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andrey21
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Find bases for the kernal of the following:
T(x1,x2,x3,x4) = (x1+x2+x3,-x3,x1+x2)
Any help would be great thank you
T(x1,x2,x3,x4) = (x1+x2+x3,-x3,x1+x2)
Any help would be great thank you
Yes, good! So you haveandrey21 said:It is the set of vectors in the domain for which T(v) = 0 correct?
andrey21 said:Ah is it x1(000)??
andrey21 said:Ah brilliant with there being no basis for the kernal is tht like saying it has a dimension of 0??
Any vector in the kernel is of that form. Strictly speaking a 'basis" is a set of vectors: {(2, 1, 0), (0, 0, 1)}.andrey21 said:I know this is an old thread but I have a similar question I wish to ask.
Find the basis for the kernel of the following:
T3(x,y,z) = (x-2y,3x-6y)
so
x-2y=0 (1)
x=2y
3x-6y=0 (2)
Therefore (2) becomes:
3(2y)-6y=0
6y=6y
y=y
So the basis for kernel is:
y(2,1,0)+z(0,0,1)
is this correct??
The kernal, also known as the null space, is the set of all vectors that when multiplied by a given matrix, result in a zero vector.
Finding the bases for the kernal is important because it allows for a better understanding of the properties and behavior of a matrix. It can also help in solving systems of linear equations and determining the rank of a matrix.
To find the bases for the kernal, you need to reduce the given matrix to its reduced row echelon form using operations such as row operations and pivoting. The resulting columns with leading ones will form the basis for the kernal.
Yes, it is possible for the bases for the kernal to be empty. This means that the kernal is a trivial space, or the zero vector space. This occurs when the given matrix is invertible.
Finding the bases for the kernal is closely related to linear independence. The bases for the kernal are the vectors that can be combined to form the zero vector. If these vectors are linearly independent, then the kernal will have a non-zero dimension.