# Determining the dimension and a basis for a vector space

I have the following problem:

Let $W$ be a vector space of all solutions to these homogenous equations:

$$\begin{matrix} x &+& 2y &+& 2z &-& s &+& 3t &=& 0 \\ x &+& 2y &+& 3z &+& s &+& t &=& 0 \\ 3x &+& 6y &+& 8z &+& s &+& 5t &=& 0\end{matrix}$$

Find the dimension of the space $W$ and determine a basis for $W$.

I tried solving the above kernel to get the solutions.

The matrix:

$$\left(\begin{matrix} 1 & 2 & 2 & -1 & 3 \\ 1 & 2 & 3 & 1 & 1 \\ 3 & 6 & 8 & 1 & 5\end{matrix}\right)$$

When performing Gauss-Jordan on it, I get the matrix rank to be $3$:

$$\left(\begin{matrix} 1 & 0 & -1 & 0 & 0 \\ 0 & 5 & 2 & 0 & 0 \\ 4 & 10 & 0 & 0 & 0\end{matrix}\right)$$

So I get lost at this point. I don't know how to get the dimension nor how to determine a basis for it.

Can anyone point out the next thing I should do and whether I started off good?

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You have not finished Gaussian elimination yet. (In this case you do want to continue, since you will obtain another zero-column.) – TMM May 3 '12 at 19:09

Subtracting the first row $4$ times from the third, we get the matrix

$$\left(\begin{matrix} 1 & 0 & -1 & 0 & 0 \\ 0 & 5 & 2 & 0 & 0 \\ 0 & 10 & 4 & 0 & 0\end{matrix}\right)$$

Subtracting the second row $2$ times from the third, we get the matrix

$$\left(\begin{matrix} 1 & 0 & -1 & 0 & 0 \\ 0 & 5 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\end{matrix}\right).$$

This means the original set of equations is equivalent to

\begin{align} x - z &= 0 \\ 5y + 2z &= 0.\end{align}

So for any $\lambda, \mu, \rho$ we get $z = \lambda$, $s = \mu$, $r = \rho$, $x = \lambda$ and $y = \frac{-2}{5} \lambda$ as a solution. In particular:

• $(\lambda, \mu, \rho) = (1,0,0)$ gives the solution $(x,y,z,s,t) = (1, \frac{-2}{5}, 1, 0, 0)$.
• $(\lambda, \mu, \rho) = (0,1,0)$ gives the solution $(x,y,z,s,t) = (0,0,0,1,0)$.
• $(\lambda, \mu, \rho) = (0,0,1)$ gives the solution $(x,y,z,s,t) = (0,0,0,0,1)$.

This means that $W$ is spanned by the three vectors:

$$W = \left\langle \left(\begin{matrix} 1 \\ -2/5 \\ 1 \\ 0 \\ 0\end{matrix}\right), \left(\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \\ 0\end{matrix}\right), \left(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 1\end{matrix}\right)\right\rangle.$$

So there are three basis vectors. Can you then determine the dimension of $W$?

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The dimension is 3. – Tool May 3 '12 at 22:50

After clearing the bottom row, beneath the pivots $1$ and $5$, you'll have all zeros there. Thus you can take any $z,s,t$ and then $x$ and $y$ will be determined by $z$. Thus your basis is $\{e_1-\tfrac25e_2+e_3,e_4,e_5\}$.

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