Linear Transformation to 1-Dimensional Vector Space and Its Kernel

Linear Transformation problems and solutions

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 nullspace of $T$ is $n-1$.

(b) Let $B=\{\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}\}$ be a basis of the nullspace $\calN(T)$ of $T$.
Let $\mathbf{w}$ be the $n$-dimensional vector that is not in $\calN(T)$. 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 $\mathbf{v}\in \calN(T)$.

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(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) $B’$ 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 $\mathbf{w}\in \Span(B)=\calN(T)$, 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 $\R^n$.

(c) Each $\mathbf{u}\in \R^n$ can be expressed as $\mathbf{u}=\mathbf{v}+\frac{T(\mathbf{u})}{T(\mathbf{w})}\mathbf{w}$.

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 $c_1, \dots, c_n \in \R$.
\[\mathbf{u}=c_1\mathbf{v}_1+ \cdots +c_{n-1} \mathbf{v}_{n-1}+c_n\mathbf{w}.\] Then $\mathbf{v}$ is in $\calN(T)$ 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
&=T(\mathbf{v})+c_{n}T(\mathbf{w}) && \text{by linearity of $T$}\\
&=\mathbf{0}+c_{n}T(\mathbf{w}) && \text{since $\mathbf{v}\in \calN(T)$}\\

Note that since $\mathbf{w}\notin \calN(T)$, 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 $\mathbf{v}\in \calN(T)$.

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