# Write a matrix given the null space

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1. Jan 30, 2017

### Zero2Infinity

1. The problem statement, all variables and given/known data

Build the matrix A associated with a linear transformation ƒ:3→ℝ3 that has the line x-4y=z=0 as its kernel.

2. Relevant equations

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

3. 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{cases} x-4y=0\\z=0\end{cases} \Leftrightarrow \begin{cases} x=4y\\z=0\end{cases}

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 3 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!

2. Jan 30, 2017

### Staff: Mentor

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

3. Jan 31, 2017

### Zero2Infinity

Sorry but I'm not convinced.
The basis of the kernel is found by solving the system I wrote, corresponding to the matrix
$$\begin{pmatrix} 1&-4&0\\0&0&1 \end{pmatrix}$$
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.

Last edited: Jan 31, 2017
4. Jan 31, 2017

### Staff: Mentor

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

5. Jan 31, 2017

### Zero2Infinity

Let's see: knowing the basis vectors I can write

\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}

Thus I get

\begin{cases}4-4a+0b=0\\1-4c+0d=0\\0-4e+0f=0\end{cases}

from which $a=1, c=\frac{1}{4}, e=0$. Repeating the procedure with

\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}

I get

\begin{cases}0+0a+b=0\\0+0c+d=0\\0+0e+f=0\end{cases}

from which $b=0, d=0, f=0$.
Concluding, I get the matrix

\begin{pmatrix}4&1&0\\1&\frac{1}{4}&0\\0&0&0\end{pmatrix}

By doing so, though, the kernel becomes 2-dimensional with $(0,0,1),(-\frac{1}{4},1,0)$ as basis.

Last edited: Jan 31, 2017
6. Jan 31, 2017

### Staff: Mentor

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.

7. Jan 31, 2017

### Zero2Infinity

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{pmatrix}1&-4&0\\0&0&1\\0&0&2\end{pmatrix}

Then this is a solution for the excercise? As you might guess I'm not very sure about the conceptual meaning of what I'm doing...

8. Jan 31, 2017

### Staff: Mentor

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.

9. Jan 31, 2017

### 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. $\mathbb{R}^3$.

10. Jan 31, 2017

### Staff: Mentor

Yes.

11. Feb 2, 2017

### pasmith

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 $A$.