Blogs, Linear Algebra

Is That a Subspace

In our last post, we looked at what a vector space is. Now that we know what a vector space is, we will look at what a subspace is. Here we will look at what a subspace is, and show how to prove that a set is a subspace of a vector space.

Subspaces

To begin with, we need a formal definition of a subspace.

Let \(V\) be a vector space over a field \(F\) with a defined addition and scalar multiplication. Then \(W\) is a subspace of \(V\) if \(W\) is a subset of \(V\) and \(W\) is also a vector space over \(F\) with the same addition and scalar multiplication.

That is, a subspace is just a vector space contained inside larger vector space. At first glance, it actually seems that showing that \(W\) is a subspace of some other vector space would require much more work than just showing that it is a vector space, because we have extra properties that need to hold. However, we will be able to use existing knowledge to save us a significant amount of work when showing that a given set \(W\) is a vector space.

Note that, we were able to show that

Let \(V\) be a vector space over \(F\) and \(W\) a nonempty subset of \(V\). Then \(W\) is a subspace of \(V\) if and only if, for every \(\mathbf{u}\) and \(\mathbf{v}\) in \(W\) and \(c,d \in F\), \(c\mathbf{u}+d\mathbf{v} \in W\).

Instead of providing a full proof of this, we will just point out that most of the properties of being a subspace are satisfied because the addition and scalar multiplication already satisfy the properties required for being a vector space. The only criteria we have left to check is that the set is closed under the required operations. We then get the closure using the additional property in the theorem.

Example 1

Let \(W\) be the set of all symmetric \(2 \times 2\) matrices. Either prove or disprove that \(W\) is a subspace of the \(M_{2,2}\) over \(\mathbb{R}\).

Before going about determining whether or not \(W\) is a subspace \(M_{2,2}\), we should recall the definitions we used above. First, recall that by \(M_{2,2}\) we mean the set of all \(2 \times 2\) matrices with real number entries. Furthermore, a matrix is symmetric if \(A=A^{T}\), that is, the matrix is equal to its transpose.

Now that we have our definitions, we need to determine whether or not \(W\) is a subspace. Since we can determine if \(W\) is a subspace by looking at \(c\mathbf{u}+d\mathbf{v}\), we want to see what these look like. To that end, we note that vectors in \(M_{2,2}\) are \(2 \times 2\) matrices, so we will denote the vectors as matrices.

Therefore, let
\begin{align*}
A&=\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix} \text{ and } \\
B&=\begin{bmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{bmatrix}.
\end{align*}
Where each \(a_{ij}\) and \(b_{ij}\) are real numbers. Furthermore, let \(c\) and \(d\) be real numbers.

Now, note that \(A=A^{T}\), we get that
\begin{align*}
A&=A^{T} \\
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix}&=
\begin{bmatrix}
a_{11} & a_{21} \\
a_{12} & a_{22}
\end{bmatrix}
\end{align*}
Therefore, \(a_{12}=a_{21}\). Similarly, \(b_{12}=b_{21}\). We now see that
\begin{align*}
cA+dB&=c\begin{bmatrix}
a_{11} & a_{12}\\
a_{12} & a_{22} \end{bmatrix}
+d\begin{bmatrix}
b_{11} & b_{12} \\
b_{12} & b_{22} \end{bmatrix} \\
&=\begin{bmatrix}
ca_{11} & ca_{12}\\
ca_{12} & ca_{22} \end{bmatrix}
+\begin{bmatrix}
db_{11} & db_{12} \\
db_{12} & db_{22} \end{bmatrix} \\
&=\begin{bmatrix}
ca_{11}+db_{11} & ca_{12}+db_{12}\\
ca_{12}+db_{12} & ca_{22}+db_{22} \end{bmatrix}.
\end{align*}

If we now look at our resulting matrix, we will note we again have a symmetric matrix. Since this matrix is again symmetric, we will have that \(W\) is a subspace of \(M_{2,2}\). We are now ready to provide a proof.

Proof

Let \(W\) be the set of all \(2 \times 2\) symmetric matrices over \(\mathbb{R}\). First note that
\begin{align*}
\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix} \in W
\end{align*}
Furthermore every matrix in \(W\) is a \(2 \times 2\) matrix, and is therefore in \(M_{2,2}\). Hence \(W\) is a nonempty subset of \(M_{2,2}\) over \(\mathbb{R}\).

Now, let \(A,B \in W\). Then
\begin{align*}
A&=\begin{bmatrix}
a_{11} & a \\
a & a_{22}
\end{bmatrix} \\
B&=\begin{bmatrix}
b_{11} & b \\
b & b_{22}
\end{bmatrix}
\end{align*}
for \(a_{11},a_{22},a,b_{11},b_{22},b \in \mathbb{R}\). Furthermore, let \(c,d \in \mathbb{R}\). We then have that
\begin{align*}
cA+dB&=c\begin{bmatrix}
a_{11} & a_{12}\\
a_{12} & a_{22} \end{bmatrix}
+d\begin{bmatrix}
b_{11} & b_{12} \\
b_{12} & b_{22} \end{bmatrix} \\
&=\begin{bmatrix}
ca_{11} & ca_{12}\\
ca_{12} & ca_{22} \end{bmatrix}
+\begin{bmatrix}
db_{11} & db_{12} \\
db_{12} & db_{22} \end{bmatrix} \\
&=\begin{bmatrix}
ca_{11}+db_{11} & ca_{12}+db_{12}\\
ca_{12}+db_{12} & ca_{22}+db_{22} \end{bmatrix}.
\end{align*}
Hence, \(cA+dB\) is a symmetric \(2 \times 2\) matrix. Therefore, \(W\) is a subspace of \(M_{2,2}\).

Example 2

Let \(W\) be the set of all order pairs of the form \((x,x^{2})\). Prove or disprove that \(W\) is a subspace of \(\mathbb{R}^{2}\).

As we try to prove or disprove that \(W\) is a subspace of \(\mathbb{R}^{2}\), we want to think about what it would look like if we took \(c\mathbf{u}+d\mathbf{v}\) for \(c,d \in \mathbb{R}\) and \(\mathbf{u}=(u,u^{2}),\mathbf{v}=(v,v^{2}) \in W\). In general these would look like
\begin{align*}
c(u, u^{2})+d(v,v^{2}) =(cu+dv, cu^{2}+dv^{2}).
\end{align*}

Therefore, we need to determine if this new element will again be in \(W\).

We know that \(c,u,d,v \in \mathbb{R}\), so we also get that \(cu+dv \in \mathbb{R}\). Therefore, we will get that this vector is in \(W\) if and only if the second component is the square of the first. That is, if
\begin{align*}
(cu+dv)^{2}=cu^{2}+dv^{2}.
\end{align*}

As we examine this equation, this would mean that we can distribute a square across sums and factor out constants. However, this is not the case. Since we get that this is not possible, we would believe that \(W\) is not a subspace. We now need to find an example.

Here, we will arbitrarily pick \(u,v,c\) and \(d\) until we come up with a counter example. For our first guess we can take \(u=1,v=0,c=1,d=0\). In this case, we get that
\begin{align*}
1(1,1)+0(0,0)=(1,1) \in W.
\end{align*}
That is, for this choice, we are still in \(W\). Since we are still in \(W\) this would not provide a counter example, so we try again.

We now let \(c=2, u=1, d=0, u=0\) and get
\begin{align*}
2(1,1)+0(0,0)=(2,2).
\end{align*}
Now, note that \(2 \neq 4=2^{2}\), hence this vector is not in \(W\). This then tells us that \(W\) is not a subspace of \(\mathbb{R}^{2}\). We are now ready for our proof.

Proof

Let \(W=\left\{(x,x^{2}): x \in \mathbb{R}\right\}\). Note that \((1,1) \in W\) since \(1=1^{2}\), \((0,0) \in W\) since \(0=0^{2}\) and \(0,2 \in \mathbb{R}\). However,
\begin{align*}
2*(1,1)+0* (0,0)=(2,2) \notin W,
\end{align*}
since \(2 \neq 4=2^{2}\). Hence, \(W\) is not a subspace of \(\mathbb{R}^{2}\).

Conclusion

Here we have seen examples of subsets of vector spaces. In one case, we were able to show the subset was a vector space. In the other case, we were able to prove that the subset was not a subspace. While there are plenty of vector spaces and subsets thereof to look out, the process of determining whether or not we have a subspace hinges around determining if a vector of the form \(c\mathbf{u}+d\mathbf{v}\) will be in the given subset or not. Looking at what these elements look like will give us insight into how to provide our proofs.

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