Definition 1.3.1.
Let \(V\) be a vector space, and let \(U\subseteq V\) be a subset. We say that \(U\) is a subspace of \(V\) if \(U\) is itself a vector space when using the addition and scalar multiplication of \(V\text{.}\)
Section 1.3 Subspaces
We begin with a motivating example. Let \(\vv\) be a nonzero vector in some vector space \(V\text{.}\) Consider the set \(S = \{c\vv\,|\, c\in \R\}\text{.}\) Given \(a\vv,b\vv\in S\text{,}\) notice that \(a\vv+b\vv=(a+b)\vv\) is also an element of \(S\text{,}\) since \(a+b\) is again a real number. Moreover, for any real number \(c\text{,}\) \(c(a\vv)=(ca)\vv\) is an element of \(S\text{.}\)
There are two important observations: one is that performing addition or scalar multiplication on elements of \(S\) produces a new element of \(S\text{.}\) The other is that this addition and multiplication is essentially that of \(\R\text{.}\) The vector \(\vv\) is just a placeholder. Addition simply involves the real number addition \(a+b\text{.}\) Scalar multiplication becomes the real number multiplication \(ca\text{.}\) So we expect that the rules for addition and scalar multiplication in \(S\) follow those in \(\R\text{,}\) so that \(S\) is like a “copy” of \(\R\) inside of \(V\text{.}\) In particular, addition and scalar multiplication in \(S\) will satisfy all the vector space axioms, so that \(S\) deserves to be considered a vector space in its own right.
A similar thing happens if we consider a set \(U=\{a\vv+b\ww\,|\, a,b\in\R\}\text{,}\) where \(\vv,\ww\) are two vectors in a vector space \(V\text{.}\) Given two elements \(a_1\vv+a_2\ww,b_1\vv+b_2\ww\text{,}\) we have
\begin{equation*}
(a_1\vv+a_2\ww)+(b_1\vv+b_2\ww) = (a_1+b_1)\vv+(a_2+b_2)\ww\text{,}
\end{equation*}
which is again an element of \(U\text{,}\) and the addition rule looks an awful lot like the addition rule \((a_1,a_2)+(b_1,b_2)=(a_1+b_1,a_2+b_2)\) in \(\R^2\text{.}\) Scalar multiplication follows a similar pattern.
In general we are often interested in subsets of vectors spaces that behave like “copies” of smaller vector spaces contained within the larger space. The technical term for this is subspace.
If we were to follow the definition, then verifying that a subset \(U\) is a subspace would involve checking all ten vector space axioms. Fortunately, this is not necessary. Since the operations are those of the vector space \(V\text{,}\) most properties follow automatically, being inherited from those of \(V\text{.}\)
Theorem 1.3.2. Subspace Test.
Let \(V\) be a vector space and let \(U\subseteq V\) be a subset. Then \(U\) is a subspace of \(V\) if and only if the following conditions are satisfied:
-
\(U\) is closed under addition. That is, for all \(\uu_1,\uu_2\in U\text{,}\) we have \(\uu_1+\uu_2\in U\text{.}\)
-
\(U\) is closed under scalar multiplication. That is, for all \(\uu\in U\) and \(c\in\R\text{,}\) \(c\uu\in U\text{.}\)
Proof.
If \(U\) is a vector space, then the second and third conditions must hold, since they are part of the axioms of a vector space. Since a vector space must be nonempty, there is some \(\uu\in U\text{,}\) from which it follows that \(\zer=0\uu\in U\text{.}\)
Conversely, if all three conditions hold, we have axioms A1, A4, and S1 by assumption. Axioms A2 and A3 hold since any vector in \(U\) is also a vector in \(V\text{;}\) the same reasoning shows that axioms S2, S3, S4, and S5 hold. Finally, axiom A5 holds because condition 3 ensures that \((-1)\uu\in U\) for any \(\uu\in U\text{,}\) and we know that \((-1)\uu=-\uu\) by Exercise 1.2.2.
In some texts, the condition that \(\zer\in U\) is replaced by the requirement that \(U\) be nonempty. Existence of \(\zer\) then follows from the fact that \(0\vv=\zer\text{.}\) However, it is usually easy to check that a set contains the zero vector, so it’s the first thing one typically looks for when confirming that a subset is nonempty.
Example 1.3.3.
For any vector space \(V\text{,}\) the set \(\{\zer\}\) is a subspace, known as the trivial subspace.
If \(V=P(\R)\) is the vector space of all polynomials, then for any natural number \(n\text{,}\) the subset \(U\) of all polynomials of degree less than or equal to \(n\) is a subspace of \(V\text{.}\) Another common type of polynomial subspace is the set of all polynomials with a given root. For example, the set \(U=\{p(x)\in P(\R)\,|\,p(1)=0\}\) is easily confirmed to be a subspace. However, a condition such as \(p(1)=2\) would not define a subspace, since this condition is not satisfied by the zero polynomial.
In \(\R^n\text{,}\) we can define a subspace using one or more homogeneous linear equations. For example, the set
\begin{equation*}
\{(x,y,z)\,|\, 2x-3y+4z=0\}
\end{equation*}
is a subspace of \(\R^3\text{.}\) A non-homogeneous equation won’t work, however, since it would exclude the zero vector. Of course, we should expect that any non-linear equation fails to define a subspace, although one is still expected to verify this by confirming the failure of one of the axioms. For example, the set \(S=\{(x,y)\,|\,x=y^2\}\) is not a subspace; although it contains the zero vector (since \(0^2=0\)), we have \((1,1)\in S\text{,}\) but \(2(1,1)=(2,2)\) does not belong to \(S\text{.}\)
Example 1.3.4.
In the vector space \(\R^3\text{,}\) we can visualize subspaces geometrically. There are precisely four types:
-
The trivial vector space \(\{\zer\}\subseteq \R^3\text{,}\) consisting of the origin alone.
-
Subspaces of the form \(\{t\vv \,|\, \vv\in\R^3, \vv\neq \zer\}\text{.}\) These are lines through the origin, in the direction of the vector \(\vv\text{.}\)
-
Subspaces of the form \(\{s\vv+t\ww \,|\, \vv,\ww\in\R^3\}\text{,}\) where \(\vv,\ww\) are both nonzero vectors that are not parallel. These are planes through the origin.Note that we must insist that \(\vv\) is not parallel to \(\ww\text{.}\) If \(\ww = c\vv\) for some scalar \(c\text{,}\) then\begin{equation*} s\vv+t\ww = s\vv+t(c\vv) = s\vv+(tc)\vv = (s+tc)\vv\text{,} \end{equation*}and every vector in our set would be a multiple of \(\vv\text{;}\) in other words, we’d once again have a line.If you encountered the cross product in your first course in linear algebra, or in a calculus course, then you can state the “non-parallel” condition by the requirement that \(\vv\times\ww\neq\zer\text{.}\) The vector \(\vv\times\ww\) is then a normal vector for the plane.
-
The entire vector space \(\R^3\subseteq\R^3\) also counts as a subspace: every vector space is a subspace of itself.
Remark 1.3.5.
Often we define subsets of a vector space by an equation. For example, instead of specifying a plane through the origin in \(\R^3\) using a pair of vectors, as we did in Example 1.3.4, we could define it using a single equation, of the form \(ax+by+cz=0\text{,}\) where \(a,b,c\in\R\text{,}\) and \((x,y,z)\) are coordinates in \(\R^3\text{.}\)
Given a vector space \(V\) and some equation (or other condition) that defines a subset, one of the things we need to be able to do is determine whether or not the subset is in fact a subspace. We do so using Subspace Test. Applying the subspace test is relatively straightforward, but we would also like to develop some intiution to help us decide whether or not a subset is likely to be a subspace, before attempting a proof.
There are a few relatively easy things we can check before we begin:
-
Does the subset contain the zero vector?This is part of the subspace test, of course, but it tends to be easy to check, and if the answer is no, then we’ve already ruled out the possibility that this subset could be a subspace.
-
Is the equation defining the subset linear?If not, your subset is probably not a subspace, and you should look for a counterexample. For example, the set \(\{(x,y)\in\R^2 \,|\, y=x^2\}\) contains zero, but it is defined by the nonlinear equation \(y=x^2\text{.}\) This tells us that our set is unlikely to be a subspace, but we still have to demonstrate this. Typically, we do so by showing that one of the two closure axioms fails. For example, we know that \((1,1)\) and \((2,4)\) belong to the subset, but \((1,1)+(2,4)=(3,5)\text{,}\) and since \(5\neq 3^2\text{,}\) the subset is not closed under addition, and therefore is not a subspace.
-
Is the equation defining the subset homogeneous?Even if an equation is linear, it may fail to define a subspace due to the special role played by the zero vector. For example, the plane in \(\R^3\) defined by the equation \(x+2y-5z=4\) is not a subspace. The fastest way to see this is to note that the equation is not satisfied by the zero vector! But both closure conditions fail as well. For example, the point \((2,1,0)\) is on the plane. But \(2(2,1,0)=(4,2,0)\) is not, since \(4+2(2)-5(0)=8\neq 4\text{.}\)
Exercise 1.3.6.
Determine whether or not the following subsets of vector spaces are subspaces.
(a)
True.
- This equation may not appear linear, but it is: if \(p(x)=ax^3+bx^2+cx+d\text{,}\) then \(p(2)=8a+4b+2c+d=0\) is a homogeneous linear equation. The zero poloynomial is zero everywhere, including at \(x=2\text{.}\) If \(p(2)=0\) and \(q(2)=0\text{,}\) then \((p+q)(2)=p(2)+q(2)=0+0=0\text{,}\) and for any scalar \(c\text{,}\) \((cp)(2)=c(p(2))=c(0)=0\text{.}\)
False.
- This equation may not appear linear, but it is: if \(p(x)=ax^3+bx^2+cx+d\text{,}\) then \(p(2)=8a+4b+2c+d=0\) is a homogeneous linear equation. The zero poloynomial is zero everywhere, including at \(x=2\text{.}\) If \(p(2)=0\) and \(q(2)=0\text{,}\) then \((p+q)(2)=p(2)+q(2)=0+0=0\text{,}\) and for any scalar \(c\text{,}\) \((cp)(2)=c(p(2))=c(0)=0\text{.}\)
(b)
True.
- We can immediately rule this out as a subspace because the zero polynomial is neither irreducible nor quadratic. Furthermore, it is not closed under addition: consider the sum of \(x^2+1\) and \(4-x^2\text{.}\)
False.
- We can immediately rule this out as a subspace because the zero polynomial is neither irreducible nor quadratic. Furthermore, it is not closed under addition: consider the sum of \(x^2+1\) and \(4-x^2\text{.}\)
The subset of \(\mathbb{P}_2\) consisting of all irreducible quadratics.
(c)
True.
- The equation is homogeneous, but it is not linear. Although this set contains the zero vector, it is not closed under addition: the vectors \((1,1,0)\) and \((0,0,1)\) belong to the set, but their sum \((1,1,1)\) does not.
False.
- The equation is homogeneous, but it is not linear. Although this set contains the zero vector, it is not closed under addition: the vectors \((1,1,0)\) and \((0,0,1)\) belong to the set, but their sum \((1,1,1)\) does not.
(d)
True.
- The defining equation can be rearranged as \(x+y-z=0\text{,}\) which you might recognize as the equation of a plane through the origin. Since \(0+0=0\text{,}\) the set contains the zero vector. To check closure under addition, suppose \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\) are in the set. This means that \(x_1+y_1=z_1\text{,}\) and \(x_2+y_2=z_2\text{.}\) For the sum \((x_1+x_2,y_1+y_2,z_1+z_2)\text{,}\) we have\begin{equation*} (x_1+x_2)+(y_1+y_2) = (x_1+y_1)+(x_2+y_2)=z_1+z_2\text{,} \end{equation*}so the sum is in the set. And for any scalar \(c\text{,}\) \(cx_1+cy_1=c(x_1+y_1)=cz_1\text{,}\) so \((cx_1,cy_1,cz_1)=c(x_1y_1z_1)\) is in the set as well.
False.
- The defining equation can be rearranged as \(x+y-z=0\text{,}\) which you might recognize as the equation of a plane through the origin. Since \(0+0=0\text{,}\) the set contains the zero vector. To check closure under addition, suppose \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\) are in the set. This means that \(x_1+y_1=z_1\text{,}\) and \(x_2+y_2=z_2\text{.}\) For the sum \((x_1+x_2,y_1+y_2,z_1+z_2)\text{,}\) we have\begin{equation*} (x_1+x_2)+(y_1+y_2) = (x_1+y_1)+(x_2+y_2)=z_1+z_2\text{,} \end{equation*}so the sum is in the set. And for any scalar \(c\text{,}\) \(cx_1+cy_1=c(x_1+y_1)=cz_1\text{,}\) so \((cx_1,cy_1,cz_1)=c(x_1y_1z_1)\) is in the set as well.
Let’s try a few more examples.
Example 1.3.7.
Determine whether or not the following subsets are subspaces.
(a)
The subset \(\{(x-y, x+y+2, 2x-3y) \,|\, x,y\in\R\}\subseteq\R^3\)
Solution.
The clue here that this is not a subspace is the presence of the 2 in the second component. Typically for a subspace, we expect to see linear expressions involving our variables, but in linear algebra, the adjective linear doesn’t imply the inclusion of constant terms the way it does in calculus. The reason, again, is the special role of zero in a vector space.
While it’s true that this set doesn’t contain the zero vector (which rules it out as a subspace), it’s not as obvious: perhaps there are values of \(x\) and \(y\) that give us \(x+y+2=0\text{,}\) and \(x-y=0, 2x-3y=0\) as well? Solving a system of equations would tell us that indeed, this is not possible.
We could also show that the closure conditions fail. Putting \(x=1, y=0\) gives the element \((1,3,2)\text{,}\) and putting \(x=0,y=1\) gives the element \((-1,3,-3)\text{.}\) Adding these, we get the vector \((0,6,-1)\text{.}\) Why is this not in the set? We would need \(x-y=0\text{,}\) so \(x=y\text{.}\) Then \(x+y+2=2x+2=6\) implies \(x=y=2\text{,}\) but \(2(2)-3(2)=-2\neq -3\text{.}\)
(b)
The subset \(\{p(x)\in \mathbb{P}_3\,|\, p(1)=p(2)\}\subseteq \mathbb{P}_3\text{.}\)
Solution.
At first glance, it may not be clear whether the condition \(p(1)=p(2)\) is linear. One approach is to write out our polynomial in terms of coefficients. If \(p(x)=ax^3+bx^2+cx+d\text{,}\) then \(p(1)=p(2)\) implies
\begin{equation*}
a+b+c+d=8a+4b+2c+d\text{,}
\end{equation*}
or \(7a+3b+c=0\text{,}\) which is a homogeneous linear equation. This isn’t yet a proof — we still have to apply the subspace test!
We can use the subspace test in terms of coefficients with the condition \(7a+3b+c=0\text{,}\) or we can use the original condition directly. First, the zero polynomial \(\zer\) satisfies \(\zer(1)=\zer(2)\text{,}\) since it’s equal to zero everywhere. Next, suppose we have polynomials \(p(x),q(x)\) with \(p(1)=p(2)\) and \(q(1)=q(2)\text{.}\) Then
\begin{equation*}
(p+q)(1)=p(1)+q(1)=p(2)+q(2)=(p+q)(2)\text{,}
\end{equation*}
and for any scalar \(c\text{,}\) \((cp)(1)=c(p(1))=c(p(2))=(cp)(2)\text{.}\) This shows that the set is closed under both addition and scalar multiplication.
(c)
The subset \(\{A\in M_{2\times 2}(\R) \,|\, \det(A)=0\}\text{.}\)
Solution.
Here, we have the condition \(\det(A)=0\text{,}\) which is homogeneous, but is it linear? If you remember a bit about the determinant, you might recall that it behaves well with respect to multiplication, but not addition, and indeed, this is going to mean that we don’t have a subspace.
To see that this is the case, consider closure under addition. The matrices \(A=\bbm 1\amp 0\\0\amp 0\ebm\) and \(B=\bbm 0\amp 0\\0\amp 1\ebm\) both have determinant \(0\text{,}\) but \(A+B=\bbm 1\amp 0\\0\amp 1\ebm\) has determinant 1. Therefore, \(A\) and \(B\) both belong to the set, but \(A+B\) does not.
In the next section, we’ll encounter perhaps the most fruitful source of subspaces: sets of linear combinations (or spans). We will see that such sets are always subspaces, so if we can identify a subset as a span, we know automatically that it is a subspace.
For example, in the last part of Exercise 1.3.6 above, if the vector \((x,y,z)\) satisfies \(x+y=z\text{,}\) then we have
\begin{equation*}
(x,y,z)=(x,y,x+y)=(x,0,x)+(0,y,y)=x(1,0,1)+y(0,1,1)\text{,}
\end{equation*}
so every vector in the set is a linear combination of the vectors \((1,0,1)\) and \((0,1,1)\text{.}\)
Exercises Exercises
1.
2.
Let \(V\) be the vector space of all \(2 \times 2\) matrices with real entries. Let \(H\) be the set of all \(2 \times 2\) idempotent matrices with real entries. Is \(H\) a subspace of the vector space \(V\text{?}\)
(b) Is \(H\) closed under addition?
(Hint: to show that \(H\) is not closed under addition, it is sufficient to find two idempotent matrices \(A\) and \(B\) such that \((A+B)^2 \ne (A+B)\text{.}\))
(c) Is \(H\) closed under scalar multiplication?
(Hint: to show that \(H\) is not closed under scalar multiplication, it is sufficient to find a real number \(r\) and an idempotent matrix \(A\) such that \((rA)^2 \ne (rA)\text{.}\))
You should be able to justify your answer by writing a complete, coherent, and detailed proof based on your answers to parts a-c.
3.
The trace of a square \(n \times n\) matrix \(A = (a_{ij})\) is the sum \(a_{11} + a_{22} + \cdots + a_{nn}\) of the entries on its main diagonal.
Let \(H\) be the set of all \(2 \times 2\) matrices with real entries that have trace \(0\text{.}\) Is \(H\) a subspace of the vector space \(V\text{?}\)
(b) Is \(H\) closed under addition?
(Hint: to show that \(H\) is not closed under addition, it is sufficient to find two trace zero matrices \(A\) and \(B\) such that \(A+B\) has nonzero trace.)
(c) Is \(H\) closed under scalar multiplication?
(Hint: to show that \(H\) is not closed under scalar multiplication, it is sufficient to find a real number \(r\) and a trace zero matrix \(A\) such that \(rA\) has nonzero trace.)
You should be able to justify your answer by writing a complete, coherent, and detailed proof based on your answers to parts a-c.
4.
The trace of a square \(n \times n\) matrix \(A = (a_{ij})\) is the sum \(a_{11} + a_{22} + \cdots + a_{nn}\) of the entries on its main diagonal.
Let \(V\) be the vector space of all \(2 \times 2\) matrices with real entries. Let \(H\) be the set of all \(2 \times 2\) matrices with real entries that have trace \(1\text{.}\) Is \(H\) a subspace of the vector space \(V\text{?}\)
(b) Is \(H\) closed under addition?
(Hint: to show that \(H\) is not closed under addition, it is sufficient to find two trace one matrices \(A\) and \(B\) such that \(A+B\) has trace not equal to one.)
(c) Is \(H\) closed under scalar multiplication?
(Hint: to show that \(H\) is not closed under scalar multiplication, it is sufficient to find a real number \(r\) and a trace one matrix \(A\) such that \(rA\) has trace not equal to one.)
You should be able to justify your answer by writing a complete, coherent, and detailed proof based on your answers to parts a-c.
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