Mixing
Determine the dimension and find a basis of a vector space
$begingroup$
$V_1 = (x_1, ..., x_7)^T in mathbb{Z}^7_{5} : x_1 + 3x_2 +x_3+2x_4+3x_5+x_6+2x_7 = 0$
$3x_1 + 4x_2 +3x_3+x_4+4x_5+2x_6+4x_7 = 0$
$2x_1 + x_2 +4x_3+4x_5+x_6+2x_7 = 0$
I am supposed to find the dimension and some basis of this vector space.
After putting these equations in matrix form and doing gaussian elimination I got this matrix, but I don't know what to do now, any help would be appreciated
$$left(begin{matrix} 1 & 3 & 1 & 2 & 3 & 1 & 2\ 0 & 0 & 2 & 1 & 3 & 3& 3\ 0 & 0 & 0 & 0 & 0 & 4 & 3end{matrix}right)$$
linear-algebra vector-spaces
asked Jan 16 at 18:50
J. LastinJ. Lastin
936
$endgroup$
add a comment |
$begingroup$
$V_1 = (x_1, ..., x_7)^T in mathbb{Z}^7_{5} : x_1 + 3x_2 +x_3+2x_4+3x_5+x_6+2x_7 = 0$
$3x_1 + 4x_2 +3x_3+x_4+4x_5+2x_6+4x_7 = 0$
$2x_1 + x_2 +4x_3+4x_5+x_6+2x_7 = 0$
I am supposed to find the dimension and some basis of this vector space.
After putting these equations in matrix form and doing gaussian elimination I got this matrix, but I don't know what to do now, any help would be appreciated
$$left(begin{matrix} 1 & 3 & 1 & 2 & 3 & 1 & 2\ 0 & 0 & 2 & 1 & 3 & 3& 3\ 0 & 0 & 0 & 0 & 0 & 4 & 3end{matrix}right)$$
linear-algebra vector-spaces
asked Jan 16 at 18:50
J. LastinJ. Lastin
936
$endgroup$
$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref.
$endgroup$
– amd
Jan 16 at 23:09
add a comment |
$begingroup$
$V_1 = (x_1, ..., x_7)^T in mathbb{Z}^7_{5} : x_1 + 3x_2 +x_3+2x_4+3x_5+x_6+2x_7 = 0$
$3x_1 + 4x_2 +3x_3+x_4+4x_5+2x_6+4x_7 = 0$
$2x_1 + x_2 +4x_3+4x_5+x_6+2x_7 = 0$
I am supposed to find the dimension and some basis of this vector space.
After putting these equations in matrix form and doing gaussian elimination I got this matrix, but I don't know what to do now, any help would be appreciated
$$left(begin{matrix} 1 & 3 & 1 & 2 & 3 & 1 & 2\ 0 & 0 & 2 & 1 & 3 & 3& 3\ 0 & 0 & 0 & 0 & 0 & 4 & 3end{matrix}right)$$
linear-algebra vector-spaces
asked Jan 16 at 18:50
J. LastinJ. Lastin
936
$endgroup$
$V_1 = (x_1, ..., x_7)^T in mathbb{Z}^7_{5} : x_1 + 3x_2 +x_3+2x_4+3x_5+x_6+2x_7 = 0$
$3x_1 + 4x_2 +3x_3+x_4+4x_5+2x_6+4x_7 = 0$
$2x_1 + x_2 +4x_3+4x_5+x_6+2x_7 = 0$
I am supposed to find the dimension and some basis of this vector space.
After putting these equations in matrix form and doing gaussian elimination I got this matrix, but I don't know what to do now, any help would be appreciated
$$left(begin{matrix} 1 & 3 & 1 & 2 & 3 & 1 & 2\ 0 & 0 & 2 & 1 & 3 & 3& 3\ 0 & 0 & 0 & 0 & 0 & 4 & 3end{matrix}right)$$
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 16 at 18:50
J. LastinJ. Lastin
936
asked Jan 16 at 18:50
J. LastinJ. Lastin
936
edited Jan 16 at 19:25
J. Lastin
asked Jan 16 at 18:50
J. LastinJ. Lastin
936
asked Jan 16 at 18:50
J. LastinJ. Lastin
936
asked Jan 16 at 18:50
J. LastinJ. Lastin
936
936
$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref.
$endgroup$
– amd
Jan 16 at 23:09
add a comment |
$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref.
$endgroup$
– amd
Jan 16 at 23:09
$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref.
$endgroup$
– amd
Jan 16 at 23:09
$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref.
$endgroup$
– amd
Jan 16 at 23:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'll assume that your Gaussian elimination is correct and explain how to proceed with the matrix you gave.
You have pivot positions corresponding to $x_{1},x_{3},$ and $x_{6}$. That means your matrix has rank $3$ so its null space, which is what we're after, has dimension $7-3=4$. Let's solve for the pivot variables in terms of the free ones. From the last equation, $4x_{6}=-3x_{7}$ can be rewritten $-x_{6}=-3x_{7}$ since we are in $mathbb{Z}_{5}$. Now multiply both sides by $-1$ to get $x_{6}=3x_{7}$.
For the middle equation, we have
$$2x_{3}=-x_{4}-3x_{5}-3x_{6}-3x_{7}$$
Multiplying both sides by $3$, and simplifying/rewriting mod $5$ gives
$$x_{3}=2x_{4}+x_{5}+x_{6}+x_{7}$$
which, substituting in what we found previously, becomes
$$x_{3}=2x_{4}+x_{5}+3x_{7}+x_{7}=2x_{4}+x_{5}+4x_{7}$$
Finally, repeat for the first equation. Starting with
$$x_{1}=-3x_{2}-x_{3}-2x_{4}-3x_{5}-x_{6}-2x_{7}$$
substitute in for $x_{3}$ and $x_{6}$ to get
begin{align}
x_{1} &= -3x_{2}-(2x_{4}+x_{5}+4x_{7})-2x_{4}-3x_{5}-3(x_{7})-2x_{7} \
&= -3x_{2}-4x_{4}-4x_{5}-4x_{7}\
&equiv 2x_{2}+x_{4}+x_{5}+x_{7}
end{align}
Now we can conclude. The solution set is all vectors of the following form
begin{equation}
begin{pmatrix}
x_{1}\
x_{2}\
x_{3}\
x_{4}\
x_{5}\
x_{6}\
x_{7}\
end{pmatrix}
=
begin{pmatrix}
2x_{2}+x_{4}+x_{5}+x_{7}\
x_{2}\
2x_{4}+x_{5}+4x_{7}\
x_{4}\
x_{5}\
3x_{7}\
x_{7}\
end{pmatrix}=
x_{2}begin{pmatrix}2\1\0\0\0\0\0\end{pmatrix}
+x_{4}begin{pmatrix}1\0\2\1\0\0\0end{pmatrix}
+x_{5}begin{pmatrix}1\0\1\0\1\0\0end{pmatrix}
+x_{7}begin{pmatrix}1\0\1\0\0\3\1end{pmatrix}
end{equation}
Therefore a basis is given by
$$begin{pmatrix}2\1\0\0\0\0\0\end{pmatrix},begin{pmatrix}1\0\2\1\0\0\0end{pmatrix},begin{pmatrix}1\0\1\0\1\0\0end{pmatrix},begin{pmatrix}1\0\1\0\0\3\1end{pmatrix}$$
and the dimension is $4$.
answered Jan 16 at 19:25
pwerthpwerth
3,243417
$endgroup$
add a comment |
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1 Answer
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$begingroup$
I'll assume that your Gaussian elimination is correct and explain how to proceed with the matrix you gave.
You have pivot positions corresponding to $x_{1},x_{3},$ and $x_{6}$. That means your matrix has rank $3$ so its null space, which is what we're after, has dimension $7-3=4$. Let's solve for the pivot variables in terms of the free ones. From the last equation, $4x_{6}=-3x_{7}$ can be rewritten $-x_{6}=-3x_{7}$ since we are in $mathbb{Z}_{5}$. Now multiply both sides by $-1$ to get $x_{6}=3x_{7}$.
For the middle equation, we have
$$2x_{3}=-x_{4}-3x_{5}-3x_{6}-3x_{7}$$
Multiplying both sides by $3$, and simplifying/rewriting mod $5$ gives
$$x_{3}=2x_{4}+x_{5}+x_{6}+x_{7}$$
which, substituting in what we found previously, becomes
$$x_{3}=2x_{4}+x_{5}+3x_{7}+x_{7}=2x_{4}+x_{5}+4x_{7}$$
Finally, repeat for the first equation. Starting with
$$x_{1}=-3x_{2}-x_{3}-2x_{4}-3x_{5}-x_{6}-2x_{7}$$
substitute in for $x_{3}$ and $x_{6}$ to get
begin{align}
x_{1} &= -3x_{2}-(2x_{4}+x_{5}+4x_{7})-2x_{4}-3x_{5}-3(x_{7})-2x_{7} \
&= -3x_{2}-4x_{4}-4x_{5}-4x_{7}\
&equiv 2x_{2}+x_{4}+x_{5}+x_{7}
end{align}
Now we can conclude. The solution set is all vectors of the following form
begin{equation}
begin{pmatrix}
x_{1}\
x_{2}\
x_{3}\
x_{4}\
x_{5}\
x_{6}\
x_{7}\
end{pmatrix}
=
begin{pmatrix}
2x_{2}+x_{4}+x_{5}+x_{7}\
x_{2}\
2x_{4}+x_{5}+4x_{7}\
x_{4}\
x_{5}\
3x_{7}\
x_{7}\
end{pmatrix}=
x_{2}begin{pmatrix}2\1\0\0\0\0\0\end{pmatrix}
+x_{4}begin{pmatrix}1\0\2\1\0\0\0end{pmatrix}
+x_{5}begin{pmatrix}1\0\1\0\1\0\0end{pmatrix}
+x_{7}begin{pmatrix}1\0\1\0\0\3\1end{pmatrix}
end{equation}
Therefore a basis is given by
$$begin{pmatrix}2\1\0\0\0\0\0\end{pmatrix},begin{pmatrix}1\0\2\1\0\0\0end{pmatrix},begin{pmatrix}1\0\1\0\1\0\0end{pmatrix},begin{pmatrix}1\0\1\0\0\3\1end{pmatrix}$$
and the dimension is $4$.
answered Jan 16 at 19:25
pwerthpwerth
3,243417
$endgroup$
add a comment |
$begingroup$
I'll assume that your Gaussian elimination is correct and explain how to proceed with the matrix you gave.
You have pivot positions corresponding to $x_{1},x_{3},$ and $x_{6}$. That means your matrix has rank $3$ so its null space, which is what we're after, has dimension $7-3=4$. Let's solve for the pivot variables in terms of the free ones. From the last equation, $4x_{6}=-3x_{7}$ can be rewritten $-x_{6}=-3x_{7}$ since we are in $mathbb{Z}_{5}$. Now multiply both sides by $-1$ to get $x_{6}=3x_{7}$.
For the middle equation, we have
$$2x_{3}=-x_{4}-3x_{5}-3x_{6}-3x_{7}$$
Multiplying both sides by $3$, and simplifying/rewriting mod $5$ gives
$$x_{3}=2x_{4}+x_{5}+x_{6}+x_{7}$$
which, substituting in what we found previously, becomes
$$x_{3}=2x_{4}+x_{5}+3x_{7}+x_{7}=2x_{4}+x_{5}+4x_{7}$$
Finally, repeat for the first equation. Starting with
$$x_{1}=-3x_{2}-x_{3}-2x_{4}-3x_{5}-x_{6}-2x_{7}$$
substitute in for $x_{3}$ and $x_{6}$ to get
begin{align}
x_{1} &= -3x_{2}-(2x_{4}+x_{5}+4x_{7})-2x_{4}-3x_{5}-3(x_{7})-2x_{7} \
&= -3x_{2}-4x_{4}-4x_{5}-4x_{7}\
&equiv 2x_{2}+x_{4}+x_{5}+x_{7}
end{align}
Now we can conclude. The solution set is all vectors of the following form
begin{equation}
begin{pmatrix}
x_{1}\
x_{2}\
x_{3}\
x_{4}\
x_{5}\
x_{6}\
x_{7}\
end{pmatrix}
=
begin{pmatrix}
2x_{2}+x_{4}+x_{5}+x_{7}\
x_{2}\
2x_{4}+x_{5}+4x_{7}\
x_{4}\
x_{5}\
3x_{7}\
x_{7}\
end{pmatrix}=
x_{2}begin{pmatrix}2\1\0\0\0\0\0\end{pmatrix}
+x_{4}begin{pmatrix}1\0\2\1\0\0\0end{pmatrix}
+x_{5}begin{pmatrix}1\0\1\0\1\0\0end{pmatrix}
+x_{7}begin{pmatrix}1\0\1\0\0\3\1end{pmatrix}
end{equation}
Therefore a basis is given by
$$begin{pmatrix}2\1\0\0\0\0\0\end{pmatrix},begin{pmatrix}1\0\2\1\0\0\0end{pmatrix},begin{pmatrix}1\0\1\0\1\0\0end{pmatrix},begin{pmatrix}1\0\1\0\0\3\1end{pmatrix}$$
and the dimension is $4$.
answered Jan 16 at 19:25
pwerthpwerth
3,243417
$endgroup$
add a comment |
$begingroup$
I'll assume that your Gaussian elimination is correct and explain how to proceed with the matrix you gave.
You have pivot positions corresponding to $x_{1},x_{3},$ and $x_{6}$. That means your matrix has rank $3$ so its null space, which is what we're after, has dimension $7-3=4$. Let's solve for the pivot variables in terms of the free ones. From the last equation, $4x_{6}=-3x_{7}$ can be rewritten $-x_{6}=-3x_{7}$ since we are in $mathbb{Z}_{5}$. Now multiply both sides by $-1$ to get $x_{6}=3x_{7}$.
For the middle equation, we have
$$2x_{3}=-x_{4}-3x_{5}-3x_{6}-3x_{7}$$
Multiplying both sides by $3$, and simplifying/rewriting mod $5$ gives
$$x_{3}=2x_{4}+x_{5}+x_{6}+x_{7}$$
which, substituting in what we found previously, becomes
$$x_{3}=2x_{4}+x_{5}+3x_{7}+x_{7}=2x_{4}+x_{5}+4x_{7}$$
Finally, repeat for the first equation. Starting with
$$x_{1}=-3x_{2}-x_{3}-2x_{4}-3x_{5}-x_{6}-2x_{7}$$
substitute in for $x_{3}$ and $x_{6}$ to get
begin{align}
x_{1} &= -3x_{2}-(2x_{4}+x_{5}+4x_{7})-2x_{4}-3x_{5}-3(x_{7})-2x_{7} \
&= -3x_{2}-4x_{4}-4x_{5}-4x_{7}\
&equiv 2x_{2}+x_{4}+x_{5}+x_{7}
end{align}
Now we can conclude. The solution set is all vectors of the following form
begin{equation}
begin{pmatrix}
x_{1}\
x_{2}\
x_{3}\
x_{4}\
x_{5}\
x_{6}\
x_{7}\
end{pmatrix}
=
begin{pmatrix}
2x_{2}+x_{4}+x_{5}+x_{7}\
x_{2}\
2x_{4}+x_{5}+4x_{7}\
x_{4}\
x_{5}\
3x_{7}\
x_{7}\
end{pmatrix}=
x_{2}begin{pmatrix}2\1\0\0\0\0\0\end{pmatrix}
+x_{4}begin{pmatrix}1\0\2\1\0\0\0end{pmatrix}
+x_{5}begin{pmatrix}1\0\1\0\1\0\0end{pmatrix}
+x_{7}begin{pmatrix}1\0\1\0\0\3\1end{pmatrix}
end{equation}
Therefore a basis is given by
$$begin{pmatrix}2\1\0\0\0\0\0\end{pmatrix},begin{pmatrix}1\0\2\1\0\0\0end{pmatrix},begin{pmatrix}1\0\1\0\1\0\0end{pmatrix},begin{pmatrix}1\0\1\0\0\3\1end{pmatrix}$$
and the dimension is $4$.
answered Jan 16 at 19:25
pwerthpwerth
3,243417
$endgroup$
I'll assume that your Gaussian elimination is correct and explain how to proceed with the matrix you gave.
You have pivot positions corresponding to $x_{1},x_{3},$ and $x_{6}$. That means your matrix has rank $3$ so its null space, which is what we're after, has dimension $7-3=4$. Let's solve for the pivot variables in terms of the free ones. From the last equation, $4x_{6}=-3x_{7}$ can be rewritten $-x_{6}=-3x_{7}$ since we are in $mathbb{Z}_{5}$. Now multiply both sides by $-1$ to get $x_{6}=3x_{7}$.
For the middle equation, we have
$$2x_{3}=-x_{4}-3x_{5}-3x_{6}-3x_{7}$$
Multiplying both sides by $3$, and simplifying/rewriting mod $5$ gives
$$x_{3}=2x_{4}+x_{5}+x_{6}+x_{7}$$
which, substituting in what we found previously, becomes
$$x_{3}=2x_{4}+x_{5}+3x_{7}+x_{7}=2x_{4}+x_{5}+4x_{7}$$
Finally, repeat for the first equation. Starting with
$$x_{1}=-3x_{2}-x_{3}-2x_{4}-3x_{5}-x_{6}-2x_{7}$$
substitute in for $x_{3}$ and $x_{6}$ to get
begin{align}
x_{1} &= -3x_{2}-(2x_{4}+x_{5}+4x_{7})-2x_{4}-3x_{5}-3(x_{7})-2x_{7} \
&= -3x_{2}-4x_{4}-4x_{5}-4x_{7}\
&equiv 2x_{2}+x_{4}+x_{5}+x_{7}
end{align}
Now we can conclude. The solution set is all vectors of the following form
begin{equation}
begin{pmatrix}
x_{1}\
x_{2}\
x_{3}\
x_{4}\
x_{5}\
x_{6}\
x_{7}\
end{pmatrix}
=
begin{pmatrix}
2x_{2}+x_{4}+x_{5}+x_{7}\
x_{2}\
2x_{4}+x_{5}+4x_{7}\
x_{4}\
x_{5}\
3x_{7}\
x_{7}\
end{pmatrix}=
x_{2}begin{pmatrix}2\1\0\0\0\0\0\end{pmatrix}
+x_{4}begin{pmatrix}1\0\2\1\0\0\0end{pmatrix}
+x_{5}begin{pmatrix}1\0\1\0\1\0\0end{pmatrix}
+x_{7}begin{pmatrix}1\0\1\0\0\3\1end{pmatrix}
end{equation}
Therefore a basis is given by
$$begin{pmatrix}2\1\0\0\0\0\0\end{pmatrix},begin{pmatrix}1\0\2\1\0\0\0end{pmatrix},begin{pmatrix}1\0\1\0\1\0\0end{pmatrix},begin{pmatrix}1\0\1\0\0\3\1end{pmatrix}$$
and the dimension is $4$.
answered Jan 16 at 19:25
pwerthpwerth
3,243417
answered Jan 16 at 19:25
pwerthpwerth
3,243417
answered Jan 16 at 19:25
pwerthpwerth
3,243417
answered Jan 16 at 19:25
pwerthpwerth
3,243417
3,243417
add a comment |
add a comment |
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$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref.
$endgroup$
– amd
Jan 16 at 23:09