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... hyper-cube has 2n vertices. The vertex coordinates are precisely the bits of the binary numbers 0 through 2n − 1. This is illustrated in Figure 1. The vertices can be organized as the columns of a matrix, where entry (i, j) contains the ith bit of the binary number j (ordered from least significant ...
... hyper-cube has 2n vertices. The vertex coordinates are precisely the bits of the binary numbers 0 through 2n − 1. This is illustrated in Figure 1. The vertices can be organized as the columns of a matrix, where entry (i, j) contains the ith bit of the binary number j (ordered from least significant ...
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... v1 ; : : :; vn 2 V are linearly independent if the only way to express 0 as a linear combination of the vi 's is with all coecients equal to 0; whenever c1v1 + + cn vn = 0, we have c1 = = cn = 0 Otherwise, we say the vectors are linearly dependent. I.e, some non-trivial linear combinati ...
... v1 ; : : :; vn 2 V are linearly independent if the only way to express 0 as a linear combination of the vi 's is with all coecients equal to 0; whenever c1v1 + + cn vn = 0, we have c1 = = cn = 0 Otherwise, we say the vectors are linearly dependent. I.e, some non-trivial linear combinati ...
SOME REMARKS ON BASES Let V denote a vector space. We say
... Now, why should we be interested in finding bases for R2 if we already have the standard basis, which seems to be the best kind of basis we could use? The reason is that the standard basis is not always the best basis to use. There is no best basis for any vector space. Sometimes we may wish to choo ...
... Now, why should we be interested in finding bases for R2 if we already have the standard basis, which seems to be the best kind of basis we could use? The reason is that the standard basis is not always the best basis to use. There is no best basis for any vector space. Sometimes we may wish to choo ...
Math 315: Linear Algebra Solutions to Assignment 5
... Since span(u, v) is a linear subspace of R3 , it should contain the zero vector, which means that in the plane equation ax + by + cz + d = 0 the coefficient d is 0. Therefore, the plain is of the form ax + by + cz = 0. We know that each of the given vectors is in the plain, so it satisfies the plain ...
... Since span(u, v) is a linear subspace of R3 , it should contain the zero vector, which means that in the plane equation ax + by + cz + d = 0 the coefficient d is 0. Therefore, the plain is of the form ax + by + cz = 0. We know that each of the given vectors is in the plain, so it satisfies the plain ...
