Let $\R^n$ be an inner product space with inner product $\langle \mathbf{x}, \mathbf{y}\rangle=\mathbf{x}^{\trans}\mathbf{y}$ for $\mathbf{x}, \mathbf{y}\in \R^n$.

A linear transformation $T:\R^n \to \R^n$ is called orthogonal transformation if for all $\mathbf{x}, \mathbf{y}\in \R^n$, it satisfies
\[\langle T(\mathbf{x}), T(\mathbf{y})\rangle=\langle\mathbf{x}, \mathbf{y} \rangle.\]

Prove that if $T:\R^n\to \R^n$ is an orthogonal transformation, then $T$ is an isomorphism.

 

We give two proofs.
The second one uses a fact about the injectivity of linear transformations.

Proof 1.

As $T$ is a linear transformation from $\R^n$ to itself, it suffices to show that $T$ is an injective linear transformation.

Suppose that $T(\mathbf{x})=T(\mathbf{y})$ for $\mathbf{x}, \mathbf{y}\in \R^n$.
We show that $\mathbf{x}=\mathbf{y}$.

We have
\begin{align*}
&\|\mathbf{x}-\mathbf{y}\|^2\\
&=(\mathbf{x}-\mathbf{y})^{\trans}(\mathbf{x}-\mathbf{y})\\
&=(\mathbf{x}^{\trans}-\mathbf{y}^{\trans})(\mathbf{x}-\mathbf{y})\\
&=\mathbf{x}^{\trans}\mathbf{x}-\mathbf{x}^{\trans}\mathbf{y}-\mathbf{y}^{\trans}\mathbf{x}+\mathbf{y}^{\trans}\mathbf{y}\\
&=\langle\mathbf{x}, \mathbf{x} \rangle-\langle\mathbf{x}, \mathbf{y} \rangle-\langle\mathbf{y}, \mathbf{x} \rangle+\langle\mathbf{y}, \mathbf{y} \rangle\\
&=\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{y}) \rangle-\langle T(\mathbf{y}), T(\mathbf{x}) \rangle+\langle T(\mathbf{y}), T(\mathbf{y}) \rangle\\
&\text{(since $T$ is an orthogonal transformation)}\\
&=\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{x}) \rangle+\langle T(\mathbf{x}), T(\mathbf{x}) \rangle\\
&\text{(since $T(\mathbf{x})=T(\mathbf{y})$)}\\
&=0.
\end{align*}
It follows that $\|\mathbf{x}-\mathbf{y}\|=0$ and hence $\mathbf{x}=\mathbf{y}$.
This proves that $T:\R^n\to \R^n$ is injective.

As $T$ is an injective linear transformation from the $n$-dimensional vector space $\R^n$ to itself, it is also surjective, and thus $T$ is an isomorphism.

Proof 2.

Recall that the linear transformation $T$ is injective if and only if the null space $\calN(T)=\{\mathbf{0}\}$, that is, $T(\mathbf{x})=\mathbf{0}$ implies that $\mathbf{x}=\mathbf{0}$.
(See the post “A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero” for the proof of this fact.)

We use this fact to show that $T$ is injective.
Suppose that $T(\mathbf{x})=\mathbf{0}$.
Then we have
\begin{align*}
\|\mathbf{x}\|^2&=\langle \mathbf{x}, \mathbf{x}\rangle\\
&=\langle T(\mathbf{x}), T(\mathbf{x})\rangle &&\text{as $T$ is orthogonal}\\
&=\langle \mathbf{0}, \mathbf{0}\rangle=0 &&\text{as $T(\mathbf{x})=\mathbf{0}$}.
\end{align*}

It follows that the length $\|\mathbf{x}\|=0$, and hence $\mathbf{x}=\mathbf{0}$.
This proves that the null space $\calN(T)=\{\mathbf{0}\}$ and $T$ is injective.

As $T$ is an injective linear transformation from $\R^n$ to itself, it is also surjective, and hence $T$ is an isomorphism.

The post An Orthogonal Transformation from $\R^n$ to $\R^n$ is an Isomorphism 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.

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

Prove that $T$ is injective (one-to-one) if and only if the nullity of $T$ is zero.

Definition (Injective, One-to-One Linear Transformation).

A linear transformation is said to be injective or one-to-one if provided that for all $\mathbf{u}_1$ and $\mathbf{u}_1$ in $U$, whenever $T(\mathbf{u}_1)=T(\mathbf{u}_2)$, then we have $\mathbf{u}_1=\mathbf{u}_2$.

Proof.

$(\implies)$: If $T$ is injective, then the nullity is zero.

Suppose that $T$ is injective.
Our objective is to show that the null space $\calN(T)=\{\mathbf{0}_U\}$.

Since $T$ is a linear transformation, it sends the zero vector $\mathbf{0}_U$ of $U$ to the zero vector $\mathbf{0}_V$ of $V$.
In fact, we have
\begin{align*}
T(\mathbf{0}_U)&=T(\mathbf{0}_U-\mathbf{0}_U)\\
&=T(\mathbf{0}_U+(-1)\mathbf{0}_U)\\
&=T(\mathbf{0}_U)+(-1)T(\mathbf{0}_U) &&\text{by linearity of $T$}\\
&=T(\mathbf{0}_U)-T(\mathbf{0}_U)=\mathbf{0}_V.
\end{align*}
Hence $\mathbf{0}_U\in \calN(T)$.

On the other hand, if $\mathbf{u}\in \calN(T)$, then we have
\[T(\mathbf{u})=\mathbf{0}_V=T(\mathbf{0}_U).\] Since $T$ is injective, it yields that $\mathbf{u}=\mathbf{0}_U$.
Therefore we obtain $\calN(T)=\{\mathbf{0}_U\}$, and the nullity of $T$ is zero.
(Recall that the nullity of $T$ is the dimension of $\calN(T)$.)

$(\impliedby)$: If the nullity is zero, then $T$ is injective.

Next, suppose that the nullity of $T$ is zero.
This is equivalent to the condition $\calN(T)=\{\mathbf{0}_U\}$.
Our goal is to show that $T: U \to V$ is injective.

Suppose that $T(\mathbf{u}_1)=T(\mathbf{u}_2)$ for some $\mathbf{u}_1, \mathbf{u}_2\in U$.
Then we have
\begin{align*}
\mathbf{0}_V&=T(\mathbf{u}_1)-T(\mathbf{u}_2)\\
&=T(\mathbf{u}_1)+(-1)T(\mathbf{u}_2)\\
&=T(\mathbf{u}_1+(-1)\mathbf{u}_2) && \text{by linearity of $T$}\\
&=T(\mathbf{u}_1-\mathbf{u}_2)
\end{align*}
It follows that the vector $\mathbf{u}_1-\mathbf{u}_2$ is in the null space $\calN(T)=\{\mathbf{0}_U\}$.
Thus, we have $\mathbf{u}_1-\mathbf{u}_2=\mathbf{0}_U$, or $\mathbf{u}_1=\mathbf{u}_2$.
So the linear transformation $T$ is injective.

Related Question.

The proof of this problem is given in the post ↴
A Linear Transformation $T: U\to V$ cannot be Injective if $\dim(U) > \dim(V)$

The post A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero first appeared on Problems in Mathematics.