, for instance in $\mathbb {R} ^{10}$.)

### Problem 17

Let $V=W_{1}\oplus \dots \oplus W_{k}$ and for each index $i$ suppose that $S_{i}$ is a linearly independent subset of $W_{i}$. Prove that the union of the $S_{i}$'s is linearly independent.

### Answer

Expand each $S_{i}$ to a basis $B_{i}$ for $W_{i}$. The concatenation of those bases $B_{1}\!{\mathbin {^{\frown }}}\!\cdots \!{\mathbin {^{\frown }}}\!B_{k}$ is a basis for $V$ and thus its members form a linearly independent set. But the union $S_{1}\cup \cdots \cup S_{k}$ is a subset of that linearly independent set, and thus is itself linearly independent.

### Problem 18

A matrix is symmetric if for each pair of indices $i$ and $j$, the $i,j$ entry equals the $j,i$ entry. A matrix is antisymmetric if each $i,j$ entry is the negative of the $j,i$ entry.
* Give a symmetric $2\!\times \!2$ matrix and an antisymmetric $2\!\times \!2$ matrix. (Remark. For the second one, be careful about the entries on the diagional.)
* What is the relationship between a square symmetric matrix and its transpose? Between a square antisymmetric matrix and its transpose?
* Show that ${\mathcal {M}}_{n\!\times \!n}$ is the direct sum of the space of symmetric matrices and the space of antisymmetric matrices.

### Answer

* Two such are these. For the antisymmetric one, entries on the diagonal must be zero.
    ${\begin{pmatrix}1&2\\2&3\end{pmatrix}}\qquad {\begin{pmatrix}0&1\\-1&0\end{pmatrix}}$
* A square symmetric matrix equals its transpose. A square antisymmetric matrix equals the negative of its transpose.
* Showing that the two sets are subspaces is easy. Suppose that $A\in {\mathcal {M}}_{n\!\times \!n}$.To express $A$ as a sum of a symmetric and an antisymmetric matrix, we observe that and note the first summand is symmetric while the second is antisymmetric. Thus ${\mathcal {M}}_{n\!\times \!n}$ is the sum of the two subspaces. To show that the sum is direct, assume a matrix $A$ is both symmetric $A={{A}^{\rm {trans}}}$ and antisymmetric $A=-{{A}^{\rm {trans}}}$. Then $A=-A$ and so all of $A$'s entries are zeroes.
    $A=(1/2)(A+{{A}^{\rm {trans}}})+(1/2)(A-{{A}^{\rm {trans}}})$

### Problem 19

Let $W_{1},W_{2},W_{3}$ be subspaces of a vector space. Prove that $(W_{1}\cap W_{2})+(W_{1}\cap W_{3})\subseteq W_{1}\cap (W_{2}+W_{3})$. Does the inclusion reverse?

### Answer

Assume that ${\vec {v}}\in (W_{1}\cap W_{2})+(W_{1}\cap W_{3})$. Then ${\vec {v}}={\vec {w}}_{2}+{\vec {w}}_{3}$ where ${\vec {w}}_{2}\in W_{1}\cap W_{2}$ and ${\vec {w}}_{3}\in W_{1}\cap W_{3}$. Note that ${\vec {w}}_{2},{\vec {w}}_{3}\in W_{1}$ and, as a subspace is closed under addition, ${\vec {w}}_{2}+{\vec {w}}_{3}\in W_{1}$. Thus ${\vec {v}}={\vec {w}}_{2}+{\vec {w}}_{3}\in W_{1}\cap (W_{2}+W_{3})$.
This example proves that the inclusion may be strict: in $\mathbb {R} ^{2}$ take $W_{1}$ to be the $x$-axis, take $W_{2}$ to be the $y$-axis, and take $W_{3}$ to be the line $y=x$. Then $W_{1}\cap W_{2}$ and $W_{1}\cap W_{3}$ are trivial and so their sum is trivial. But $W_{2}+W_{3}$ is all of $\mathbb {R} ^{2}$ so $W_{1}\cap (W_{2}+W_{3})$ is the $x$-axis.

### Problem 20

The example of the $x$-axis and the $y$-axis in $\mathbb {R} ^{2}$ shows that $W_{1}\oplus W_{2}=V$ does not imply that $W_{1}\cup W_{2}=V$. Can $W_{1}\oplus W_{2}=V$ and $W_{1}\cup W_{2}=V$ happen?

### Answer

It happens when at least one of $W_{1},W_{2}$ is trivial. But that is the only way it can happen.
To prove this, assume that both are non-trivial, select nonzero vectors ${\vec {w}}_{1},{\vec {w}}_{2}$ from each, and consider ${\vec {w}}_{1}+{\vec {w}}_{2}$. This sum is not in $W_{1}$ because ${\vec {w}}_{1}+{\vec {w}}_{2}={\vec {v}}\in W_{1}$ would imply that ${\vec {w}}_{2}={\vec {v}}-{\vec {w}}_{1}$ is in $W_{1}$, which violates the assumption of the independence of the subspaces. Similarly, ${\vec {w}}_{1}+{\vec {w}}_{2}$ is not in $W_{2}$. Thus there is an element of $V$ that is not in $W_{1}\cup W_{2}$.
    This exercise is recommended for all readers.

### Problem 21

Our model for complementary subspaces, the $x$-axis and the $y$-axis in $\mathbb {R} ^{2}$, has one property not used here. Where $U$ is a subspace of $\mathbb {R} ^{n}$ we define the orthogonal complement of $U$ to be
    $U^{\perp }=\{{\vec {v}}\in \mathbb {R} ^{n}\,{\big |}\,{\vec {v}}\cdot {\vec {u}}=0{\text{ for all }}{\vec {u}}\in U\}$
(read "$U$ perp").
* Find the orthocomplement of the $x$-axis in $\mathbb {R} ^{2}$.
* Find the orthocomplement of the $x$-axis in $\mathbb {R} ^{3}$.
* Find the orthocomplement of the $xy$-plane in $\mathbb {R} ^{3}$.
* Show that the orthocomplement of a subspace is a subspace.
* Show that if $W$ is the orthocomplement of $U$ then $U$ is the orthocomplement of $W$.
* Prove that a subspace and its orthocomplement have a trivial intersection.
* Conclude that for any $n$ and subspace $U\subseteq \mathbb {R} ^{n}$ we have that $\mathbb {R} ^{n}=U\oplus U^{\perp }$.
* Show that $\dim(U)+\dim(U^{\perp })$ equals the dimension of the enclosing space.

### Answer

* The set is easily seen to be the $y$-axis.
    $\{{\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}\,{\big |}\,{\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}\cdot {\begin{pmatrix}x\\0\end{pmatrix}}=0{\text{ for all }}x\in \mathbb {R} \}$
* The $yz$-plane.
* The $z$-axis.
* Assume that $U$ is a subspace of some $\mathbb {R} ^{n}$. Because $U^{\perp }$ contains the zero vector, since that vector is perpendicular to everything, we need only show that the orthocomplement is closed under linear combinations of two elements. If ${\vec {w}}_{1},{\vec {w}}_{2}\in U^{\perp }$ then ${\vec {w}}_{1}\cdot {\vec {u}}=0$ and ${\vec {w}}_{2}\cdot {\vec {u}}=0$ for all ${\vec {u}}\in U$. Thus $(c_{1}{\vec {w}}_{1}+c_{2}{\vec {w}}_{2})\cdot {\vec {u}}=c_{1}({\vec {w}}_{1}\cdot {\vec {u}})+c_{2}({\vec {w}}_{2}\cdot {\vec {u}})=0$ for all ${\vec {u}}\in U$ and so $U^{\perp }$ is closed under linear combinations.
* The only vector orthogonal to itself is the zero vector.
* This is immediate.
* To prove that the dimensions add, it suffices by Corollary 4.13 and Lemma 4.15 to show that $U\cap U^{\perp }$ is the trivial subspace $\{{\vec {0}}\}$. But this is one of the prior items in this problem.
    This exercise is recommended for all readers.

### Problem 22

Consider Corollary 4.13. Does it work both ways— that is, supposing that $V=W_{1}+\dots +W_{k}$, is $V=W_{1}\oplus \dots \oplus W_{k}$ if and only if $\dim(V)=\dim(W_{1})+\dots +\dim(W_{k})$?

### Answer

Yes. The left-to-right implication is Corollary 4.13. For the other direction, assume that $\dim(V)=\dim(W_{1})+\dots +\dim(W_{k})$. Let $B_{1},\dots ,B_{k}$ be bases for $W_{1},\dots ,W_{k}$. As $V$ is the sum of the subspaces, any ${\vec {v}}\in V$ can be written ${\vec {v}}={\vec {w}}_{1}+\cdots +{\vec {w}}_{k}$ and expressing each ${\vec {w}}_{i}$ as a combination of vectors from the associated basis $B_{i}$ shows that the concatenation $B_{1}\!{\mathbin {^{\frown }}}\!\cdots \!{\mathbin {^{\frown }}}\!B_{k}$ spans $V$. Now, that concatenation has $\dim(W_{1})+\dots +\dim(W_{k})$ members, and so it is a spanning set of size $\dim(V)$. The concatenation is therefore a basis for $V$. Thus $V$ is the direct sum.

### Problem 23

We know that if $V=W_{1}\oplus W_{2}$ then there is a basis for $V$ that splits into a basis for $W_{1}$ and a basis for $W_{2}$. Can we make the stronger statement that every basis for $V$ splits into a basis for $W_{1}$ and a basis for $W_{2}$?

### Answer

No. The standard basis for $\mathbb {R} ^{2}$ does not split into bases for the complementary subspaces the line $x=y$ and the line $x=-y$.

### Problem 24

We can ask about the algebra of the "$+$" operation.
* Is it commutative; is $W_{1}+W_{2}=W_{2}+W_{1}$?
* Is it associative; is $(W_{1}+W_{2})+W_{3}=W_{1}+(W_{2}+W_{3})$?
* Let $W$ be a subspace of some vector space. Show that $W+W=W$.
* Must there be an identity element, a subspace $I$ such that $I+W=W+I=W$ for all subspaces $W$?
* Does left-cancelation hold:if $W_{1}+W_{2}=W_{1}+W_{3}$ then $W_{2}=W_{3}$? Right cancelation?

### Answer

* Yes, $W_{1}+W_{2}=W_{2}+W_{1}$ for all subspaces $W_{1},W_{2}$ because each side is the span of $W_{1}\cup W_{2}=W_{2}\cup W_{1}$.
* This one is similar to the prior one— each side of that equation is the span of $(W_{1}\cup W_{2})\cup W_{3}=W_{1}\cup (W_{2}\cup W_{3})$.
* Because this is an equality between sets, we can show that it holds by mutual inclusion. Clearly $W\subseteq W+W$. For $W+W\subseteq W$ just recall that every subset is closed under addition so any sum of the form ${\vec {w}}_{1}+{\vec {w}}_{2}$ is in $W$.
* In each vector space, the identity element with respect to subspace addition is the trivial subspace.
* Neither of left or right cancelation needs to hold. For an example, in $\mathbb {R} ^{3}$ take $W_{1}$ to be the $xy$-plane, take $W_{2}$ to be the $x$-axis, and take $W_{3}$ to be the $y$-axis.

### Problem 25

Consider the algebraic properties of the direct sum operation.
* Does direct sum commute: does $V=W_{1}\oplus W_{2}$ imply that $V=W_{2}\oplus W_{1}$?
* Prove that direct sum is associative:$(W_{1}\oplus W_{2})\oplus W_{3}=W_{1}\oplus (W_{2}\oplus W_{3})$.
* Show that $\mathbb {R} ^{3}$ is the direct sum of the three axes (the relevance here is that by the previous item, we needn't specify which two of the three axes are combined first).
* Does the direct sum operation left-cancel:does $W_{1}\oplus W_{2}=W_{1}\oplus W_{3}$ imply $W_{2}=W_{3}$? Does it right-cancel?
* There is an identity element with respect to this operation. Find it.
* Do some, or all, subspaces have inverses with respect to this operation:is there a subspace $W$ of some vector space such that there is a subspace $U$ with the property that $U\oplus W$ equals the identity element from the prior item?

### Answer

* They are equal because for each, $V$ is the direct sum if and only if each ${\vec {v}}\in V$ can be written in a unique way as a sum ${\vec {v}}={\vec {w}}_{1}+{\vec {w}}_{2}$ and ${\vec {v}}={\vec {w}}_{2}+{\vec {w}}_{1}$.
* They are equal because for each, $V$ is the direct sum if and only if each ${\vec {v}}\in V$ can be written in a unique way as a sum of a vector from each ${\vec {v}}=({\vec {w}}_{1}+{\vec {w}}_{2})+{\vec {w}}_{3}$ and ${\vec {v}}={\vec {w}}_{1}+({\vec {w}}_{2}+{\vec {w}}_{3})$.
* Any vector in $\mathbb {R} ^{3}$ can be decomposed uniquely into the sum of a vector from each axis.
* No. For an example, in $\mathbb {R} ^{2}$ take $W_{1}$ to be the $x$-axis, take $W_{2}$ to be the $y$-axis, and take $W_{3}$ to be the line $y=x$.
* In any vector space the trivial subspace acts as the identity element with respect to direct sum.
* In any vector space, only the trivial subspace has a direct-sum inverse (namely, itself). One way to see this is that dimensions add, and so increase.