Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.
Let
\[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\] where
\begin{align*}
p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\
p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3.
\end{align*}

(a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.

(b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.

(The Ohio State University, Linear Algebra Midterm)
 

Solution.

(a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.

Let $B=\{1, x, x^2, x^3\}$ be the standard basis for $\calP_3$.
With respect to the basis $B$, the coordinate vectors of the given polynomials are
\begin{align*}
\mathbf{v}_1:=[p_1(x)]_B&=\begin{bmatrix}
1 \\
3 \\
2 \\
-1
\end{bmatrix}, &\mathbf{v}_2:=[p_2(x)]_B=\begin{bmatrix}
0 \\
1 \\
0 \\
1
\end{bmatrix}\\[6pt] \mathbf{v}_3:=[p_3(x)]_B&=\begin{bmatrix}
0 \\
1 \\
1 \\
-1
\end{bmatrix}, &\mathbf{v}_4:=[p_4(x)]_B=\begin{bmatrix}
3 \\
8 \\
0 \\
8
\end{bmatrix}.
\end{align*}
Let $T=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}$ be the set of these coordinate vectors.

We find a basis for $\Span(T)$ among the vectors in $T$ by the leading 1 method.
We form the matrix whose column vectors are the vectors in $T$ and apply elementary row operations as follows.

\begin{align*}
\begin{bmatrix}
1 & 0 & 0 & 3 \\
3 &1 & 1 & 8 \\
2 & 0 & 1 & 0 \\
-1 & 1 & -1 & 8
\end{bmatrix}
\xrightarrow{\substack{R_2-3R_1\\R_3-2R_1\\R_4+R_1}}
\begin{bmatrix}
1 & 0 & 0 & 3 \\
0 &1 & 1 & -1 \\
0 & 0 & 1 & -6 \\
0 & 1 & -1 & 11
\end{bmatrix}
\xrightarrow{R_4-R_2}\\[6pt] \begin{bmatrix}
1 & 0 & 0 & 3 \\
0 &1 & 1 & -1 \\
0 & 0 & 1 & -6 \\
0 & 0 & -2 & 12
\end{bmatrix}
\xrightarrow{R_4+2R_2}
\begin{bmatrix}
1 & 0 & 0 & 3 \\
0 &1 & 0 & 5 \\
0 & 0 & 1 & -6 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}

The first three columns of the reduced row echelon form matrix contain the leading 1’s.
Thus, $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a basis for $\Span(T)$.
It follows that $Q:=\{p_1(x), p_2(x), p_3(x)\}$ is a basis for $\Span(S)$.

(b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.

Note that $p_4(x)$ is not in the basis $Q$.

The fourth column of the matrix in reduced row echelon form of part (a) gives the coefficients of the linear combination:
\[p_4(x)=3p_1(x)+5p_2(x)-6p_3(x).\]

Thus, the coordinate vector of $p_4(x)$ with respect to the basis $Q$ is
\[[p_4(x)]_Q=\begin{bmatrix}
3 \\
5 \\
-6
\end{bmatrix}.\]

Comment.

This is one of the midterm 2 exam problems for Linear Algebra (Math 2568) in Autumn 2017.

In part (b), some students stopped after obtaining the linear combination $p_4(x)=3p_1(x)+5p_2(x)-6p_3(x)$.
You must read the problem carefully. You are asked to find the coordinate vector of $p_4(x)$ with respect to $Q$.

List of Midterm 2 Problems for Linear Algebra (Math 2568) in Autumn 2017

1. Vector Space of 2 by 2 Traceless Matrices
2. Find an Orthonormal Basis of the Given Two Dimensional Vector Space
3. Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent?
4. Find Bases for the Null Space, Range, and the Row Space of a $5\times 4$ Matrix
5. Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$
6. Determine the Dimension of a Mysterious Vector Space From Coordinate Vectors
7. Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less ←The current problem

The post Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less first appeared on Problems in Mathematics.

Let $C[-2\pi, 2\pi]$ be the vector space of all continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the functions \[f(x)=\sin^2(x) \text{ and } g(x)=\cos^2(x)\] in $C[-2\pi, 2\pi]$.

Prove or disprove that the functions $f(x)$ and $g(x)$ are linearly independent.

(The Ohio State University, Linear Algebra Midterm)
 

Proof.

To determine whether $f(x)$ and $g(x)$ are linearly independent or not, consider the linear combination
\[c_1f(x)+c_2g(x)=0,\] equivalently
\[c_1\sin^2(x)+c_2 \cos^2(x)=0, \tag{*}\] where $c_1, c_2$ are scalars.

If the only scalars satisfying the above equality are $c_1=0, c_2=0$, then $f(x)$ and $g(x)$ are linearly independent, otherwise they are linearly dependent.

Note that this is an equality as functions.
That is, this equality must hold for any $x$ in the interval $[-2\pi, 2\pi]$.

Let $x=0$. Then as $\sin(0)=0$ and $\cos(0)=1$, we obtain $c_2=0$ from (*).
Next, let $x=\pi/2$. Then as $\sin(\pi/2)=1$ and $\cos(\pi/2)=0$, we obtain $c_1=0$ from (*).

Therefore, we must have $c_1=c_2=0$, and hence the functions $f(x)=\sin^2(x)$ and $g(x)=\cos^2(x)$ are linearly independent.

Comment.

This is one of the midterm 2 exam problems for Linear Algebra (Math 2568) in Autumn 2017.

Here is the most common mistake.
The linear combination $c_1\sin^2(x)+c_2 \cos^2(x)$ is a function defined over the interval $[-2\pi, 2\pi]$ and we are assuming it is the zero function.

So saying that “if $c_1=1, c_2=0$, then $c_1\sin^2(x)+c_2 \cos^2(x)$ is zero at $x=0$, hence $f(x)$ and $g(x)$ are linearly independent” is totally wrong.

What you are claiming here is that the function $\sin^2(x)$ is zero at $x=0$, hence it is the zero function.
This is clearly wrong as $\sin^2(x)$ is not the zero function.

List of Midterm 2 Problems for Linear Algebra (Math 2568) in Autumn 2017

1. Vector Space of 2 by 2 Traceless Matrices
2. Find an Orthonormal Basis of the Given Two Dimensional Vector Space
3. Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent?←The current problem
4. Find Bases for the Null Space, Range, and the Row Space of a $5\times 4$ Matrix
5. Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$
6. Determine the Dimension of a Mysterious Vector Space From Coordinate Vectors
7. Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less

The post Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent? 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.

Let $V$ be the vector space over $\R$ of all real valued function on the interval $[0, 1]$ and let
\[W=\{ f(x)\in V \mid f(x)=f(1-x) \text{ for } x\in [0,1]\}\] be a subset of $V$. Determine whether the subset $W$ is a subspace of the vector space $V$.

Proof.

We claim that $W$ is a subspace of $V$.
To show the claim, we need to check that the following subspace criteria.

The zero vector of $V$ is the zero function $\theta(x)=0$.
Since we have
\[\theta(x)=0=\theta(1-x)\] for any $x\in [0, 1]$, the zero vector $\theta$ is in $W$, hence condition 1 is met.

Let $f(x), g(x)$ be arbitrary elements in $W$. Then these functions satisfy
\[f(x)=f(1-x) \text{ and } g(x)=g(1-x)\] for any $x\in [0,1]$.
We want to show that the sum $h(x):=f(x)+g(x)$ is in $W$. This follows since we have
\[h(x)=f(x)+g(x)=f(1-x)+g(1-x)=h(1-x).\] Thus, condition 2 is satisfied.

Finally, we check condition 3. Let $c$ be a scalar and let $f(x)$ be an element in $W$.
Then we have
\[f(x)=f(1-x).\] It follows from this that
\[cf(x)=cf(1-x),\] and this shows that the scalar product $cf(x)$ is in $W$.

Therefore condition 3 holds, and we have proved the subspace criteria for $W$. Thus $W$ is a subspace of the vector space $V$.

The post Determine Whether a Set of Functions $f(x)$ such that $f(x)=f(1-x)$ is a Subspace first appeared on Problems in Mathematics.

Let $P_2$ be the vector space of all polynomials of degree two or less.
Consider the subset in $P_2$
\[Q=\{ p_1(x), p_2(x), p_3(x), p_4(x)\},\] where
\begin{align*}
&p_1(x)=1, &p_2(x)=x^2+x+1, \\
&p_3(x)=2x^2, &p_4(x)=x^2-x+1.
\end{align*}

(a) Use the basis $B=\{1, x, x^2\}$ of $P_2$, give the coordinate vectors of the vectors in $Q$.

(b) Find a basis of the span $\Span(Q)$ consisting of vectors in $Q$.

(c) For each vector in $Q$ which is not a basis vector you obtained in (b), express the vector as a linear combination of basis vectors.

(The Ohio State University Linear Algebra Exam Problem)
 

Solution.

(a) Use the basis $B=\{1, x, x^2\}$ of $P_2$, give the coordinate vectors of the vectors in $Q$

To obtain the coordinate vectors with respect to the given basis $B=\{1, x, x^2\}$, we fist express a given vector as a linear combination of basis vectors. ( You need to take care of the order of the basis vectors.)

For example, for the vector $p_4(x)$, we have the linear combination
\[p_4(x)=1\cdot 1+(-1)\cdot x+1\cdot x^2.\] Thus reading the coefficient of the basis vectors $1, x, x^2$, we see that the coordinate vector of $p_4(x)$ with respect to the basis $B$ is
\[[p_4(x)]_B=\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}.\]

By the similar argument, the coordinate vectors are
\begin{align*}
&[p_1(x)]_B=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, &[p_2(x)]_B=\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix},\\
& [p_3(x)]_B=\begin{bmatrix}
0 \\
0 \\
2
\end{bmatrix}, &[p_4(x)]_B=\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}.
\end{align*}

(b) Find a basis of the span $\Span(Q)$ consisting of vectors in $Q$

Let
\begin{align*}
T:&=\{[p_1(x)]_B, [p_2(x)]_B, [p_3(x)]_B, [p_4(x)]_B\}\\
&=\left\{\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
0 \\
0 \\
2
\end{bmatrix}, \begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}\right\}.
\end{align*}
Then a basis of $\Span(Q)$ consisting of vectors in $Q$ corresponds to a basis of $\Span(T)$ consisting of vectors in $T$.

Consider the matrix whose columns are the vectors in $T$. We reduce it by elementary row operations as follows.
\begin{align*}
\begin{bmatrix}
1 & 1 & 0 & 1 \\
0 &1 & 0 & -1 \\
0 & 1 & 2 & 1
\end{bmatrix}
\xrightarrow[R_3-R_2]{R_1-R_2}
\begin{bmatrix}
1 & 0 & 0 & 2 \\
0 &1 & 0 & -1 \\
0 & 0 & 2 & 2
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_3}
\begin{bmatrix}
1 & 0 & 0 & 2 \\
0 &1 & 0 & -1 \\
0 & 0 & 1 & 1
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form and the first three columns contain the leading 1’s.
Therefore the set consisting of the first three vectors in $T$
\[\left\{ \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
0 \\
0 \\
2
\end{bmatrix}\right\}\] is a basis for $\Span(T)$.
(This is what I call the leading 1 method.)

Therefore by the coordinate vector correspondence, it follows that the set
\[\{p_1(x), p_2(x), p_3(x)\}\] is a basis for $\Span(Q)$ consisting of vectors in $Q$.

(c) For each vector in $Q$ which is not a basis vector you obtained in (b), express the vector as a linear combination of basis vectors

From the solution of part (b), we see that the vector $p_4(x)$ is the only vector in $Q$ which is not a basis vector.
Thus we want to express $p_4(x)$ as a linear combination of $p_1(x), p_2(x), p_3(x)$.

Consider a linear combination
\[a_1p_1(x)+a_2 p_2(x)+a_3p_3(x)+a_4p_4(x)=0 \tag{*}.\] By taking coordinate vectors, we obtain
\[a_1\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}+a_2\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}+a_3\begin{bmatrix}
0 \\
0 \\
2
\end{bmatrix}+a_4\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}=\begin{bmatrix}0\\0\\0 \end{bmatrix}.\]

The augmented matrix is
\[\left[\begin{array}{rrrr|r}
1 & 1 & 0 & 1 &0\\
0 &1 & 0 & -1 &0\\
0 & 1 & 2 & 1 &0
\end{array} \right].\] Note that this is the matrix we considered in part (b) except for the last zero column.
Thus by the same elementary row operations, we obtain the reduced row echelon form of this matrix
\[\left[\begin{array}{rrrr|r}
1 & 0 & 0 & 2 &0\\
0 &1 & 0 & -1 &0 \\
0 & 0 & 1 & 1&0
\end{array} \right].\]

Therefore we obtain the solutions
\begin{align*}
a_1&=-2a_4\\
a_2&=a_4\\
a_3&=-a_4\\
\end{align*}
and $a_4$ is a free variable.

Setting $a_4=1$ we obtain $a_1=-2, a_2=1, a_3=-1$ from the above solutions.
We plug these values in the linear combination (*) and obtain
\[-2p_1(x)+p_2(x)-p_3(x)+p_4(x)=0.\] Solving this for $p_4(x)$, the vector $p_4(x)$ can be expressed as
\[p_4(x)=2p_1(x)-p_2(x)+p_3(x).\] Click here if solved 19

The post Vector Space of Polynomials and a Basis of Its Subspace first appeared on Problems in Mathematics.

Let $P_n(\R)$ be the vector space over $\R$ consisting of all degree $n$ or less real coefficient polynomials. Let
\[U=\{ p(x) \in P_n(\R) \mid p(1)=0\}\] be a subspace of $P_n(\R)$.

Find a basis for $U$ and determine the dimension of $U$.

Solution.

Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ be an element in $U$.
Since $p(1)=0$, we have $a_n+a_{n-1}+\cdots+a_1+a_0=0$. Thus we have
\begin{align*}
p(x)&=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x-(a_n+a_{n-1}+\cdots+a_1)\\
&=a_n(x^n-1)+a_{n-1}(x^{n-1}-1)+\cdots a_1(x-1).
\end{align*}

Let us define
\[q_n(x)=x^n-1, q_{n-1}(x)=x^{n-1}-1, \dots, q_1(x)=x-1 \in U.\] Then the above computation shows that we have
\[p(x)=a_nq_n(x)+a_{n-1}q_{n-1}(x)+\cdots a_1q_1(x).\] In other words, any element $p(x)\in U$ is a linear combination of $q_1(x), \dots, q_n(x)$.
Hence $B:=\{q_1(x), \dots, q_n(x)\}$ is a spanning set for $U$.

We show that $B$ is a linearly independent set.
Consider a linear combination
\[c_1q_1(x)+\cdots+c_n q_n(x)=\theta(x):=0x^{n}+0x^{n-1}+\cdots+0x+0.\]

Then we have
\begin{align*}
c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x-(c_1+c_2+\cdots+c_n)=0.
\end{align*}
Comparing the coefficients of both sides, we obtain
\[c_1=c_2=\cdots=c_n=0\] and thus $q_1(x), \dots, q_n(x)$ are linearly independent, and hence $B=\{q_1(x), \dots, q_n(x)\}$ is a basis for the subspace $U$. Since the dimension is the number of vectors in a basis, the dimension of $U$ is $n$.

The post Find a Basis and Determine the Dimension of a Subspace of All Polynomials of Degree $n$ or Less first appeared on Problems in Mathematics.

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

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 $W$.

The post Subspaces of the Vector Space of All Real Valued Function on the Interval first appeared on Problems in Mathematics.