Finding a matrix from a given null space

In summary, The matrix A must have 4 columns and 4 rows, and its nullity is 2, meaning its rank is 2. The vectors v1 and v2 are given in the exercise, but no information is given about bases. To find a simpler solution, it is necessary to have a basis {v1, v2, e1, e2, e3, e4}, and the matrix A must satisfy the requirement Av1=Av2=0. The simplest solution is to have A act as the identity on the remaining two dimensions. The basis can be found by row-reducing the matrix, and the matrix A can be obtained by multiplying it by (r, p)(p, c)=(
  • #1
archaic
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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.
 
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  • #2
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}##.
 
  • #3
@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.
 
  • #4
So you know that [itex]Av_1 = Av_2 = 0[/itex]. That leaves two other independent dimensions on which [itex]A[/itex] can do anything. Simplest would be to have it act as the identity on those.
 
  • #5
archaic said:
@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.
 
  • #6
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##.
 
  • #7
archaic said:
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?
 
  • #8
I am pretty familiar with linear algebra, but what is M sub 41?
 
  • #9
mpresic3 said:
I am pretty familiar with linear algebra, but what is M sub 41?
A matrix with 4 rows and 1 column.
 
  • #10
vela said:
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. :)
fresh_42 said:
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..
 
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  • #11
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)##.
 
  • #12
archaic said:
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?
 
  • #13
vela said:
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.
 
  • #14
archaic said:
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.
 
  • #15
vela said:
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##. 😳
 
  • #16
archaic said:
@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##.
 
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Related to Finding a matrix from a given null space

1. What is a null space in a matrix?

The null space of a matrix is the set of all vectors that, when multiplied by the matrix, result in a zero vector. In other words, it is the solution space for the homogeneous equation Ax = 0, where A is the given matrix.

2. How can I find a matrix from a given null space?

To find a matrix from a given null space, you can use the basis of the null space to construct the columns of the matrix. The basis vectors will form the columns of the matrix, and the number of basis vectors will determine the size of the matrix.

3. Can a null space have more than one basis?

Yes, a null space can have multiple bases. This is because there can be more than one set of linearly independent vectors that can span the null space of a matrix.

4. How do I know if a matrix is unique from its null space?

A matrix is unique from its null space if and only if its null space only contains the zero vector. This means that the only solution to the homogeneous equation Ax = 0 is the trivial solution, where all the variables are equal to zero.

5. Can a null space be empty?

Yes, a null space can be empty if the matrix is a zero matrix or if all the columns are linearly dependent. This means that there are no vectors that can satisfy the homogeneous equation Ax = 0, and therefore, the null space is empty.

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