Question

For the given sets of vectors, find the dimension of the space spanned by them and a basis for this space.

(a) $\left(1, -1, 0, 0 \right)\mathbf{,}\mathbf{ }\mathbf{ }\left(0, -2, 5, 1 \right)\mathbf{,}\mathbf{ }\mathbf{ }\left(1, -3, 5, 1 \right)\mathbf{,}\mathbf{ }\mathbf{ }\left(2, -4, 5, 1 \right)$

(b) $$\begin{gathered} \mathbf{(}\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{4}\mathbf{ }\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{5}\mathbf{,}\mathbf{ }\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{5}\mathbf{ }\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{5}\mathbf{,}\mathbf{ }\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{-}\mathbf{5}\mathbf{,}\mathbf{ }\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{3}\mathbf{ }\mathbf{)}\mathbf{,}\mathbf{ } \\ \mathbf{(}\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{0}\mathbf{ }\mathbf{)}\mathbf{ } \end{gathered}$$

(c)$\left(0, 10, -1, 1,10 \right)\mathbf{,}\mathbf{ }\mathbf{ }\left(2, -2, -4, 0, -3 \right)\mathbf{,}\mathbf{ }\mathbf{ }\left(4, 2, 0, 4, 5 \right)\mathbf{,}\mathbf{ }\left(3, 2, 0, 3, 4 \right)\mathbf{,}\left(5, -4, 5, 6, 2 \right)\mathbf{ }$

Short answer

(a) There are only two linearly independent vectors $\left(1, -1, 0, 0 \right), \left(0, -2, 5, 1 \right)$.The dimension of the space is 2.

(b) There are three linearly independent vectors$\left(1, 1, 3, 5, -3, 5 \right), \left(0, 1, 2, 0, 0, 4 \right), \left(0, 0, -1, 0, 3, 0 \right)$.The dimension of the space is 3.

(c)There are 4 linearly independent vectors $\left(1, 0, 0, 0,-3 \right), \left(0, 2, 0, 0, 1 \right), \left(0, 0, 1, 0, -1 \right), \left(0, 0, 0, 1, 4 \right)$.The dimension of the space is 4.

Step-by-step solution

(a)To find the dimension and the basis

The given vectors $\left(1, -1, 0, 0 \right), \left(0, -2, 5, 1 \right), \left(1, -3, 5, 1 \right), \left(2, -4, 5, 1 \right)$.

We will write these vectors in the form of matrix such that the rows of the matrix are the components of the vectors.

Therefore,

$\left[\begin{array}{cccc}1& -1& 0& 0\\ 0& -2& 5& 1\\ 1& -3& 5& 1\\ 2& -4& 5& 1\end{array}\right]$

.

Now, reducing this matrix by elementary row transformation, we get,

role="math" localid="1664257491039" $$\begin{gathered} R_3\to R_3-R_1, R_4\to R_4-2R_1 \\ \left[\begin{array}{cccc}1& -1& 0& 0\\ 0& -2& 5& 1\\ 0& -2& 5& 1\\ 0& -2& 5& 1\end{array}\right] \\ {R}_3\to R_3-R_2, R_4\to R_4-R_2 \\ \left[\begin{array}{cccc}1& -1& 0& 0\\ 0& -2& 5& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right] \end{gathered}$$

Here, there are only two linearly independent vectors$\left(1, -1, 0, 0 \right), \left(0, -2, 5, 1 \right)$.

Hence, the dimension of the space is 2.

(b)To find the dimension and basis

The given vectors are$\left(0, 1, 2, 0, 0, 4 \right), \left(1, 1, 3, 5, -3, 5 \right), \left(1, 0, 0, 5, 0, 1 \right), \left(-1, 1, 3, -5, -3, 3 \right), \left(0, 0, 1, 0, -3, 0 \right)$

We will write these vectors in the form of matrix such that the rows of the matrix are the components of the vectors.

Therefore,

role="math" localid="1664259316113" $$\begin{gathered} \left[\begin{array}{cccccc}0& 1& 2& 0& 0& 0\\ 1& 1& 3& 5& -3& 5\\ 1& 0& 0& 5& 0& 1\\ -1& 1& 3& -5& -3& 3\\ 0& 0& 1& 0& -3& 0\end{array}\right] \\ \begin{array}{}\\ \\ \\ \\ \end{array} \end{gathered}$$

Now, reducing this matrix by elementary row transformation, we get,

Interchanging the first and the second row, we get,

role="math" localid="1664259412796" $$\begin{gathered} \left[\begin{array}{cccccc}1& 1& 3& 5& -3& 5\\ 0& 1& 2& 0& 0& 0\\ 1& 0& 0& 5& 0& 1\\ -1& 1& 3& -5& -3& 3\\ 0& 0& 1& 0& -3& 0\end{array}\right] \\ {R}_3\to R_3-R_1, R_4\to R_4+R_1 \\ \left[\begin{array}{cccccc}1& 1& 3& 5& -3& 5\\ 0& 1& 2& 0& 0& 4\\ 0& -1& -3& 0& 3& -4\\ 0& 2& 6& 0& -6& 8\\ 0& 0& 1& 0& -3& 0\end{array}\right] \\ {R}_4\to R_4+2R_3, R_5\to R_5+R_3 \\ \left[\begin{array}{cccccc}1& 1& 3& 5& -3& 5\\ 0& 1& 2& 0& 0& 4\\ 0& 0& -1& 0& 3& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right] \end{gathered}$$

Here, there are three linearly independent vectors.

$\left(1, 1, 3, 5, -3, 5 \right), \left(0, 1, 2, 0, 0, 4 \right), \left(0, 0, -1, 0, 3, 0 \right)$

Hence, the dimension of the space is 3.

(c)To find the dimension and basis

The given vectors are $\left(0, 10, -1, 1,10 \right), \left(2, -2, -4, 0, -3 \right), \left(4, 2, 0, 4, 5 \right), \left(3, 2, 0, 3, 4 \right),\left(5, -4, 5, 6, 2 \right)$.

We will write these vectors in the form of matrix such that the rows of the matrix are the components of the vectors.

Therefore,

$\left[\begin{array}{ccccc}0& 10& -1& 1& 10\\ 2& -2& -4& 0& -3\\ 4& 2& 0& 4& 5\\ 3& 2& 0& 3& 4\\ 5& -4& 5& 6& 2\end{array}\right]$

Interchanging the first and the second row and dividing first row by 2, we get,

role="math" localid="1664261537157" $$\begin{gathered} \left[\begin{array}{ccccc}1& -1& -2& 0& \frac{-3}{2}\\ 0& 10& -1& 1& 10\\ 4& 2& 0& 4& 5\\ 3& 2& 0& 3& 4\\ 5& -4& 5& 6& 2\end{array}\right] \\ {R}_3\to R_3-4R_1, R_4\to R_4-3R_1, R_5\to R_5-5R_1 \\ \left[\begin{array}{ccccc}1& -1& -2& 0& \frac{-3}{2}\\ 0& 10& -1& 1& 10\\ 0& 6& 8& 4& 11\\ 0& 5& 6& 3& \frac{17}{2}\\ 0& 1& 15& 6& \frac{19}{2}\end{array}\right] \end{gathered}$$

Interchanging the second and the fifth row, we get,

$$\begin{gathered} \left[\begin{array}{ccccc}1& -1& -2& 0& \frac{-3}{2}\\ 0& 1& 15& 6& \frac{19}{2}\\ 0& 6& 8& 4& 11\\ 0& 5& 6& 3& \frac{17}{2}\\ 0& 10& -1& 1& 10\end{array}\right] \\ \left[\begin{array}{ccccc}1& 0& 13& 6& 8\\ 0& 1& 15& 6& \frac{19}{2}\\ 0& 0& -82& -32& -46\\ 0& 5& -69& -27& -39\\ 0& 10& -151& -59& -85\end{array}\right] \\ \left[\begin{array}{ccccc}1& 0& 0& \frac{38}{41}& \frac{29}{41}\\ 0& 1& 0& \frac{6}{41}& \frac{89}{41}\\ 0& 0& 1& \frac{16}{41}& \frac{23}{41}\\ 0& 0& 0& \frac{-3}{41}& \frac{-12}{41}\\ 0& 0& 1& \frac{-3}{41}& \frac{-12}{41}\end{array}\right] \\ \left[\begin{array}{ccccc}1& 0& 0& 0& -3\\ 0& 1& 0& 0& \frac{1}{2}\\ 0& 0& 1& 0& -1\\ 0& 0& 0& 1& 4\\ 0& 0& 0& 0& 0\end{array}\right] \end{gathered}$$

Here, there are 4 linearly independent vectors .

$\left(1, 0, 0, 0,-3 \right), \left(0, 2, 0, 0, 1 \right), \left(0, 0, 1, 0, -1 \right), \left(0, 0, 0, 1, 4 \right)$

Hence, the dimension of the space is 4.