Finding a matrix from a given null space

[archaic]

Homework Statement

Find a matrix $A$ such that its null space is $\mathrm{span}(v_1,v_2)$, where $v_i\in\mathcal{M}_{41}$.

Relevant Equations

rank + nullity = number of columns

I have solved the exercise, so I'm not giving the vectors explicitly. I just want to know if there is a quicker way than mine.
We know that $A$ must have $4$ columns and $4$ lines, and we also know that its nullity is $2$, thus its rank is $2$.
I took the simplest matrix that can have a rank $2$, namely $L_1 = (1,a,b,c)$, $L_2=(0,1,d,e)$, and the rest is zero.
Then, I multiplied by $tv_1+sv_2$, factorized $t$ and $s$ in each line, and made their coefficients $0$.
The last step was to solve for $a,\,b,\,c,\,d$ and $e$ with the equations I put. $c$ is a free variable, so I took a convenient value.

 

[fresh_42]

I think there is an easier way, but that depends on some assumptions which have to be made in order to complete your description.

What are $v_1$ and $v_2$?
According to which basis is the matrix built?

Hence I assume we have a basis $\{v_1,v_2,v_3,v_4\}$ according to which the matrices and vectors are expressed. This means $v_1=(1,0,0,0)^\tau$ and $v_2=(0,1,0,0)^\tau$. Now if $A=(a_{ij})$, then the requirement is $Av_1=Av_2=0$.

If all this is the case, you simply have to do those two multiplications and get the required conditions in terms of $a_{ij}$.

 

[archaic]

@fresh_42 The vectors given in the exercise are $v_1=(-1,1,4,3)^\tau$ and $v_2=(2,0,6,-2)^\tau$. Nothing is said about bases.

 

[pasmith]

So you know that $Av_1 = Av_2 = 0$. That leaves two other independent dimensions on which $A$ can do anything. Simplest would be to have it act as the identity on those.

 

[fresh_42]

@fresh_42 The vectors given in the exercise are $v_1=(-1,1,4,3)^\tau$ and $v_2=(2,0,6,-2)^\tau$. Nothing is said about bases.

Solving the linear equation system won't be shorter in that case, but probably faster than calculating the orthogonal complement.

 

[WWGD]

The transformation is not fully-determined. You can find a basis from $\{v_1,v_2,e_1,e_2,e_3,e_4\}$ by row-reducing until having 4 columns. Then define $Tv_1=Tv_2=0$and send remaining basis vectors to themselves. The extension is not unique but the kernel will be spanned by $v_1,v_2$.

 

[vela]

We know that $A$ must have $4$ columns and $4$ lines, and we also know that its nullity is $2$, thus its rank is $2$.

I don't know if I'm missing something obvious because no one else has mentioned anything, but why does $A$ have to have 4 rows?

 

[mpresic3]

I am pretty familiar with linear algebra, but what is M sub 41?

 

[fresh_42]

I am pretty familiar with linear algebra, but what is M sub 41?

A matrix with 4 rows and 1 column.

 

[archaic]

I don't know if I'm missing something obvious because no one else has mentioned anything, but why does $A$ have to have 4 rows?

When you multiply two matrices, the number of columns of the first should be equal to the number of rows of the second, and the result is a matrix with the same number of rows as the first and the same number of columns as the second; $(r, p)(p,c)=(r,c)$.

@pasmith @fresh_42 @WWGD tell me if this is what you guys mean.
I followed what WWGD said and found out a basis for $\{v_1,v_2,e_1,e_2,e_3,e_4\}$ using row reduction. The transpose of the resulting matrix (without the zero lines) is$$\begin{pmatrix}1&0&0&0\\ 0&1&0&0\\ 3&7&1&0\\ -1&2&\frac{2}{7}&1\end{pmatrix}$$I let $A=(a_{ij})$, and multiplied it by the above matrix. I got$$
\begin{pmatrix}a_{11}+3a_{13}-a_{14}&a_{12}+7a_{13}+2a_{14}&\frac{7a_{13}+2a_{14}}{7}&a_{14}\\ a_{21}+3a_{23}-a_{24}&a_{22}+7a_{23}+2a_{24}&\frac{7a_{23}+2a_{24}}{7}&a_{24}\\ a_{31}+3a_{33}-a_{34}&a_{32}+7a_{33}+2a_{34}&\frac{7a_{33}+2a_{34}}{7}&a_{34}\\ a_{41}+3a_{43}-a_{44}&a_{42}+7a_{43}+2a_{44}&\frac{7a_{43}+2a_{44}}{7}&a_{44}\end{pmatrix}
$$Since the first two column vectors in the first matrix span the same space as $\{v_1,v_2\}$, I want the first two columns in the last matrix to be equal to zero, so I put the first two equations in each line equal to zero, the last two to whatever, and then solve.
Correct?
Edit: Not whatever; if I put everything equal to zero, then I'll get the zero matrix. I did what WWGD proposed again. :)

Solving the linear equation system won't be shorter in that case, but probably faster than calculating the orthogonal complement.

Eh, sorry, I didn't yet see orthogonal complements..

 

[archaic]

Although the above is correct in sense I get $Av_1=Av_2=0$, I think that I still need to show that $\mathrm{null\,space}(A)=\mathrm{span}\{v_1,v_2\}$, right? From $Av_1=Av_2=0$, I can only see that $\mathrm{span}\{v_1,v_2\}\subseteq\mathrm{null\,space}(A)$.

 

[vela]

When you multiply two matrices, the number of columns of the first should be equal to the number of rows of the second, and the result is a matrix with the same number of rows as the first and the same number of columns as the second; $(r, p)(p,c)=(r,c)$.

That doesn't answer my question. $A$ is the first matrix. It needs 4 columns, but you claimed above it also needed 4 rows. Why does it need 4 rows?

 

[archaic]

That doesn't answer my question. $A$ is the first matrix. It needs 4 columns, but you claimed above it also needed 4 rows. Why does it need 4 rows?

Well, since we're talking of the null space of a matrix, the result of the multiplication is the zero column vector. $Ax=0$ as a representation of a homogeanous system of linear equations.

 

[vela]

Well, since we're talking of the null space of a matrix, the result of the multiplication is the zero column vector. $Ax=0$ as a representation of a homogeanous system of linear equations.

I don't see what that has to do with the number of rows in $A$. In any case, $A$ doesn't require four rows. The way you solved the problem, as described in your original post, the two rows of zeros didn't really do anything, did they? You effectively found a 2x4 matrix with the required null space.

You could have saved yourself a little work by multiplying your matrix by $v_1$ and $v_2$ separately (instead of using the linear combination) to get the four linear equations you ended up with.

 

[archaic]

I don't see what that has to do with the number of rows in $A$. In any case, $A$ doesn't require four rows. The way you solved the problem, as described in your original post, the two rows of zeros didn't really do anything, did they? You effectively found a 2x4 matrix with the required null space.

You could have saved yourself a little work by multiplying your matrix by $v_1$ and $v_2$ separately (instead of using the linear combination) to get the four linear equations you ended up with.

Right, right, you can have it like that and end up with a 2x1 zero vector, but, since we have four unknows, I just assumed a 4x1 zero vector, so that A is a 4x4 square matrix. Thank you! The exercise didn't mention anything about matrix dimension, so I guess that that would be accepted too.

Yep. For some reason, I was adamant on doing it with the linear combination, but eh $A(av_1+bv_2)=aAv_1+bAv_2$.

 

[fresh_42]

@pasmith @fresh_42 @WWGD tell me if this is what you guys mean.

I would have chosen $A=(v_1,v_2,0,0)$ instead of a regular matrix. And if you want a regular one, then choose at least $0$ over $2/7$.