Do you have to have three vectors to span R3?
![Do you have to have three vectors to span R3?](https://i.ytimg.com/vi/9R6zQfd6vlQ/hq720.jpg?sqp=-oaymwEcCOgCEMoBSFXyq4qpAw4IARUAAIhCGAFwAcABBg==&rs=AOn4CLAKPQdcghUJ6TJg8gFOA47LYpVvbA)
Do you have to have three vectors to span R3?
So vectors are linearly independent and form a basis for R 3. and span it. Also, it is not necessary that exactly three vectors span R 3. It might be 2 or 4 or any number of vectors.
Do you span R3 if there is always a solution?
If there is always a solution, then the vectors span R3; if there is a choice of a, b, c for which the system is inconsistent, then the vectors do not span R3. You can use the same set of elementary row operations I used in 1, with the augmented matrix leaving the last column indicated as expressions of a, b, and c.
Can a set of three vectors span a field?
Besides, any set of three vectors v 1, v 2, v 3 can span a vectors space over a field, provided they are linearly independent or the matrix of those vectors can be reduced to Echelon form or reduced Echelon form. For example, consider a set of vectors v 1 = ( 1, 2, − 1), v 2 = ( 0, 3, 1) v 3 = ( 1, − 5, 3). Which is in Echelon form.
Can you generate a vector space with three vectors?
Of course three vectors can generate a vector space over a certain field. One example is the standard basis for R 3 that comprise of e 1 = ( 1, 0, 0), e 2 = ( 0, 1, 0) e 3 = ( 0, 0, 1).