$V_1 = (x_1, ..., x_7)^T \in \mathbb{Z}^7_{5} : x_1 + 3x_2 +x_3+2x_4+3x_5+x_6+2x_7 = 0$

$3x_1 + 4x_2 +3x_3+x_4+4x_5+2x_6+4x_7 = 0$

$2x_1 + x_2 +4x_3+4x_5+x_6+2x_7 = 0$

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 $$\left(\begin{matrix} 1 & 3 & 1 & 2 & 3 & 1 & 2\\ 0 & 0 & 2 & 1 & 3 & 3& 3\\ 0 & 0 & 0 & 0 & 0 & 4 & 3\end{matrix}\right)$$


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\\0\end{pmatrix} +x_{5}\begin{pmatrix}1\\0\\1\\0\\1\\0\\0\end{pmatrix} +x_{7}\begin{pmatrix}1\\0\\1\\0\\0\\3\\1\end{pmatrix} \end{equation}

Therefore a basis is given by $$\begin{pmatrix}2\\1\\0\\0\\0\\0\\0\\\end{pmatrix},\begin{pmatrix}1\\0\\2\\1\\0\\0\\0\end{pmatrix},\begin{pmatrix}1\\0\\1\\0\\1\\0\\0\end{pmatrix},\begin{pmatrix}1\\0\\1\\0\\0\\3\\1\end{pmatrix}$$ and the dimension is $4$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.