Kernel and matrix of linear transformation $(x_1,x_2,x_3,x_4)^T\mapsto(x_2-x_1,x_3-x_2,x_4-x_3)$

Question

Let $f:\mathbb{R}^4\rightarrow\mathbb{R}^3,x = \left(\begin{array},x_1 \\x_2\\x_3\\x_4\end{array}\right)\mapsto\left(\begin{array},x_2-x_1\\x_3-x_2\\x_4-x_3\end{array}\right)$be a linear transformation.

Give its matrix $A\in\mathbb{R}^{3\times4}$ with $f(x)=A\cdot x$, $\ker(f):=\{x\in\mathbb{R}^4|f(x)=0\}$ and $\dim(\ker(f))$.

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I'm pretty lost on how to do this right now, the topic is going a bit too fast for me. Could anybody be so kind to show me how to proceed? I apologize for the bad post but I'm really rather clueless.

Answer 1

Here is a post of mine here on Math Stack Exchange where you can read up a bit on the matrix of a linear transformation.

I'm sure you're well aware that, in order to find the matrix of a linear transformation $T \colon U \to V$, where $U$ and $V$ are vector spaces over the same field $F$, we need to first select an ordered basis for $U$ and an ordered basis for $V$. Thus the matrix of a linear transformation is relative to a specific choice of ordered bases for both the domain and the co-domain space.

Now for your problem:

Let us choose the following ordered basis for $\mathbb{R}^4$: $$ \left( \ \left[ \begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix} \right], \left[ \begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix} \right], \left[ \begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix} \right], \left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix} \right] \ \right), \tag{1} $$ and let us choose the following ordered basis for $\mathbb{R}^3$: $$ \left( \ \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right], \left[ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right], \left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] \ \right). \tag{2} $$

These are of course the standard (ordered) bases for $\mathbb{R}^4$ and $\mathbb{R}^3$, respectively.

We note that $$ \begin{align} T \left( \left[ \begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix} \right] \right) &= \left[ \begin{matrix} -1 \\ 0 \\ 0 \end{matrix} \right] = (-1) \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right] + 0 \left[ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right] + 0 \left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right], \\ T \left( \left[ \begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix} \right] \right) &= \left[ \begin{matrix} 1 \\ -1 \\ 0 \end{matrix} \right] = 1 \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right] + (-1) \left[ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right] + 0 \left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right], \\ T \left( \left[ \begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix} \right] \right) &= \left[ \begin{matrix} 0 \\ 1 \\ -1 \end{matrix} \right] = 0 \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right] + 1 \left[ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right] + (-1) \left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right], \\ T \left( \left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix} \right] \right) &= \left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] = 0 \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right] + 0 \left[ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right] + 1 \left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right]. \end{align} $$

Now the matrix of $T$ with respect to the standard ordered bases for $\mathbb{R}^4$ and $\mathbb{R}^3$ will be the one formed by putting the co-efficients in the above representations of each one of the images under $T$ of the basis vectors in (1) above as a linear combination of the basis vectors in (2) above into the columns.

Therefore the required matrix of the linear transformation $T$ is the matrix $[T]$ given by $$[T] \colon= \left[ \begin{matrix} -1 \ & 1 \ & 0 \ & 0 \\ 0 \ & -1 \ & 1 \ & 0 \\ 0 \ & 0 \ & -1 \ & 1 \end{matrix} \right]. $$

Now for any vector $$ \mathbf{x} \colon= \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right] $$ in $\mathbb{R}^4$, we find that $$ \begin{align} [T] \mathbf{x} &= \left[ \begin{matrix} -1 \ & 1 \ & 0 \ & 0 \\ 0 \ & -1 \ & 1 \ & 0 \\ 0 \ & 0 \ & -1 \ & 1 \end{matrix} \right] \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right] \\ &= \left[ \begin{matrix} -x_1 + x_2 \\ -x_2 + x_3 \\ -x_3 + x_4 \end{matrix} \right] \\ &= \left[ \begin{matrix} x_2 -x_1 \\ x_3 -x_2 \\ x_4 - x_3 \end{matrix} \right] \\ &= T( \mathbf{x} ). \end{align} $$

A vector $$ \mathbf{x} \colon= \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right] $$ in $\mathbb{R}^4$ is in the kernel of $T$ if and only if $$ T( \mathbf{0} ) = \mathbf{0}_{\mathbb{R}^3}, $$ that is, $$ \left[ \begin{matrix} x_2 -x_1 \\ x_3 -x_2 \\ x_4 - x_3 \end{matrix} \right] = \left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right], $$ which is the case if and only if $$ x_4 = x_3 = x_2 = x_1. $$

Hence the kernel of $T$ is given by $$ \ker (T) = \left\{ \ \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right] \in \mathbb{R}^4 \ \colon \ x_1 = x_2 = x_3 = x_4 \ \right\}, $$ which can also be expressed as $$ \ker (T) = \left\{ \ \alpha \left[ \begin{matrix} 1 \\ 1 \\ 1 \\ 1 \end{matrix} \right] \ \colon \ \alpha \in \mathbb{R} \ \right\}. $$

Hope this helps.

Answer 2

The kernel of a linear transformation is the set of all inputs such that they output the zero vector.

Suppose that $f\left(\begin{bmatrix}x_1\\ x_2\\ x_3 \\ x_4\end{bmatrix}\right)=\begin{bmatrix}0\\0\\0\end{bmatrix}$. What can we say about $x_1,x_2,x_3,x_4$?

Remember how $f$ is defined, that is to say remember that $f\left(\begin{bmatrix}x_1\\ x_2\\ x_3 \\ x_4\end{bmatrix}\right)=\begin{bmatrix}x_2-x_1\\x_3-x_2\\x_4-x_3\end{bmatrix}$ and for this to be equal to the zero vector implies...

that $x_2-x_1=0$, that $x_3-x_2=0$, and that $x_4-x_3=0$.

Rearranging this a bit further implies that...

$x_1=x_2$ and $x_2=x_3$ and $x_3=x_4$, implying that they are all equal. That is to say, if we assumed $\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}$ was in the kernel that implies that it must have been of the form $k\cdot \begin{bmatrix}1\\1\\1\\1\end{bmatrix}$ for some constant $k$

As for the dimension of the kernel... this is in effect asking how many vectors must be included in a basis that describes the kernel. The previous work should give you a strong idea about that.

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As for finding a matrix representation of $f$., consider what happens if you apply $f$ to each of the standard basis elements.

What is $f\left(\begin{bmatrix}1\\0\\0\\0\end{bmatrix}\right)$?

What is $f\left(\begin{bmatrix}0\\1\\0\\0\end{bmatrix}\right)$?

What is $f\left(\begin{bmatrix}0\\0\\1\\0\end{bmatrix}\right)$?

What is $f\left(\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\right)$?

We can combine this information in a useful way to come up with the matrix representation. See if you can figure out how.

Answer 3

Solving $f(x)=0$ you’ll get $x_1=x_2=x_3=x_4$, so exactly any multiple of $(1,1,1,1)$ is an element of the kern. Hence its dimension is ...