031 Review Part 2, Problem 2

Find the dimension of the subspace spanned by the given vectors. Are these vectors linearly independent?

${\begin{bmatrix}1\\0\\2\end{bmatrix}},{\begin{bmatrix}3\\1\\1\end{bmatrix}},{\begin{bmatrix}-2\\-1\\1\end{bmatrix}},{\begin{bmatrix}5\\2\\2\end{bmatrix}}$

Foundations:
1.  ${\text{dim Col }}A$  is the number of pivots in  $A.$
2. A set of vectors  $\{{\vec {v_{1}}},{\vec {v_{2}}},\ldots ,{\vec {v_{n}}}\}$  is linearly independent if
the only solution to  $x_{1}{\vec {v_{1}}}+x_{2}{\vec {v_{2}}}+\cdots +x_{n}{\vec {v_{n}}}={\vec {0}}$  is the trivial solution.

Solution:

Step 1:
We begin by putting these vectors together in a matrix. So, we have
${\begin{bmatrix}1&3&-2&5\\0&1&-1&2\\2&1&1&2\end{bmatrix}}.$
Now, we row reduce this matrix. We get
${\begin{array}{rcl}\displaystyle {\left[{\begin{array}{cccc}1&3&-2&5\\0&1&-1&2\\2&1&1&2\end{array}}\right]}&\sim &\displaystyle {\left[{\begin{array}{cccc}1&3&-2&5\\0&1&-1&2\\0&-5&5&-8\end{array}}\right]}\\&&\\&\sim &\displaystyle {\left[{\begin{array}{cccc}1&3&-2&5\\0&1&-1&2\\0&0&0&2\end{array}}\right]}\end{array}}$

Step 2:
Now, we have 3 pivots in this matrix. So, the dimension of the column space of the matrix we started with is 3.
Hence, the dimension of the subspace spanned by these vectors is  $3.$
When we row reduced the matrix, we had a column that did not contain a pivot.
This means we have a free variable in the system corresponding to  $Ax=0.$
So, these vectors are not linearly independent.

Final Answer:
The dimension is  $3$  and the vectors are not linearly independent.

Return to Review Problems