Subspaces of the Vector Space of All Real Valued Function on the Interval

Linear algebra problems and solutions

Problem 134

Let $V$ be the vector space over $\R$ of all real valued functions defined on the interval $[0,1]$. Determine whether the following subsets of $V$ are subspaces or not.

(a) $S=\{f(x) \in V \mid f(0)=f(1)\}$.

(b) $T=\{f(x) \in V \mid f(0)=f(1)+3\}$.

 
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Hint.

To show that a subset $W$ of a vector space $V$ is a subspace, we need to check that

  1. the zero vector in $V$ is in $W$
  2. for any two vectors $u,v \in W$, we have $u+v \in W$
  3. for any scalar $c$ and any vector $u \in W$, we have $cu \in W$.

Solution.

(a) Is $S=\{f(x) \in V \mid f(0)=f(1)\}$ a subspace?

We show that $S$ is a subspace of the vector space $V$ by checking conditions (1)-(3) given in the hint above.
First note that the zero vector in $V$ is the zero function $\theta(x)$, that is, $\theta(x)=0$ for any $x \in [0,1]$.
Since we have $\theta(0)=0=\theta(1)$, the zero function $\theta(x)\in S$.
Condition (1) is met.

Now, take any $f(x), g(x) \in S$. By the defining relation of $S$, we have
\[f(0)=f(1), \quad g(0)=g(1).\] Consider the addition $(f+g)(x)$. We have
\[(f+g)(0)=f(0)+g(0)=f(1)+g(1)=(f+g)(1)\] and it follows that $(f+g)(x) \in S$.
Thus $S$ satisfies condition (2).

To check the condition (3), take any scalar $c \in \R$ and $f(x) \in S$.
Since $f(x)\in S$, we have $f(0)=f(1)$. The scalar multiplication $(cf)(x)$ satisfies
\[(cf)(0)=c\cdot f(0)=c\cdot f(1)= (cf)(0).\] Thus $(cf)(x) \in S$.

Therefore, the subset $S$ satisfies conditions (1)-(3). Hence $S$ is a subspace of $V$.

(b) Is $T=\{f(x) \in V \mid f(0)=f(1)+3\}$ a subspace?

We claim that $T$ is not a subspace of the vector space $V$.
For example, the subset $T$ does not satisfy condition (1).

The zero vector of $V$ is the zero function $\theta(x)$.
Then we have
\[\theta(0)=0 \neq 0+3=\theta(1)+3,\] and hence the zero vector $\theta(x) \in V$ is not in $W$.


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