As we continue to study vector spaces, we now turn our attention to mapping between vector spaces. We defined mappings that respected both the vector addition and scalar multiplication as linear transformation. These mappings between vector spaces will allow to give a comparison between vector spaces beyond that of just subspaces. Since it will be extremely helpful to see the connection between different spaces, we want to be discuss some of the properties of these mappings. Here, we focus on finding the kernel, range, rank and nullity of a linear transformation. Using these, we will be able to determine if a mapping is one-to-one, onto, both or none.

**A mapping from \(\mathbb{R}^{3} \to \mathbb{R}^{3}\)**

Let \(T: \mathbb{R}^{3} \to \mathbb{R}^{3}\) be given by \(T(\mathbf{v})=A\mathbf{v}\) where \(A\) is the matrix

\begin{align*}

\begin{bmatrix}

2 & 1 & 4 \\

2 & -1 & 2 \\

4 & 0 & 6

\end{bmatrix}.

\end{align*}

Then find \(ker(T)\), \(range(T)\), \(rank(T)\), \(nullity(T)\) and the whether or not \(T\) is one-to-one, onto, both or none.

**Kernel of \(T\)**

Recall that the kernel of a linear transformation is the set of vectors in the domain that get mapped to the zero vector. When dealing with transformations defined by matrix multiplication, we were able to show that this kernel was precisely the null space of the matrix. We were able to find null spaces in Row, Column and Null Space, so we will again work through this process for the given matrix \(A\).

We first note that \(A\) can be row-reduced to the matrix

\begin{align*}

\begin{bmatrix}

1 & 0 & \frac{3}{2} \\

0 & 1 & 1 \\

0 & 0 & 0

\end{bmatrix}

\end{align*}

Now, we can see that there will be one free variable, so we let \(x_{3}=s\). This then gives us that \(x_{1}=-\frac{3}{2}s\) and \(x_{2}=-s\). Combining this together we find that

\begin{align*}

ker(T)=\left\{(-\frac{3}{2}s,-s,s): s \in \mathbb{R}\right\}.

\end{align*}

Furthermore, we can find a basis for this space by letting \(s=1\), so we have

\begin{align*}

ker(T)&=\left\{(-\frac{3}{2}s,-s,s): s \in \mathbb{R}\right\} \\

&=span(\left\{(-\frac{3}{2},-1,1)\right\}).

\end{align*}

Note that since the basis of \(ker(T)\) has one vector, \(dim(ker(T))=nullity(T)=1\). Furthermore, since the \(nullity(T) \neq 0\), we get that the linear transformation is not one-to-one.

**Range of \(T\)**

The range of \(T\) is the set of all vectors that get mapped to under \(T\). We saw that, when dealing with transformations defined by matrix multiplication, these vectors were precisely the vectors that could be written as a linear combination of the columns of \(A\). That is, \(range(T)=\)column space\((T)\).

In order to find the column space of \(A\), we note that we have already row-reduced the matrix \(A\) above. Therefore, the basis of the column space is just the set of vectors which correspond to columns with leading \(1\)s in the row-reduced matrix. We now see that

\begin{align*}

range(T)=span(\left\{(2,2,4),(1,-1,0)\right\}).

\end{align*}

Here we note that the basis of the \(range(T)\) has 2 vectors in it, so \(dim(range(T))=rank(T)=2\). Furthermore, since \(dim(\mathbb{R}^{3})=3 > 2 =rank(T)\), we get that the codomain is larger than the range, so this is not an onto mapping.

**Conclusion**

We have now seen that the linear transformation given by the multiplication by this matrix has \(rank(T)=2\), \(nullity(T)=1\) and is therefore neither one-to-one nor onto. In order to see this, we were able to generalize the work we’ve done in the past on matrices to work on linear transformations.

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