Are Linear Transformations of Derivatives and Integrations Linearly Independent?
Problem 463
Let $W=C^{\infty}(\R)$ be the vector space of all $C^{\infty}$ real-valued functions (smooth function, differentiable for all degrees of differentiation).
Let $V$ be the vector space of all linear transformations from $W$ to $W$.
The addition and the scalar multiplication of $V$ are given by those of linear transformations.
Let $T_1, T_2, T_3$ be the elements in $V$ defined by
\begin{align*}
T_1\left(\, f(x) \,\right)&=\frac{\mathrm{d}}{\mathrm{d}x}f(x)\\[6pt]
T_2\left(\, f(x) \,\right)&=\frac{\mathrm{d}^2}{\mathrm{d}x^2}f(x)\\[6pt]
T_3\left(\, f(x) \,\right)&=\int_{0}^x \! f(t)\,\mathrm{d}t.
\end{align*}
Then determine whether the set $\{T_1, T_2, T_3\}$ are linearly independent or linearly dependent.
Sponsored Links
Proof.
We prove that the set $\{T_1, T_2, T_3\}$ are linearly independent.
Suppose that we have a linear combination
\[c_1T_1+c_2T_2+c_3 T_3=0,\tag{*}\]
for some real numbers $c_1, c_2, c_3$.
Note that $0$ is the zero linear transformation $0(f(x))=0$ for all $x$, and it is the zero vector in the vector space $V$.
The equality (*) is an equality as linear transformations.
We need to show that $c_1=c_2=c_3=0$.
Let us consider the function $f(x)=1$ in $C^{\infty}(\R)$.
Then we have
\begin{align*}
T_1\left( 1 \right)&=0\\
T_2\left( 1 \right)&=0\\
T_3\left( 1 \right)&=\int_{0}^x \! 1\,\mathrm{d}t=x.
\end{align*}
These yield that
\begin{align*}
0=c_1T_1(1)+c_2T_2(1)+c_3 T_3(1)=c_3x,
\end{align*}
and hence $c_3=0$.
Next, consider the function $f(x)=x$.
We have
\begin{align*}
T_1\left( x \right)=1\\
T_2\left( x \right)=0
\end{align*}
and
\begin{align*}
0=c_1T_1(x)+c_2T_2(x)=c_1.
\end{align*}
Thus $c_1=0$.
Finally, consider the function $f(x)=x^2$.
We compute
\begin{align*}
T_2\left( x^2\right)=2,
\end{align*}
and the equality (*) together with $c_1=c_3=0$ gives
\begin{align*}
0=c_2T_2\left( x^2 \right)=2c_2.
\end{align*}
So $c_2=0$, and we have proved that $c_1=c_2=c_3=0$.
Thus we conclude that the set $\{T_1, T_2, T_3\}$ is linearly independent.
Add to solve later
Sponsored Links
More from my site
Subspace Spanned By Cosine and Sine Functions
Let $\calF[0, 2\pi]$ be the vector space of all real valued functions defined on the interval $[0, 2\pi]$.
Define the map $f:\R^2 \to \calF[0, 2\pi]$ by
\[\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix} \,\right) \,\right)(x):=\alpha \cos x + \beta […]
Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis
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 […]
Differentiation is a Linear Transformation
Let $P_3$ be the vector space of polynomials of degree $3$ or less with real coefficients.
(a) Prove that the differentiation is a linear transformation. That is, prove that the map $T:P_3 \to P_3$ defined by
\[T\left(\, f(x) \,\right)=\frac{d}{dx} f(x)\]
for any $f(x)\in […]
Differentiating Linear Transformation is Nilpotent
Let $P_n$ be the vector space of all polynomials with real coefficients of degree $n$ or less.
Consider the differentiation linear transformation $T: P_n\to P_n$ defined by
\[T\left(\, f(x) \,\right)=\frac{d}{dx}f(x).\]
(a) Consider the case $n=2$. Let $B=\{1, x, x^2\}$ be a […]
Exponential Functions Form a Basis of a Vector Space
Let $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let
\[V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}\]
be a subset in $C[-1, 1]$.
(a) Prove that $V$ is a subspace of $C[-1, 1]$.
(b) […]- Matrix Representations for Linear Transformations of the Vector Space of Polynomials
Let $P_2(\R)$ be the vector space over $\R$ consisting of all polynomials with real coefficients of degree $2$ or less.
Let $B=\{1,x,x^2\}$ be a basis of the vector space $P_2(\R)$.
For each linear transformation $T:P_2(\R) \to P_2(\R)$ defined below, find the matrix representation […]
- Taking the Third Order Taylor Polynomial is a Linear Transformation
The space $C^{\infty} (\mathbb{R})$ is the vector space of real functions which are infinitely differentiable. Let $T : C^{\infty} (\mathbb{R}) \rightarrow \mathrm{P}_3$ be the map which takes $f \in C^{\infty}(\mathbb{R})$ to its third order Taylor polynomial, specifically defined […]
- Find a Polynomial Satisfying the Given Conditions on Derivatives
Find a cubic polynomial
\[p(x)=a+bx+cx^2+dx^3\]
such that $p(1)=1, p'(1)=5, p(-1)=3$, and $ p'(-1)=1$.
Solution.
By differentiating $p(x)$, we obtain
\[p'(x)=b+2cx+3dx^2.\]
Thus the given conditions are
[…]
What if we use f(x)=e^x then we have c1*e^x+c2*e^x+c3*e^x=0 => c1=-1, c2=1/2, c3=1/2, and the are linearly dependent.
For each specific $f(x)$, we get some restrictions on $c_1, c_2, c_3$. The choice $f(x) = e^x$ gives the restriction $c_1+c_2+c_3 = 0$ but this is not strong enough to show that $c_1=c_2=c_3=0$. Also this does not mean that $T_1, T_2, T_3$ are dependent. This just say that for your choice of $f(x)$, the coefficient must be like this. You could use other functions $f(x)$ to get other restrictions and solve them to get $c_1+c_2+c_3 = 0$. In my case, I used three functions $f(x)= 1, x, x^2$ and get three equations which yields $c_1=c_2=c_3=0$.