Write a matrix given the null space

[Zero2Infinity]

Homework Statement

Build the matrix A associated with a linear transformation ƒ:ℝ³→ℝ³ that has the line x-4y=z=0 as its kernel.

Homework Equations

I don't see any relevant equation to be specified here .

The Attempt at a Solution

First of all, I tried to find a basis for the null space by solving the homogeneous linear system:
\begin{equation}
\begin{cases} x-4y=0\\z=0\end{cases} \Leftrightarrow \begin{cases} x=4y\\z=0\end{cases}
\end{equation}
The solutions are thus of the kind (4y,y,0)=y(4,1,0). A basis is hence given by the single vector (4,1,0).
Due to the rank-nullity theorem I know that dim(Im) = dim(ℝ³)-dim(ker(f))=3-1=2. Then, I need two more vectors in order to obtain a basis of ℝ³ that is composed by (4,1,0).
If I chose two vector such as that the matrix of the coordinates has the wanted kernel, I've concluded. My problem is that I don't see how to consciously pick them.
Can anyone help me? Thank you very much!

 

[Mark44]

Homework Statement

Build the matrix A associated with a linear transformation ƒ:ℝ³→ℝ³ that has the line x-4y=z=0 as its kernel.

Homework Equations

I don't see any relevant equation to be specified here .

The Attempt at a Solution

First of all, I tried to find a basis for the null space by solving the homogeneous linear system:
\begin{equation}
\begin{cases} x-4y=0\\z=0\end{cases} \Leftrightarrow \begin{cases} x=4y\\z=0\end{cases}
\end{equation}
The solutions are thus of the kind (4y,y,0)=y(4,1,0). A basis is hence given by the single vector (4,1,0).

There is another basis vector for the kernel: <0, 0, 1>

Due to the rank-nullity theorem I know that dim(Im) = dim(ℝ³)-dim(ker(f))=3-1=2. Then, I need two more vectors in order to obtain a basis of ℝ³ that is composed by (4,1,0).
If I chose two vector such as that the matrix of the coordinates has the wanted kernel, I've concluded. My problem is that I don't see how to consciously pick them.
Can anyone help me? Thank you very much!

 

[Zero2Infinity]

Sorry but I'm not convinced.
The basis of the kernel is found by solving the system I wrote, corresponding to the matrix
\begin{equation} \begin{pmatrix} 1&-4&0\\0&0&1 \end{pmatrix} \end{equation}
This system ha $\infty^1$ solutions of the type $(4y,y,0)$, i.e. a basis is $(4,1,0)$ and the kernel is one-dimensional.

 

[fresh_42]

Why don't you use $f(\vec{x})=A\vec{x} = \vec{0}\,$?

 

[Zero2Infinity]

Let's see: knowing the basis vectors I can write
\begin{equation}
\begin{pmatrix}4&a&b\\1&c&d\\0&e&f\end{pmatrix} \begin{pmatrix}1\\-4\\0\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}
\end{equation}
Thus I get
\begin{equation}
\begin{cases}4-4a+0b=0\\1-4c+0d=0\\0-4e+0f=0\end{cases}
\end{equation}
from which $a=1, c=\frac{1}{4}, e=0$. Repeating the procedure with
\begin{equation}
\begin{pmatrix}4&a&b\\1&c&d\\0&e&f\end{pmatrix} \begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}
\end{equation}
I get
\begin{equation}
\begin{cases}0+0a+b=0\\0+0c+d=0\\0+0e+f=0\end{cases}
\end{equation}
from which $b=0, d=0, f=0$.
Concluding, I get the matrix
\begin{equation}
\begin{pmatrix}4&1&0\\1&\frac{1}{4}&0\\0&0&0\end{pmatrix}
\end{equation}
By doing so, though, the kernel becomes 2-dimensional with $(0,0,1),(-\frac{1}{4},1,0)$ as basis.

 

[fresh_42]

Of course. You have only one linear independent vector to span the image, but you need two. The matrix $A'$ of the linear equations in post #3 already had two (row) vectors to span the image. Why don't you expand it with a third row? What does $A' x = 0$ tell you? I mean, reread your own comment.

 

[Zero2Infinity]

Of course. You have only one linear independent vector to span the image, but you need two. The matrix $A'$ of the linear equations in post #3 already had two (row) vectors to span the image. Why don't you expand it with a third row? What does $A' x = 0$ tell you? I mean, reread your own comment.

First of all, thank you for your patience.
If I add a vector linearly dependent to the other two I get a matrix which kernel is $(4,1,0)$. For example
\begin{equation}
\begin{pmatrix}1&-4&0\\0&0&1\\0&0&2\end{pmatrix}
\end{equation}
Then this is a solution for the exercise? As you might guess I'm not very sure about the conceptual meaning of what I'm doing...

 

[fresh_42]

You get a matrix which kernel is $\{(x,y,z)\,\vert \,(1,-4,0)\cdot \vec{x} = 0 \wedge (0,0,1) \cdot \vec{x} = 0 \}$, i.e. the linear equations you started with. I would have chosen $0$ as the last entry, but $2$ is o.k., too. The first two row vectors are now perpendicular to $(4,1,0)$, which you were looking for in post #1, and also perpendicular to each other. Therefore together they build a nice little basis of orthogonal vectors in $\mathbb{R}^3$: two in the image of $f$ and one in its kernel.

 

[Zero2Infinity]

Thank you again Sir!
Just to be sure: $(1,-4,0),(0,0,1)$ is a orthogonal basis of the $Im(f)$ subspace, $(4,1,0)$ is the basis for the one-dimensional nullspace of $f$ and so, by virtue of the rank-nullity theorem, I have an orthogonal basis of the domain of $f$, i.e. $\mathbb{R}^3$.

 

[fresh_42]

Yes.

 

[pasmith]

Homework Statement

Build the matrix A associated with a linear transformation ƒ:ℝ³→ℝ³ that has the line x-4y=z=0 as its kernel.

Homework Equations

I don't see any relevant equation to be specified here .

The Attempt at a Solution

First of all, I tried to find a basis for the null space by solving the homogeneous linear system:
\begin{equation}
\begin{cases} x-4y=0\\z=0\end{cases} \Leftrightarrow \begin{cases} x=4y\\z=0\end{cases}
\end{equation}
The solutions are thus of the kind (4y,y,0)=y(4,1,0). A basis is hence given by the single vector (4,1,0).
Due to the rank-nullity theorem I know that dim(Im) = dim(ℝ³)-dim(ker(f))=3-1=2. Then, I need two more vectors in order to obtain a basis of ℝ³ that is composed by (4,1,0).
If I chose two vector such as that the matrix of the coordinates has the wanted kernel, I've concluded. My problem is that I don't see how to consciously pick them.
Can anyone help me? Thank you very much!

You don't need to.

A linear transformation which does the trick is \mathbf{x} \mapsto \mathbf{x} \times (4, 1, 0)^{T}. This can be written as \mathbf{x} \mapsto A\mathbf{x} for some matrix A.