# Linear Transformation to 1-Dimensional Vector Space and Its Kernel

## Problem 329

Let $n$ be a positive integer. Let $T:\R^n \to \R$ be a non-zero linear transformation.

Prove the followings.

**(a)** The nullity of $T$ is $n-1$. That is, the dimension of the kernel of $T$ is $n-1$.

(The kernel of $T$ is also called the null space of $T$.)

**(b)** Let $B=\{\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}\}$ be a basis of the kernel $\ker(T)$ of $T$.

Let $\mathbf{w}$ be the $n$-dimensional vector that is not in $\ker(T)$. Then

\[B’=\{\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}, \mathbf{w}\}\]
is a basis of $\R^n$.

**(c)** Each vector $\mathbf{u}\in \R^n$ can be expressed as

\[\mathbf{u}=\mathbf{v}+\frac{T(\mathbf{u})}{T(\mathbf{w})}\mathbf{w}\]
for some vector $\mathbf{v}\in \ker(T)$.

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## Proof.

### (a) The nullity of $T$ is $n-1$.

Let $A$ be the matrix representation of the linear transformation $T:\R^n \to \R$.

Then $A$ is a $1 \times n$ matrix and nonzero since $T$ is a non-zero linear transformation.

Thus, the rank of $A$ is $1$. So the rank of $T$ is $1$.

Then the rank-nullity theorem

\[\text{rank of $T$}+\text{ nullity of $T$}=n\]
yields that the nullity of $T$ is $n-1$.

### $B’$ is a basis of $\R^n$.

We claim that the $n$ vectors $\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}, \mathbf{w}$ are linearly independent. Suppose that we have

\[c_1\mathbf{v}_1+ \cdots +c_{n-1} \mathbf{v}_{n-1}+c_n \mathbf{w}=\mathbf{0},\]
for some $c_1, \dots, c_n \in \R$.

Observe that if $c_n\neq 0$, then we have

\[\mathbf{w}=\frac{-c_1}{c_n}\mathbf{v}_1+\cdots+\frac{-c_{n-1}}{c_n}\mathbf{v}_{n-1},\]
and thus $\mathbf{w}\in \Span(B)=\ker(T)$, a contradiction.

Hence $c_n=0$. Then the linear combination becomes

\[c_1\mathbf{v}_1+ \cdots +c_{n-1} \mathbf{v}_{n-1}=\mathbf{0},\]
and since $B=\{\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}\}$ is linearly independent (as being a basis), we must have $c_1=\cdots=c_{n-1}=0$.

As a result, all $c_i$ are zero. So the vectors $\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}, \mathbf{w}$ are linearly independent.

Since $\R^n$ is an $n$-dimensional vector space and $B’$ consists of $n$ linearly independent vectors, the set $B’$ is a basis of $\R^n$.

### Expression of a vector in $\R^n$.

Let $\mathbf{u}\in \R^n$. Since $B’$ isa basis of $\R^n$, we can write

\[\mathbf{u}=c_1\mathbf{v}_1+ \cdots +c_{n-1} \mathbf{v}_{n-1}+c_n \mathbf{w},\]
for some $c_1, \dots, c_n \in \R$.

Let

\[\mathbf{u}=c_1\mathbf{v}_1+ \cdots +c_{n-1} \mathbf{v}_{n-1}+c_n\mathbf{w}.\]
Then $\mathbf{v}$ is in $\ker(T)$ and we have

\[\mathbf{u}=\mathbf{v}+c_{n}\mathbf{w}. \tag{*}\]
We determine $c_n$ as follows.

Applying the linear transformation $T$, we obtain

\begin{align*}

T(\mathbf{u})&=T(\mathbf{v}+c_{n}\mathbf{w})\\

&=T(\mathbf{v})+c_{n}T(\mathbf{w}) && \text{by linearity of $T$}\\

&=\mathbf{0}+c_{n}T(\mathbf{w}) && \text{since $\mathbf{v}\in \ker(T)$}\\

&=c_{n}T(\mathbf{w}).

\end{align*}

Note that since $\mathbf{w}\notin \ker(T)$, the real number $T(\mathbf{w})$ is not zero.

Hence, we obtain

\[c_n=\frac{T(\mathbf{u})}{T(\mathbf{w})}.\]
Combining this with (*), it follows that

\[\mathbf{u}=\mathbf{v}+\frac{T(\mathbf{u})}{T(\mathbf{w})}\mathbf{w}\]
with $\mathbf{v}\in \ker(T)$.

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