10 Examples of Subsets that Are Not Subspaces of Vector Spaces

Linear algebra problems and solutions

Problem 338

Each of the following sets are not a subspace of the specified vector space. For each set, give a reason why it is not a subspace.
(1) \[S_1=\left \{\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \in \R^3 \quad \middle | \quad x_1\geq 0 \,\right \}\] in the vector space $\R^3$.


(2) \[S_2=\left \{\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \in \R^3 \quad \middle | \quad x_1-4x_2+5x_3=2 \,\right \}\] in the vector space $\R^3$.


(3) \[S_3=\left \{\, \begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2 \quad \middle | \quad y=x^2 \quad \,\right \}\] in the vector space $\R^2$.


(4) Let $P_4$ be the vector space of all polynomials of degree $4$ or less with real coefficients.
\[S_4=\{ f(x)\in P_4 \mid f(1) \text{ is an integer}\}\] in the vector space $P_4$.


(5) \[S_5=\{ f(x)\in P_4 \mid f(1) \text{ is a rational number}\}\] in the vector space $P_4$.


(6) Let $M_{2 \times 2}$ be the vector space of all $2\times 2$ real matrices.
\[S_6=\{ A\in M_{2\times 2} \mid \det(A) \neq 0\} \] in the vector space $M_{2\times 2}$.


(7) \[S_7=\{ A\in M_{2\times 2} \mid \det(A)=0\} \] in the vector space $M_{2\times 2}$.

(Linear Algebra Exam Problem, the Ohio State University)


(8) Let $C[-1, 1]$ be the vector space of all real continuous functions defined on the interval $[a, b]$.
\[S_8=\{ f(x)\in C[-2,2] \mid f(-1)f(1)=0\} \] in the vector space $C[-2, 2]$.


(9) \[S_9=\{ f(x) \in C[-1, 1] \mid f(x)\geq 0 \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$.


(10) Let $C^2[a, b]$ be the vector space of all real-valued functions $f(x)$ defined on $[a, b]$, where $f(x), f'(x)$, and $f^{\prime\prime}(x)$ are continuous on $[a, b]$. Here $f'(x), f^{\prime\prime}(x)$ are the first and second derivative of $f(x)$.
\[S_{10}=\{ f(x) \in C^2[-1, 1] \mid f^{\prime\prime}(x)+f(x)=\sin(x) \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$.

 
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Solution.

Recall the following subspace criteria.
A subset $W$ of a vector space $V$ over the scalar field $K$ is a subspace of $V$ if and only if the following three criteria are met.

  1. The subset $W$ contains the zero vector of $V$.
  2. If $u, v\in W$, then $u+v\in W$.
  3. If $u\in W$ and $a\in K$, then $au\in W$.

Thus, to prove a subset $W$ is not a subspace, we just need to find a counterexample of any of the three criteria.
  

Solution (1). $S_1=\{ \mathbf{x} \in \R^3 \mid x_1\geq 0 \}$

The subset $S_1$ does not satisfy condition 3. For example, consider the vector
\[\mathbf{x}=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}.\] Then since $x_1=1\geq 0$, the vector $\mathbf{x}\in S_1$. Then consider the scalar product of $\mathbf{x}$ and the scalar $-1$. Then we have
\[(-1)\cdot\mathbf{x}=\begin{bmatrix}
-1 \\
0 \\
0
\end{bmatrix},\] and the first entry is $-1$, hence $-\mathbf{x}$ is not in $S_1$. Thus $S_1$ does not satisfy condition 3 and it is not a subspace of $\R^3$.
(You can check that conditions 1, 2 are met.)
  

Solution (2). $S_2= \{ \mathbf{x}\in \R^3\mid x_1-4x_2+5x_3=2 \}$

The zero vector of the vector space $\R^3$ is
\[\mathbf{0}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.\] Since the zero vector $\mathbf{0}$ does not satisfy the defining relation $x_1-4x_2+5x_3=2$, it is not in $S_2$. Hence condition 1 is not met, hence $S_2$ is not a subspace of $\R^3$.
(You can check that conditions 2, 3 are not met as well.)
  

Solution (3). $S_3=\{\mathbf{x}\in \R^2 \mid y=x^2 \quad \}$

Consider vectors
\[\begin{bmatrix}
1 \\
1
\end{bmatrix} \text{ and } \begin{bmatrix}
-1 \\
1
\end{bmatrix}.\] These are vectors in $S_3$ since both vectors satisfy the defining relation $y=x^2$.

However, their sum
\[\begin{bmatrix}
1 \\
1
\end{bmatrix} + \begin{bmatrix}
-1 \\
1
\end{bmatrix}
=
\begin{bmatrix}
0 \\
1
\end{bmatrix}\] is not in $S_3$ since $1\neq 0^2$.
Hence condition 2 is not met, and thus $S_3$ is not a subspace of $\R^2$.
(You can check that condition 1 is fulfilled yet condition 3 is not met.)
  

Solution (4). $S_4=\{ f(x)\in P_4 \mid f(1) \text{ is an integer}\}$

Consider the polynomial $f(x)=x$. Since the degree of $f(x)$ is $1$ and $f(1)=1$ is an integer, it is in $S_4$. Consider the scalar product of $f(x)$ and the scalar $1/2\in \R$.
Then we evaluate the scalar product at $x=1$ and we have
\begin{align*}
\frac{1}{2}f(1)=\frac{1}{2},
\end{align*}
which is not an integer.
Thus $(1/2)f(x)$ is not in $S_4$, hence condition 3 is not met. Thus $S_4$ is not a subspace of $P_4$.
(You can check that conditions 1, 2 are met.)
  

Solution (5). $S_5=\{ f(x)\in P_4 \mid f(1) \text{ is a rational number}\}$

Let $f(x)=x$. Then $f(x)$ is a degree $1$ polynomial and $f(1)=1$ is a rational number.
However, the scalar product $\sqrt{2} f(x)$ of $f(x)$ and the scalar $\sqrt{2} \in \R$ is not in $S_5$ since
\[\sqrt{2}f(1)=\sqrt{2},\] which is not a rational number. Hence condition 3 is not met and $S_5$ is not a subspace of $P_4$.
(You can check that conditions 1, 2 are met.)
  

Solution (6). $S_6=\{ A\in M_{2\times 2} \mid \det(A) \neq 0\}$

The zero vector of the vector space $M_{2 \times 2}$ is the $2\times 2$ zero matrix $O$.
Since the determinant of the zero matrix $O$ is $0$, it is not in $S_6$. Thus, condition 1 is not met and $S_6$ is not a subspace of $M_{2 \times 2}$.
(You can check that conditions 2, 3 are not met as well.)
  

Solution (7). $S_7=\{ A\in M_{2\times 2} \mid \det(A)=0\}$

Consider the matrices
\[A=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 0\\
0& 1
\end{bmatrix}.\] The determinants of $A$ and $B$ are both $0$, hence they belong to $S_7$.
However, their sum
\[A+B=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}\] has the determinant $1$, hence the sum $A+B$ is not in $S_7$.
So condition 2 is not met and $S_7$ is not a subspace of $M_{2 \times 2}$.
(You can check that conditions 1, 3 are met.)
  

Solution (8). $S_8=\{ f(x)\in C[-2,2] \mid f(-1)f(1)=0\}$

Consider the continuous functions
\[f(x)=x-1 \text{ and } g(x)=x+1.\] (These are polynomials, hence they are continuous.)
We have
\begin{align*}
&f(-1)f(1)=(-2)\cdot(0)=0 \text{ and }\\
&g(-1)g(1)=(0)\cdot 2=0.
\end{align*}
So these functions are in $S_8$.

However, their sum $h(x):=f(x)+g(x)$ does not belong to $S_8$ since we have
\begin{align*}
h(-1)h(1)&=\big(f(-1)+g(-1)\big) \big(f(1)+g(1) \big)\\
&=(-2+0)(0+2)=-4\neq 0.
\end{align*}
Therefore, condition 2 is not met and $S_8$ is not a subspace of $C[-1, 1]$.
(You can check that conditions 1, 3 are met.)
  

Solution (9). $S_9=\{ f(x) \in C[-1, 1] \mid f(x)\geq 0 \text{ for all } -1\leq x \leq 1\}$

Let $f(x)=x^2$, an open-up parabola.
Then $f(x)$ is continuous and non-negative for $-1 \leq x \leq 1$. Hence $f(x)=x^2$ is in $S_9$.
However, the scalar product $(-1)f(x)$ of $f(x)$ and the scalar $-1$ is not in $S_9$ since, say,
\[(-1)f(1)=-1\] is negative.
So condition 3 is not met and $S_9$ is not a subspace of $C[-1, 1]$.
(You can check that conditions 1, 2 are met.)
  

Solution (10). $S_{10}=\{ f(x) \in C^2[-1, 1] \mid f^{\prime\prime}(x)+f(x)=\sin(x) \text{ for all } -1\leq x \leq 1\}$

The zero vector of the vector space $C^2[-1, 1]$ is the zero function $\theta(x)=0$.
The second derivative of the zero function is still the zero function.
Thus,
\[\theta^{\prime\prime}(x)+\theta(x)=0\] and since $\sin(x)$ is not the zero function, $\theta(x)$ is not in $S_{10}$.
Hence $S_{10}$ is not a subspace of $C^2[-1, 1]$.

(You can check that conditions 2, 3 are not met as well.
For example, consider the function $f(x)=-\frac{1}{2}x\cos(x)\in S_{10}$.)

Linear Algebra Midterm Exam 2 Problems and Solutions


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  1. 04/06/2017

    […] Problem 2 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces” […]

  2. 04/06/2017

    […] Problem 2 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces” […]

  3. 04/07/2017

    […] Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces” […]

  4. 04/07/2017

    […] Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces” […]

  5. 07/21/2017

    […] Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces” […]

  6. 07/26/2017

    […] out the post ↴ 10 Examples of Subsets that Are Not Subspaces of Vector Spaces for various subsets in vector spaces that are not […]

  7. 07/27/2017

    […] Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces” […]

  8. 08/11/2017

    […] Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces” […]

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