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The span of Vectors Calculator + Online Solver With Free Steps

A Span of Vectors Calculator is a simple online tool that computes the set of all linear combinations of two vectors or more.

By employing this calculator, you can consistently show the distribution of a vector function.

What Is a Span of Vectors Calculator?

The Span of Vectors Calculator is a calculator that returns a list of all linear vector combinations. For instance, if $ v_1 = [11,5,-7,0]^T $  and $ v_1 = [2,13,0,-7]^T $, the set of all vectors of the form $ s \cdot v^1+t \cdot v^2 $ for certain scalars ‘s’ and ‘t’ is the span of v1 and v2.

A subspace of $ \mathbb{R}^n $ is given by the span of a set of vectors in that space. Any non-trivial subdomain can be expressed as the span of any one of an infinite number of vector set combinations.

If λi = 0 exists as the only solution to the vector expression {λ1.V1 +…..+ λm.Vm}, a collection of vectors {V1, . . . , Vm} are linearly independent. It is only linearly dependent when a series of vectors are not linearly independent.

The rows of A, for instance, are not linearly independent because

\[  -\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} + \begin{bmatrix} -2 \\ 3 \\ -1 \\ 5 \end{bmatrix} + \begin{bmatrix} 3 \\ -1 \\ 4 \\ -1 \end{bmatrix} = 0 \]

To check if a group of vectors is linearly independent or not, represent them as columns of a matrix C and compute Cx=0.
The vectors are linearly dependent if there are any nontrivial solutions; else, these vectors are linearly independent.

How To Use a Span of Vectors Calculator?

You can use the calculator by carefully following the step-by-step instructions; you can use the Span of Vectors Calculator to get the desired results. Therefore, you can adhere to the guidelines to obtain the desired result.

Step 1

Enter the values of Vector 1 and Vector 2 in the provided entry boxes.

Step 2

Press the “Span Me” button to calculate the Span of Vectors for the given vectors and to view the detailed, step-by-step solution for the Span of Vector Calculation.

How Does a Span of Vectors Calculator Work?

The Span of Vectors Calculator works by determining all the possible linear combinations of multiple vectors. 

If S is linearly dependent, given a group of vectors S ={v1, v2, … , vn}, then zero is a significant linear combination of vectors in S. That is, if  and only if $ c_1 \cdot v_1 + c_2 \cdot v_2 +… + c_n \cdot v_n = 0 $, then there are constants { c1,…, cn } with at least one of the constants nonzero.

If S is assumed to be linearly dependent, then:

     \[ v_i  =  c_1 \cdot v_1 + c_2 \cdot v_2 + … + c_{i-1} \cdot v_{i-1} + c_{i+1} \cdot v_{i+1} + … + c_n \cdot v_n  \]

$ V_i $ is subtracted from both sides to give us:

         \[ c_1 \cdot v_1 + c_2 \cdot v_2 + … + c_{i-1} \cdot v_{i-1} + c_{i+1} \cdot v_{i+1} + … + c_n \cdot v_n   =  0  \]

The nonzero value of ci in the equation above causes 0 to be a nontrivial linear combination of vectors in S.

Let’s now consider:

      \[   c_1 \cdot v_1 + c_2 \cdot v_2 +  … + c_{i-1} \cdot v_{i-1} +  c_i \cdot v_i  + c_{i+1} \cdot v_{i+1} + … + c_n \cdot v_n   =  0  \]

With nonzero ci. Let $ a_j = – \frac{c_j}{c_i} $ be the result from multiplying both sides of the equation by ci:

       \[  -a_1 v_1 – a_2 v_2 – … – a_{i-1} v_{i-1} + v_i- a_{i+1} v_{i+1}  – … – a_n v_n =  0  \]

Lastly, reposition each term to the right side of the equation:

     \[   vi  =   a_1 v_1 + a_2 v_2 + … + a_{i-1} v_{i-1} + v_i + a_{i+1} v_{i+1}  + … + a_n v_n   \]

The line across the origin determined by x1 is the span of a single nonzero vector x1 in R3 (or R2).

The collection of all x1’s potential linear combinations, or all x1’s of the type 1×1, where $ \alpha \cdot 1 \in \mathbb{R} $, is known as spam. The line across the origin given by x1 is called the span of x1, which is the set of all multiples of x1.

List of Some Linear Combinations

Here are some examples of vector combinations:

Vector 0: span(0) equals 0.

span(v) = 1 vector, which is a line.

If two vectors v1 and v2 are not collinear, then span(v1, v2) = $ \mathbb{R}^2 $.

span(v1, v2, v3…) = $ \mathbb{R}^2 $ for three or more vectors. All vectors, excluding two, are redundant.

Solved Examples

Let’s explore some examples better to understand the working of the Vector Function Grapher Calculator.

Example 1

Demonstrate that the set 

   S =  {(1, 1, 0), (0, 1, 1), (1, 1, 2)} 

spans $ \mathbb{R}^3 $ and represents the vector (2,4,8) as a linear combination of vectors in S.

Solution

A vector in $ \mathbb{R}^3 $ has the following form:

        v  =  (x, y, z)

Therefore, we must demonstrate that every such v may be expressed as:

    \[ (x,y,z)  =  c_1(1, 1, 0) + c_2(0, 1, 1) + c_3(1, 1, 2)  \]

    \[ (x,y,z) =  (c_2 + c_3, c_1 + c_3, c_1 + c_2) \]

This is compatible with the set of equations:

    \[ c_2 + c_3  =  x \]

    \[ c_1 +  c_3  =  y \]

    \[ c_1 + c_2   =  z \]

It can be expressed as a matrix:

\[  \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\  1 & 1 & 2  \end{bmatrix}  \begin{bmatrix} c_1 \\ c_2 \\ c_3  \end{bmatrix} = \begin{bmatrix} x \\ y \\ z  \end{bmatrix} \]

We can write this as: 

 A . c  =  b

Notice that: 

   det(A)  =  2

Hence A is nonsingular and:

   c  =  A$^{-1}$ . b 

So, there is a nontrivial solution. We discover that (2,4,8) can be written as a linear combination of vectors in S.

\[  A^{-1} = \begin{bmatrix} .5 & -1 & .5 \\  .5 & 1 & -.5 \\ -.5 & 0 & -.5 \end{bmatrix}  \] 

Then:

\[ c  = \begin{bmatrix} .5 & -1 & .5 \\  .5 & 1 & -.5 \\ -.5 & 0 & -.5 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \\ 8  \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 3  \end{bmatrix}\] 

We have:

(2,4,8)  =  1(0,1,1) + 1(1,0,1) + 3(1,1,0)

Example 2

Show that S does not span $ P_2 $ if $ v_1 = t + 2 $ and $ v_2 = t_2 + 1 $ and $ S  =  {v_1, v_2} $.

Solution

A general element of $ P_2 $ is of the form 

       \[  v   =  at^2  + bt + c \]

We set 

        \[ v  =  c_1 v_11 + c_2 v_2 \]

or

     \[ at^2  + bt + c  =  c_2 (t + 2) + c_2 (t^2 + 1)  =  c_2 \cdot t^2 + c_1 \cdot t + c_1 + c_2 \]

Equating coefficients gives

a  =  c1

b  =  c1

c  =  c1 + c2 

Notice that if:

     a  =  1    b  =  1    c  =  1

There is no solution to this.  Hence ‘S’ does not span ‘V.’

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