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- Thread starter LongApple
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The column space of ##B## is the set of all ##Bx##, where each ##x## is a ##3 \times 1## vector whose elements are the coefficients of a particular linear combination of the columns of ##B##.

If ##B = LU##, then ##Bx = LUx = L(Ux) = Ly##, so any linear combination of the columns of ##B## is also a linear combination of the columns of ##L##. This means that ##C(B)## (the column space of ##B##) is a subspace of ##C(L)## (the column space of ##L##).

In this case, ##C(B)## is a**proper** subspace of ##C(L)##, which we can see by arguing as follows. By inspection, the columns of ##L## are linearly independent, so ##L## is invertible. This means that the null space of ##B## is the same as the null space of ##U##, because invertibility of ##L## implies that ##LUx = 0## if and only if ##Ux = 0##. Since ##U## has two linearly independent columns, ##U## has rank 2, so the dimension of ##N(U) = N(B)## is 1. Therefore ##C(B)## has dimension 2, whereas ##C(L)## has dimension 3.

Moreover, by examining ##U## we can see that ##C(U)## consists exactly of vectors of the form ##(a,b,0)^T##, since the third row of ##U## is zero. This means that only the first two columns of ##L##, namely ##(1,2,-1)^T## and ##(0,1,0)^T## contribute to the column space of ##B##. Since these two columns are linearly independent and ##C(B)## has dimension 2, they form a basis for ##C(B)##.

If ##B = LU##, then ##Bx = LUx = L(Ux) = Ly##, so any linear combination of the columns of ##B## is also a linear combination of the columns of ##L##. This means that ##C(B)## (the column space of ##B##) is a subspace of ##C(L)## (the column space of ##L##).

In this case, ##C(B)## is a

Moreover, by examining ##U## we can see that ##C(U)## consists exactly of vectors of the form ##(a,b,0)^T##, since the third row of ##U## is zero. This means that only the first two columns of ##L##, namely ##(1,2,-1)^T## and ##(0,1,0)^T## contribute to the column space of ##B##. Since these two columns are linearly independent and ##C(B)## has dimension 2, they form a basis for ##C(B)##.

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