Let $P_3$ be the vector space over $\R$ of all degree three or less polynomial with real number coefficient.
Let $W$ be the following subset of $P_3$.
\[W=\{p(x) \in P_3 \mid p'(-1)=0 \text{ and } p^{\prime\prime}(1)=0\}.\] Here $p'(x)$ is the first derivative of $p(x)$ and $p^{\prime\prime}(x)$ is the second derivative of $p(x)$.

Show that $W$ is a subspace of $P_3$ and find a basis for $W$.

Proof.

Subspace criteria

To show that the subset $W$ of the vector space $P_3$ is a subspace, we need to check that

Check Condition 1

First note that the zero vector in $P_3$ is the zero polynomial, which we denote $\theta(x)$.
Thus we have $\theta(x)=0$ for any $x$.

The derivative of the zero function is still the zero function, we have
\[\theta'(x)=0, \text{ and } \theta^{\prime\prime}(x)=0.\]

Therefore the zero vector $\theta(x)$ satisfies the following defining relations for $W$
\[\theta'(-1)=0 \text{ and } \theta^{\prime\prime}(1)=0\] and thus the zero vector $\theta(x)$ is in $W$.
Thus condition 1 is met.

Check Condition 2

To check condition 2, let $f(x), g(x) \in W$.
Then $f(x)$ and $g(x)$ satisfy
\[f'(-1)=0, f^{\prime\prime}(1)=0, \text{ and } g'(-1)=0, g^{\prime\prime}(1)=0.\] Let $h(x):=f(x)+g(x)$. We want to show that the sum $h(x)=f(x)+g(x)$ is in $W$.

We have
\[h'(x)=(f(x)+g(x))’=f'(x)+g'(x)\] and similarly we have
\[h^{\prime\prime}(x)=f^{\prime\prime}(x)+g^{\prime\prime}(x).\]

Thus we obtain
\[h'(-1)=f'(-1)+g'(-1)=0+0=0 \text{ and } h^{\prime\prime}(1)=f^{\prime\prime}(1)+g^{\prime\prime}(1)=0+0=0.\] Therefore $h(x)$ satisfies the defining relations of $W$ and hence $h(x)=f(x)+g(x) \in W$.
So condition 2 is satisfied.

Check Condition 3

Condition 3 can be checked as follows. Let $f(x) \in W$ and let $r\in R$.
We want to show that the scalar product $k(x):=rf(x)$ is in $W$.

Since $f(x)$ is in $W$, we have
\[f'(-1)=0, f^{\prime\prime}(1)=0.\]

Note that we have
\[k'(x)=(rf(x))’=rf'(x), \text{ and } k^{\prime\prime}(x)=rf^{\prime\prime}(x).\]

Thus we see that
\[k'(-1)=rf'(-1)=r\cdot 0=0 \text{ and } k^{\prime\prime}(1)=rf^{\prime\prime}(1)=r\cdot 0=0.\]

Thus $k(x)$ satisfies the defining relations of $W$ and hence the scalar product $k(x)=rf(x)$ is in $W$.
Therefore condition 3 and thus all conditions are met, and we conclude that $W$ is a subspace of $P_3$.

Find a basis for the subspace $W$.

To find a basis, we observe the following.

Any vector in $W$ is a polynomial
\[p(x)=a_0+a_1x+a_2x^2+a_3x^3\] satisfying
\[p'(-1)=0 \text{ and } p^{\prime\prime}(1)=0.\]

Since we have
\[p'(x)=a_1+2a_2x+3a_3x^2 \text{ and } p^{\prime\prime}(x)=2a_2+6a_3x,\] the above conditions become
\[a_1-2a_2+3a_3=0 \text{ and } 2a_2+6a_3=0.\] To find solutions for these equations consider the augmented matrix
\[\left[\begin{array}{rrr|r}
1 & -2 & 3 & 0 \\
0 &2 & 6 & 0
\end{array} \right].\] We apply the elementary row operations to this matrix and obtain the following reduced row echelon matrix.
\[\left[\begin{array}{rrr|r}
1 & 0 & 9 & 0 \\
0 &1 & 3 & 0
\end{array} \right].\] Thus, solutions are\begin{align*}
a_1&=-9a_3\\
a_2&=-3a_3.
\end{align*}

Therefore, any polynomial $p(x)$ in $W$ can be written as
\begin{align*}
p(x)&=a_0-9a_3x-3a_3x^2+a_3x^3\\
&=a_0(1)+a_3(-9x-3x^2+x^3).
\end{align*}

In particular we see that the polynomials $q_1(x):=1$ and $q_2(x):=-9x-3x^2+x^3$ are in $W$.
Since any vector $p(x) \in W$ is a linear combination of $q_1(x)$ and $q_2(x)$, the set $\{q_1(x), q_2(x)\}$ is a spanning set for $W$.

We check that $q_1(x)$ and $q_2(x)$ are linearly independent.
If we have a linear combination
\[c_1q_1(x)+c_2q_2(x)=0(=\theta(x)),\] then we have
\[c_1-9c_2x-3c_2x^2+c_2x^3=0.\]

Thus we see that $c_1=c_2=0$ and the vectors $q_1(x)$ and $q_2(x)$ are linearly independent.
Therefore $\{q_1(x), q_2(x)\}$ is a linearly independent spanning set for $W$, and thus it is a basis for $W$.

In summary we found a basis
\[\{1, -9x-3x^2+x^3\}.\]

Remark: The dimension of $W$ is $2$.

The post Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis first appeared on Problems in Mathematics.

Let $B=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ be a basis for a vector space $V$ over a scalar field $K$. Then show that any vector $\mathbf{v}\in V$ can be written uniquely as
\[\mathbf{v}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3,\] where $c_1, c_2, c_3$ are scalars.
 

Proof.

Since $B$ is a basis for $V$, any vector $\mathbf{v} \in V$ is a linear combination of basis vectors in $B$.
Thus, there exist scalars $c_1, c_2, c_3 \in K$ such that
\[\mathbf{v}=c_1\mathbf{v}_+c_2\mathbf{v}_2+c_3\mathbf{v}_3.\] Hence such an expression as a linear combination of basis vectors exists.

We now show that such representation of $\mathbf{v}$ is unique.
Suppose we have another representaion
\[\mathbf{v}=d_1\mathbf{v}_+d_2\mathbf{v}_2+d_3\mathbf{v}_3\] for some scalars $d_1, d_2, d_3 \in K$.

Then we have
\begin{align*}
c_1\mathbf{v}_+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{v}=d_1\mathbf{v}_+d_2\mathbf{v}_2+d_3\mathbf{v}_3.
\end{align*}
Thus, we obtain
\[(c_1-d_1)\mathbf{v}_+(c_2-d_2)\mathbf{v}_2+(c_3-d_3)\mathbf{v}_3=\mathbf{0}.\] Since $B$ is a basis, the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent.

Hence the coefficients of the above linear combination must be all zero, and thus we obtain
\[c_1-d_1=0, \quad c_2-d_2=0, \quad c_3-d_3=0,\] equivalently, we have
\[c_1=d_1, \quad c_2=d_2, \quad c_3=d_3.\] Therefore two representations of the vector $\mathbf{v}$ are the same, and thus the representation of $\mathbf{v}$ as a linear combination of basis vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ is unique.

The post Any Vector is a Linear Combination of Basis Vectors Uniquely first appeared on Problems in Mathematics.

Let $S$ be the following subset of the 3-dimensional vector space $\R^3$.
\[S=\left\{ \mathbf{x}\in \R^3 \quad \middle| \quad \mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}, x_1, x_2, x_3 \in \Z \right\}, \] where $\Z$ is the set of all integers.
Determine whether $S$ is a subspace of $\R^3$.

Proof.

We claim that $S$ is not a subspace of $\R^3$.
If $S$ is a subspace, then $S$ is closed under scalar multiplication.
But this is not the case for $S$.

For example, consider $\mathbf{x}=\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}$.
Since all entries are integers, this is an element of $S$.
Let us compute the scalar multiplication of this vector $\mathbf{x}$ and the scalar $1/2 \in \R$. We have
\[\frac{1}{2}\cdot \mathbf{x}= \frac{1}{2}\cdot\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}=\begin{bmatrix}
1/2 \\
1/2 \\
1/2
\end{bmatrix}.\] Since $1/2$ is not an integer, the scalar multiplication $\frac{1}{2}\mathbf{x}$ is not in $S$. Therefore the subset $S$ cannot be a subspace of $R^3$.

More Examples of Non-Subspaces

Check out the post ↴
10 Examples of Subsets that Are Not Subspaces of Vector Spaces
for various subsets in vector spaces that are not subspaces.

The post Non-Example of a Subspace in 3-dimensional Vector Space $\R^3$ first appeared on Problems in Mathematics.

Suppose that $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_r$ are linearly dependent $n$-dimensional real vectors.

For any vector $\mathbf{v}_{r+1} \in \R^n$, determine whether the vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_r, \mathbf{v}_{r+1}$ are linearly independent or linearly dependent.

Solution.

We claim that the vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_r, \mathbf{v}_{r+1}$ are linearly dependent.
Since the vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_r$ are linearly dependent, there exist scalars (real numbers) $a_1, a_2, \dots, a_r$ such that
\[a_1 \mathbf{v}_1+a_2\mathbf{v}_2+\cdots +a_r\mathbf{v}_r=\mathbf{0} \tag{*}\] and not all of $a_1, \dots, a_r$ are zero, that is, $(a_1, \dots, a_r) \neq (0, \dots, 0)$.

Consider the equation
\[x_1\mathbf{v}_1+x_2 \mathbf{v}_2+\cdots +x_r \mathbf{v}_r+x_{r+1} \mathbf{v}_{r+1}=\mathbf{0}.\] If this equation has a nonzero solution $(x_1, \dots, x_r, x_{r+1})$, then the vectors $\mathbf{v}_1, \dots, \mathbf{v}_{r+1}$ are linearly dependent.

In fact,
\[(x_1,x_2,\dots, x_r, x_{r+1})=(a_1, a_2, \dots, a_r, 0)\] is a nonzero solution of the above equation.
To see this, first note that since not all of $a_1, a_2, \dots, a_r$ are zero, we have
\[(a_1, a_2, \dots, a_r, 0)\neq (0, 0, \dots, 0, 0).\]

Plug these values in the equation, we have
\begin{align*}
&a_1 \mathbf{v}_1+a_2\mathbf{v}_2+\cdots +a_r\mathbf{v}_r+0\mathbf{v}_{r+1}\\
&=a_1 \mathbf{v}_1+a_2\mathbf{v}_2+\cdots +a_r\mathbf{v}_r=\mathbf{0} \text{ by (*).}
\end{align*}
Therefore, we conclude that the vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_r, \mathbf{v}_{r+1}$ are linearly dependent.

The post If Vectors are Linearly Dependent, then What Happens When We Add One More Vectors? first appeared on Problems in Mathematics.