Let $C([-1, 1])$ denote the vector space of real-valued functions on the interval $[-1, 1]$. Define the vector subspace
\[W = \{ f \in C([-1, 1]) \mid f(0) = 0 \}.\]

Define the map $T : C([-1, 1]) \rightarrow W$ by $T(f)(x) = f(x) – f(0)$. Determine if $T$ is a linear map. If it is, determine its nullspace and range.

Proof.

$T$ is a lineawr transformation

First we must check that $T$ takes values in $W$, as the problem implies but does not prove. Consider $f \in C([-1 , 1])$. The function $T(f)$ lies in $W$ if and only if $T(f)(0) = 0$. We quickly check that it satisfies this condition:in $C([-1, 1])$ lies
\[ T(f)(0) = f(0) – f(0) = 0 . \]

Now that we know $T$ takes values in $W$, we show that it is a linear transformation. We will prove this by showing that $T$ satisfies both of the axioms for linear transformations. First, suppose that $f, g \in C([-1, 1])$. Then
\begin{align*}
T(f+g)(x) &= (f+g)(x) – (f+g)(0) \\
&= f(x) + g(x) – f(0) – g(0) \\
&= f(x) – f(0) + g(x) – g(0) \\
&= T(f)(x) + T(g)(x) . \end{align*}

Now for a scalar $c \in \mathbb{R}$ we have
\begin{align*}
T( cf )(x) &= (cf)(x) – (cf)(0) \\
&= c f(x) – c f(0) \\
&= c ( f(x) – f(0) ) \\
&= c T(f)(x) . \end{align*}

Thus we have proven that $T$ is a linear transformation.

The nullspace of $T$

Next, we will prove that the nullspace of $T$ is
\[\mathcal{N}(T) = \{ f \in C([-1 , 1]) \mid f(x) \mbox{ is a constant function } \}.\] Suppose that $f \in \calN(T)$, that is, $f$ satisfies
\[0 = T(f)(x) = f(x) – f(0).\] Then $f(x) = f(0)$ for all $x \in [-1, 1]$. This means that $f$ is a constant function. On the other hand, if $f(x)$ is a constant function, then $T(f)(x) = f(x) – f(0) = 0$. We see that $f$ lies in the nullspace of $T$ if and only if it is a constant function.

The range of $T$

Next, we want to find the range of $T$. We claim that
\[\mathcal{R}(T) = W.\]

Suppose that $f \in W$, that is, it is a function such that $f(0) = 0$. Then
\[T(f)(x) = f(x) – f(0) = f(x),\] and so we see that $f \in \mathcal{R}(T)$.
Conversely, suppose that $f \in \mathcal{R}(T)$, so that $f = T(g)$ for some $g \in C([-1, 1])$. Then
\[f(0) = T(g)(0) = g(0) – g(0) = 0,\] and so $f \in W$. Thus every $f \in C([-1, 1])$ lies in the range of $T$ if and only if $f(0) = 0$.

\end{proof}

The post Find the Nullspace and Range of the Linear Transformation $T(f)(x) = f(x)-f(0)$ first appeared on Problems in Mathematics.

Let $U$ and $V$ be vector spaces over a scalar field $\F$.
Define the map $T:U\to V$ by $T(\mathbf{u})=\mathbf{0}_V$ for each vector $\mathbf{u}\in U$.

(a) Prove that $T:U\to V$ is a linear transformation.
(Hence, $T$ is called the zero transformation.)

(b) Determine the null space $\calN(T)$ and the range $\calR(T)$ of $T$.

Proof.

(a) Prove that $T:U\to V$ is a linear transformation.

Let $\mathbf{u}_1, \mathbf{u}_2\in U$ and $r$ be a scalar, that is, $r\in \F$.
It follows from the definition of $T$ that
\begin{align*}
T(\mathbf{u}_1)&=\mathbf{0}_V, \quad T(\mathbf{v}_2)=\mathbf{0}_V, \\
T(\mathbf{u}_1+\mathbf{u}_2)&=\mathbf{0}_V, \quad T(r\mathbf{u}_1)=\mathbf{0}_V
\end{align*}
since $\mathbf{u}_1+\mathbf{u}_2, r\mathbf{u}_1\in U$.
Hence we have
\begin{align*}
T(\mathbf{u}_1+\mathbf{u}_2)&=\mathbf{0}_V=\mathbf{0}_V+\mathbf{0}_V=T(\mathbf{u}_1)+T(\mathbf{u}_2)\\
T(r\mathbf{u}_1)&=\mathbf{0}_V=r\mathbf{0}_V=rT(\mathbf{u}_1).
\end{align*}
Since these equalities holds for all $\mathbf{u}_1, \mathbf{u}_2\in U$, and $r\in \F$, the map $T:U\to V$ is a linear transformation.

(b) Determine the null space $\calN(T)$ and the range $\calR(T)$ of $T$.

The null space $\calN(T)$ of $T$ is, by definition,
\begin{align*}
\calN(T)=\{\mathbf{u}\in U \mid T(\mathbf{u})=\mathbf{0}_V\}.
\end{align*}
Since $T(\mathbf{u})=\mathbf{0}_V$ for every $\mathbf{u}\in U$, we obtain
\[\calN(T)=U.\]

The range $\calR(T)$ of $T$ is, by definition,
\[\calR(T)=\{\mathbf{v} \in V \mid \text{there exists } \mathbf{u}\in U \text{ such that } T(\mathbf{u})=\mathbf{v}\}.\]

Since every vector of $U$ is mapped into $\mathbf{0}_V$, we have
\[\calR(T)=\{\mathbf{0}_V\}.\] Click here if solved 77

The post The Range and Null Space of the Zero Transformation of Vector Spaces first appeared on Problems in Mathematics.

Let $U$ and $V$ be finite dimensional vector spaces over a scalar field $\F$.
Consider a linear transformation $T:U\to V$.

Prove that if $\dim(U) > \dim(V)$, then $T$ cannot be injective (one-to-one).

Hints.

You may use the folowing facts.

For the proof of this fact, see the post ↴
A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero

We give two proofs. The first one uses the rank-nullity theorem.
The second one avoids using the theorem.

Proof 1.

By the rank-nullity theorem, we have
\[\nullity(T)+\rk(T)=\dim(U).\] Note that the rank of $T$ is the dimension of the range $\calR(T)$, which is a subspace of the vector space of $V$.
Hence it yields that
\[\rk(T)=\dim(\calR(T)) \leq \dim(V).\]

It follows that
\begin{align*}
\nullity(T)=\dim(U)-\rk(T) > \dim(U)-\dim(V) > 0,
\end{align*}
where the last inequality follows from the assumption.

This implies that the nullity is nonzero, hence $T$ is not injective.
(See the hints above.)

Proof 2.

In the second proof, we prove $T$ is injective without using the rank-nullity theorem.

Seeking a contradiction, assume that $T: U\to V$ is injective.
Let $\{\mathbf{u}_1, \dots, \mathbf{u}_n\}$ be a basis of the vector space $U$, where $n=\dim(U)$.
Put $\mathbf{v}_1=T(\mathbf{u}_1), \dots, \mathbf{v}_n=T(\mathbf{u}_n)$.

We claim that the vectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ are linearly independent in $V$.
To see this, consider the linear combination
\[c_1\mathbf{v}_1+\cdots+c_n \mathbf{v}_n=\mathbf{0}_V,\] where $c_1, \dots, c_n$ are scalars in $\F$.
Then we have
\begin{align*}
\mathbf{0}_V&=c_1\mathbf{v}_1+\cdots+c_n \mathbf{v}_n\\
&=c_1T(\mathbf{u}_1)+\cdots+c_n T(\mathbf{u}_n)\\
&=T(c_1\mathbf{u}_1+\cdots + c_n \mathbf{u}_n).
\end{align*}
Since $T$ is assumed to be injective, we must have
\[c_1\mathbf{u}_1+\cdots + c_n \mathbf{u}_n=\mathbf{0}_U.\] Because $\mathbf{u}_1, \dots, \mathbf{u}_n$ are basis vectors, they are linearly independent.
It follows that
\[c_1=\cdots=c_n=0,\] and hence the vectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ are linearly independent.

As the vector space $V$ contains $n$ linearly independent vectors, we see that
\[\dim(V) \geq n=\dim(U),\] which contradicts the assumption that $\dim(U) > \dim(V)$.
Therefore, $T$ cannot be injective.

The post A Linear Transformation $T: U\to V$ cannot be Injective if $\dim(U) > \dim(V)$ first appeared on Problems in Mathematics.