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How do you find the dimension of the subspace of R4 consisting of the vectors a plus 2b plus c b-2c 2a plus 2b plus c 3a plus 5b plus c?
The dimension of a space is defined as the number of vectors in its basis. Assuming your vectors are 1,2,1,0 0,1,-2,0 2,2,1,0 and 3,5,1,0 (extra zeros because you are in R4) then you must first check to see if they are linearly indepent. If all the vectors are linearly independent then the subspace defined by those vectors has a dimension 4, as there are 4 vectors in the basis.
What is coplanar vector?
In geometry a vector is used to make the equations easier to understand and to figure out. Velocity and force are examples of vectors. For a vector to be coplanar there must be two or more and they must be linearly dependent. Coplanar vectors have proportional components and their rank is 2.
What is the condition for being 3 vectors in a plane?
The general idea is that 3 vectors are in a plane iff they are not linearly independent. This can be checked in several ways:guessing a way to represent one of them as a linear combination of the other two - if it can be done, then they are coplanar;if they are three-dimensional, simply by calculating the determinant of the matrix whose columns are the vectors - if it's zero, they are coplanar, otherwise, they aren't;otherwise, you may calculate the determinant of their gramian matrix, that is, a matrix whose ij-th entry is the dot product if the i-th and j-th of the three vectors (e.g. it's 1-2-nd entry would be the dot product of first and second of them); they are coplanar iff the determinant is zero.
Is every bijection a strictly monotonic function?
No. For example, consider the discontinuous bijection that increases linearly from [0,0] to [1,1], decreases linearly from (1,2) to (2,1), increases linearly from [2,2] to [3,3], decreases linearly from (3,4) to (4,3), etc.
What is arthropod vectors?
Vectors of the arthropod.
Show that only N orthogonal vectors can be formed from N linearly independent vectors?
shut up now
If a set of vectors spans R3 then the set is linearly independent?
No it is not. It's possible to have to have a set of vectors that are linearly dependent but still Span R^3. Same holds true for reverse. Linear Independence does not guarantee Span R^3. IF both conditions are met then that set of vectors is called the Basis for R^3. So, for a set of vectors, S, to be a Basis it must be:(1) Linearly Independent(2) Span S = R^3.This means that both conditions are independent.
How do you find the dimension of the subspace of R4 consisting of the vectors a plus 2b plus c b-2c 2a plus 2b plus c 3a plus 5b plus c?
The dimension of a space is defined as the number of vectors in its basis. Assuming your vectors are 1,2,1,0 0,1,-2,0 2,2,1,0 and 3,5,1,0 (extra zeros because you are in R4) then you must first check to see if they are linearly indepent. If all the vectors are linearly independent then the subspace defined by those vectors has a dimension 4, as there are 4 vectors in the basis.
How can you tell if a matrix is invertible?
An easy exclusion criterion is a matrix that is not nxn. Only a square matrices are invertible (have an inverse). For the matrix to be invertible, the vectors (as columns) must be linearly independent. In other words, you have to check that for an nxn matrix given by {v1 v2 v3 ••• vn} with n vectors with n components, there are not constants (a, b, c, etc) not all zero such that av1 + bv2 + cv3 + ••• + kvn = 0 (meaning only the trivial solution of a=b=c=k=0 works).So all you're doing is making sure that the vectors of your matrix are linearly independent. The matrix is invertible if and only if the vectors are linearly independent. Making sure the only solution is the trivial case can be quite involved, and you don't want to do this for large matrices. Therefore, an alternative method is to just make sure the determinant is not 0. Remember that the vectors of a matrix "A" are linearly independent if and only if detA�0, and by the same token, a matrix "A" is invertible if and only if detA�0.
1 1 Check whether the following set of vectors is LD or LI?
(i) They are linearly dependent since the 2nd vector is twice the 1st vector. All 3 vectors lie in the x-z plane, so they don't span 3D space. (ii) They are linearly independent. Note that the cross-product of the first two is (-1,1,1). If the third vector is not perpendicular to the above cross-product, then the third vector does not lie in the plane defined by the first two vectors. (-1,1,1) "dot" (1,1,-1) = -1+1-1 = -1, not zero, so 3rd vector is not perpendicular to the cross product of the other two.
What is coplanar vector?
In geometry a vector is used to make the equations easier to understand and to figure out. Velocity and force are examples of vectors. For a vector to be coplanar there must be two or more and they must be linearly dependent. Coplanar vectors have proportional components and their rank is 2.
What is the rank of 44 matrix whoose first 2 rows are linearly independent and any 3 columns are dependent?
2
What is the condition for being 3 vectors in a plane?
The general idea is that 3 vectors are in a plane iff they are not linearly independent. This can be checked in several ways:guessing a way to represent one of them as a linear combination of the other two - if it can be done, then they are coplanar;if they are three-dimensional, simply by calculating the determinant of the matrix whose columns are the vectors - if it's zero, they are coplanar, otherwise, they aren't;otherwise, you may calculate the determinant of their gramian matrix, that is, a matrix whose ij-th entry is the dot product if the i-th and j-th of the three vectors (e.g. it's 1-2-nd entry would be the dot product of first and second of them); they are coplanar iff the determinant is zero.
What is the difference between a scalar quantity and a vector?
Scalars are quantities that have magnitude only; they are independent of direction. Vectors have both magnitude and direction. vectors need bold letters to show them.
What is an independent system of linear equations?
An independent system of linear equations is a set of vectors in Rm, where any other vector in Rm can be written as a linear combination of all of the vectors in the set. The vector equation and the matrix equation can only have the trivial solution (x=0).
Is every bijection a strictly monotonic function?
No. For example, consider the discontinuous bijection that increases linearly from [0,0] to [1,1], decreases linearly from (1,2) to (2,1), increases linearly from [2,2] to [3,3], decreases linearly from (3,4) to (4,3), etc.
Is linearly a word?
Yes.
