# 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.

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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.

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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.

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@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.

Gold Member
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##.

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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?

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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..

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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)##.

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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.

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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##. 😳

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@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##.

• archaic