Let $R$ be the ring of all continuous functions on the interval $[0, 2]$.
Let $I$ be the subset of $R$ defined by
\[I:=\{ f(x) \in R \mid f(1)=0\}.\]

Then prove that $I$ is an ideal of the ring $R$.
Moreover, show that $I$ is maximal and determine $R/I$.

Hint.

Consider the map $\phi:R\to \R$ defined by
\[\phi(f)=f(1),\] for every $f(x)\in R$.

Proof.

Let us consider the map $\phi$ from $R$ to the field of real numbers $\R$ defined by
\[\phi(f)=f(1),\] for each $f(x)\in R$. Namely, the map $\phi$ is the evaluation at $x=1$.

We claim that $\phi:R \to \R$ is a ring homomorphism. In fact we have for any $f(x), g(x)\in R$,
\begin{align*}
\phi(fg)&=(fg)(1)=f(1)g(1)=\phi(f)\phi(g)\\
\phi(f+g)&=(f+g)(1)=f(1)+g(1)=\phi(f)+\phi(g),
\end{align*}
hence $\phi$ is a ring homomorphism.

Next, consider the kernel of $\phi$. We have
\begin{align*}
\ker(\phi)&=\{ f(x)\in R\mid \phi(f)=0\}\\
&=\{f(x) \in R \mid f(1)=0\}=I.
\end{align*}
Since the kernel of a ring homomorphism is an ideal, it follows that $I=\ker(\phi)$ is an ideal of $R$.

Next, we claim that $\phi$ is surjective. To see this, let $r\in \R$ be an arbitrary real number.
Define the constant function $f(x)=r$. Then $f(x)$ is an element in $R$ as it is continuous function on $[0, 2]$.
We have
\begin{align*}
\phi(f)=f(1)=r,
\end{align*}
and this proves that $\phi$ is surjective.

Since $\phi: R\to \R$ is a surjective ring homomorphism, the first isomorphism theorem yields that
\[R/\ker(\phi) \cong \R.\] Since $\ker(\phi)=I$ as we saw above, we have
\[R/I \cong \R.\] Thus, the quotient ring $R/I$ is isomorphic to the field $\R$.
It follows from this that $I$ is a maximal ideal of $R$.

(Recall the fact that an ideal $I$ of a commutative ring $R$ is maximal if and only if $R/I$ is a field.)

Related Question.

For a proof, see the post ↴
Polynomial Ring with Integer Coefficients and the Prime Ideal $I=\{f(x) \in \Z[x] \mid f(-2)=0\}$

The post A Maximal Ideal in the Ring of Continuous Functions and a Quotient Ring first appeared on Problems in Mathematics.

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 $C[-1, 1]$.

(11) Let $S_{11}$ be the set of real polynomials of degree exactly $k$, where $k \geq 1$ is an integer, in the vector space $P_k$.

(12) Let $V$ be a vector space and $W \subset V$ a vector subspace. Define the subset $S_{12}$ to be the complement of $W$,
\[ V \setminus W = \{ \mathbf{v} \in V \mid \mathbf{v} \not\in W \}.\]

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.

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 $f(x)=-\frac{1}{2}x\cos(x)\in S_{10}$.)

Solution (11). Let $S_{11}$ be the set of real polynomials of degree exactly $k$.

The set $S_{11}$ is not a vector subspace of $\mathbf{P}_k$. One reason is that the zero function $\mathbf{0}$ has degree $0$, and so does not lie in $S_{11}$. The set $S_{11}$ is also not closed under addition. Consider the two polynomials $f(x) = x^k + 1$ and $g(x) = – x^k + 1$. Both of these polynomials lie in $S_{11}$, however $f(x) + g(x) = 2$ has degree $0$ and so does not lie in $S_{11}$.

Solution (12). The complement

The complement $S_{12}= V \setminus W$ is not a vector subspace. Specifically, if $\mathbf{0} \in V$ is the zero vector, then we know $\mathbf{0} \in W$ because $W$ is a subspace. But then $\mathbf{0} \not\in V \setminus W$, and so $V \setminus W$ cannot be a vector subspace.

Linear Algebra Midterm Exam 2 Problems and Solutions

• True of False Problems and Solutions: True or False problems of vector spaces and linear transformations
• Problem 1 and its solution (current problem): See (7) in the post “10 examples of subsets that are not subspaces of vector spaces”
• Problem 2 and its solution: Determine whether trigonometry functions $\sin^2(x), \cos^2(x), 1$ are linearly independent or dependent
• Problem 3 and its solution: Orthonormal basis of null space and row space
• Problem 4 and its solution: Basis of span in vector space of polynomials of degree 2 or less
• Problem 5 and its solution: Determine value of linear transformation from $R^3$ to $R^2$
• Problem 6 and its solution: Rank and nullity of linear transformation from $R^3$ to $R^2$
• Problem 7 and its solution: Find matrix representation of linear transformation from $R^2$ to $R^2$
• Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space

The post 12 Examples of Subsets that Are Not Subspaces of Vector Spaces first appeared on Problems in Mathematics.

(a) Let $C[-1,1]$ be the vector space over $\R$ of all real-valued continuous functions defined on the interval $[-1, 1]$.
Consider the subset $F$ of $C[-1, 1]$ defined by
\[F=\{ f(x)\in C[-1, 1] \mid f(0) \text{ is an integer}\}.\] Prove or disprove that $F$ is a subspace of $C[-1, 1]$.

(b) Let $n$ be a positive integer.
An $n\times n$ matrix $A$ is called skew-symmetric if $A^{\trans}=-A$.
Let $M_{n\times n}$ be the vector space over $\R$ of all $n\times n$ real matrices.
Consider the subset $W$ of $M_{n\times n}$ defined by
\[W=\{A\in M_{n\times n} \mid A \text{ is skew-symmetric}\}.\] Prove or disprove that $W$ is a subspace of $M_{n\times n}$.

Proof.

(a) Continuous functions taking integer values at $0$.

We claim that the subset $F$ is not a subspace of the vector space $C[-1, 1]$.
If $F$ is a subspace, then $F$ is closed under scalar multiplication.
For instance, consider the scalar multiplication of the function $f(x)=x+1$ and the scalar $r=1/2$.
Note that since $f(0)=1$ is an integer, the function is in $F$.
However, the scalar multiplication
\[rf(x)=\frac{1}{2}(x+1)\] is not in $F$ because
\[rf(0)=\frac{1}{2}\] is not an integer.
Therefore, $F$ is not closed under scalar multiplication, hence $F$ is not a subspace of $C[-1, 1]$.

(b) Set of all $n\times n$ skew-symmetric matrices

We prove that $W$ is a subspace of $V$ by showing the following subspace criteria.

1. The zero vector $\mathbf{0}\in M_{n\times n}$ is in $W$.
2. If $A, B\in W$, then $A+B\in W$.
3. If $A\in W, r\in \R$, then $rA\in W$.

First, note that the zero vector in the vector space $M_{n\times n}$ is the $n\times n$ zero matrix $O$. Since we have
\[O^{\trans}=O=-O,\] the zero matrix $O$ is skew-symmetric, hence the zero vector $O$ is in $W$. So condition 1 is met.

To prove condition 2, let $A, B$ be arbitrary elements in $W$. Thus, $A$ and $B$ are skew-symmetric matrices:
\[A^{\trans}=-A \text{ and } B^{\trans}=-B. \tag{*}\] We want to prove that the addition $A+B$ is in $W$, namely, we want to show $A+B$ is skew-symmetric.

We have
\begin{align*}
(A+B)^{\trans}=A^{\trans}+B^{\trans}\stackrel{(*)}{=}-A+(-B)=-(A+B),
\end{align*}
and this implies that $A+B$ is skew-symmetric. Hence $A+B \in W$ and condition 2 is satisfied.

Finally, we check condition 3. Let $A\in W$ and $r\in \R$. We want to show that the scalar product $rA\in W$, that is, $rA$ is skew-symmetric.
We have
\begin{align*}
(rA)^{\trans} &= rA^{\trans} \\
&= r(-A) \qquad \qquad (\text{$A$ is skew-symmetric})\\
&=-rA,
\end{align*}
which yields that $rA$ is skew-symmetric. Hence $rA \in W$ and condition 3 is met as well.

Thus, we have checked all the subspace criteria. Hence $W$ is a subspace of $M_{n\times n}$.

Comment.

These are Quiz 8 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

• Quiz 1. Gauss-Jordan elimination / homogeneous system.
• Quiz 2. The vector form for the general solution / Transpose matrices.
• Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent
• Quiz 4. Inverse matrix/ Nonsingular matrix satisfying a relation
• Quiz 5. Example and non-example of subspaces in 3-dimensional space
• Quiz 6. Determine vectors in null space, range / Find a basis of null space
• Quiz 7. Find a basis of the range, rank, and nullity of a matrix
• Quiz 8. Determine subsets are subspaces: functions taking integer values / set of skew-symmetric matrices
• Quiz 9. Find a basis of the subspace spanned by four matrices
• Quiz 10. Find orthogonal basis / Find value of linear transformation
• Quiz 11. Find eigenvalues and eigenvectors/ Properties of determinants
• Quiz 12. Find eigenvalues and their algebraic and geometric multiplicities
• Quiz 13 (Part 1). Diagonalize a matrix.
• Quiz 13 (Part 2). Find eigenvalues and eigenvectors of a special matrix

The post Quiz 8. Determine Subsets are Subspaces: Functions Taking Integer Values / Set of Skew-Symmetric Matrices first appeared on Problems in Mathematics.

Let $C[-\pi, \pi]$ be the vector space of all continuous functions defined on the interval $[-\pi, \pi]$.

Show that the subset $\{\cos(x), \sin(x)\}$ in $C[-\pi, \pi]$ is linearly independent.

Proof.

Note that the zero vector in the vector space $C[-\pi, \pi]$ is the zero function
\[\theta(x):=0.\]

Let us consider a linear combination
\[a_1\cos(x)+a_2\sin(x)=\theta(x)=0 \tag{*}.\] If this linear combination has only the zero solution $a_1=a_2=0$, then the set $\{\cos(x), \sin(x)\}$ is linearly independent.

The equality (*) should be true for any values of $x\in [-\pi, \pi]$.
Setting $x=0$, we obtain from (*) that
\[a_1=0\] since $\cos(0)=1, \sin(0)=0$.

We also set $x=\pi/2$ and we obtain
\[a_2=0\] since $\cos(\pi/2)=0, \sin(\pi/2)=1$.

Therefore, we have $a_1=a_2=0$ and we conclude that the set $\{\cos(x), \sin(x)\}$ is linearly independent.

The post Cosine and Sine Functions are Linearly Independent first appeared on Problems in Mathematics.

Let $R$ be the ring of all continuous functions on the interval $[0,1]$.
Let $I$ be the set of functions $f(x)$ in $R$ such that $f(1/2)=f(1/3)=0$.

Show that the set $I$ is an ideal of $R$ but is not a prime ideal.

Proof.

We first show that $I$ is an ideal of $R$.
Let $f(x), g(x)\in R$ and $r\in R$.
Then the function $f(x)+rg(x)$ is a continuous function on $[0, 1]$ and we have
\begin{align*}
(f(x)+rg(x))(1/2)=f(1/2)+rg(1/2)=0+r\cdot 0=0
\end{align*}
since $f(1/2)=g(1/2)=0$.
Similarly, we have $(f(x)+rg(x))(1/3)=0$.
Hence $f(x)+rg(x)\in I$, and thus $I$ is an ideal of $R$.

Next, we show that the ideal $I$ is not a prime ideal.
Let us define
\[f(x)=x-\frac{1}{3} \text{ and } g(x)=x-\frac{1}{2}.\] Then the functions $f(x), g(x)$ are continuous on $[0, 1]$, hence they are elements in $R$.

Since we have
\[f \left(\frac{1}{2}\right)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\neq 0 \text{ and } g\left(\frac{1}{3} \right)=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}\neq 0,\] the functions $f(x), g(x)$ are not in the ideal $I$.
However, the product $f(x)g(x)$ is in $I$ since we have
\begin{align*}
f\left(\frac{1}{2}\right)g \left(\frac{1}{2}\right)=\left(\frac{1}{2}-\frac{1}{3} \right)\left(\frac{1}{2}-\frac{1}{2}\right)=0\\
f\left(\frac{1}{3}\right)g \left(\frac{1}{3}\right)=\left(\frac{1}{3}-\frac{1}{3} \right)\left(\frac{1}{3}-\frac{1}{2}\right)=0.
\end{align*}

In summary, the functions $f(x)$ and $g(x)$ are not in $I$ but the product $f(x)(g)$ is in $I$. Thus the ideal $I$ is not a prime ideal.

The post Non-Prime Ideal of Continuous Functions first appeared on Problems in Mathematics.

Let $C[3, 10]$ be the vector space consisting of all continuous functions defined on the interval $[3, 10]$. Consider the set
\[S=\{ \sqrt{x}, x^2 \}\] in $C[3,10]$.

Show that the set $S$ is linearly independent in $C[3,10]$.

Proof.

Note that the zero vector in $C[3,10]$ is the zero function $\theta(x)=0$.
Let us consider a linear combination
\[a_1\sqrt{x}+a_2x^2=\theta(x)=0.\]

We want to show that $a_1=a_2=0$.
Since this equality holds for any value of $x$ between $3$ and $10$, letting $x=4$ and $x=9$ yields the system of linear equations
\begin{align*}
2a_1+4a_2=0\\
3a_1+81a_2=0.
\end{align*}
Solving this system, we obtain $a_1=a_2=0$, and hence the set $S$ is linearly independent.

Comment.

We could have taken any two values between $3$ and $10$ for $x$ instead of $4$ and $9$, but the choice $x=4$ and $x=9$ made solving the system easier.
Also, we took two values for $x$ because we had two unknowns $a_1$, $a_2$, and thus we needed two equations to determine these unknown.

For a similar problem, show that the set $\{\sqrt{x}, x^2, x\}$ in the vector space $C[1, 10]$ is linearly independent.
In this case, there will be three unknowns that you want to show to be zero.
So you need to take three values for $x$.
A natural choice will be $x=1, 4, 9$.

The post Linear Independent Continuous Functions first appeared on Problems in Mathematics.