Question

Find out whether the given vectors are dependent or independent; if they are dependent, find a linearly independent subset. Write each of the given vectors as a linear combination of the independent vectors. $$(0,1,1),(-1,5,3),(1,0,2),(2,-15,1)$$

Short answer

The vectors are linearly dependent. A linearly independent subset is $(0,1,1),(-1,5,3),(1,0,2)$. The dependent vector $(2,-15,1)$ can be written as $-2(0,1,1)-3(-1,5,3)+5(1,0,2)$.

Step-by-step solution

Arrange the Vectors Into a Matrix

Place the vectors $(0,1,1),(-1,5,3),(1,0,2),(2,-15,1)$ into a matrix form.$$A=\left(\begin{array}{cccccccccc}0& -1& 1& 21& 5& 0& -151& 3& 2& 1\end{array}\right)$$

Row Reduce the Matrix

Use Gaussian elimination to row reduce the matrix to its row echelon form to determine the rank of the matrix. This will help identify the linearly independent vectors.

Perform Row Operations

Perform the following row operations to reduce the matrix: $$\left(\begin{array}{cccccccccc}0& -1& 1& 21& 5& 0& -151& 3& 2& 1\end{array}\right):R1\leftrightarrow R2\to \left(\begin{array}{cccccccccc}1& 5& 0& -150& -1& 1& 21& 3& 2& 1\end{array}\right)$$ $$R3=R3-R1\to \left(\begin{array}{cccccccccc}1& 5& 0& -150& -1& 1& 20& -2& 2& 16\end{array}\right)$$ $$R3=R3+2R2\to \left(\begin{array}{cccccccccc}1& 5& 0& -150& -1& 1& 20& 0& 4& 20\end{array}\right)$$

Interpret the Reduced Matrix

Since the row-reduced form has three non-zero rows, the rank of the matrix is 3. This suggests three linearly independent vectors. To identify them, look at the positions of the leading 1s in the row echelon form.

Identify the Linearly Independent Vectors

The leading 1s are in columns 1, 2, and 3. Therefore, the vectors $(0,1,1),(-1,5,3),(1,0,2)$ are linearly independent.

Write Dependent Vector as a Linear Combination

Since the vector $(2,-15,1)$ is dependent on the others, express it as a linear combination of $(0,1,1),(-1,5,3),(1,0,2)$. Solve the equation $c1(0,1,1)+c2(-1,5,3)+c3(1,0,2)=(2,-15,1)$. This gives: $$\{\begin{array}{l}c1(0)+c2(-1)+c3(1)=2c1(1)+c2(5)+c3(0)=-15c1(1)+c2(3)+c3(2)=1\end{array}$$ Solve this system of equations to get $c1=-2,c2=-3,c3=5$. Thus, \ (2,-15,1) = -2(0,1,1) -3(-1,5,3) + 5(1,0,2) \.

Key concepts

Linear Independence

When we talk about linear independence, we refer to a set of vectors that do not lie along the same line or plane. More formally, a set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others.
In our exercise, we check for linear independence by arranging vectors into a matrix and using Gaussian elimination to row reduce the matrix. If the matrix has a row-reduced form with a rank equal to the number of vectors, the vectors are linearly independent.

Gaussian Elimination

Gaussian elimination is a process used to solve systems of linear equations, find the rank of a matrix, and determine linear independence of vectors. It involves performing row operations to transform the matrix into its row echelon form.
In the exercise, this helps to systematically find leading 1s in each row, which in turn reveals the rank of the matrix and identifies which vectors form a linearly independent set.
Row operations include:

• Swapping rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row to another row

Rank of a Matrix

The rank of a matrix is the maximum number of linearly independent row or column vectors within the matrix. Thus, it indicates the dimension of the vector space spanned by the row vectors or column vectors.
In the exercise, after performing Gaussian elimination, we observe the row-reduced form of the matrix and the number of non-zero rows corresponds to the rank, which is 3 in this case. This tells us that there are three linearly independent vectors within the given set.

Linear Combination

A linear combination involves expressing a vector as a sum of scalar multiples of other vectors. In the context of the exercise, once we have identified the linearly independent vectors, we can write any dependent vector as a linear combination of these independent vectors.
For instance, vector $(2,-15,1)$ can be written as a linear combination of the independent vectors $(0,1,1),(-1,5,3),(1,0,2)$. Solving the system of equations, we find that: $$(2,-15,1)=-2(0,1,1)-3(-1,5,3)+5(1,0,2)$$.
This confirms that $(2,-15,1)$ is dependent on the other three vectors.