Finding basis and dimension
$begingroup$
Let $W= operatorname{span}{(1,-2,5,-3), (2,3,1,-4), (3,8,-3,-5)}$. In order to find the linearly independent vectors, I wrote the vectors column wise and reduced the matrix to Echelon form and found that rank is 2. Since leading elements were found only in columns 1 and 2, I concluded that only the first two vectors are linearly independent and hence these two vectors form basis and therefore dim(W)= 2. Am I right?
What if I had written the vectors row wise? Will it still be the same?
linear-algebra
edited Dec 2 '18 at 11:46
gimusi
1
asked Dec 2 '18 at 11:42
user46697user46697
191111
$endgroup$
add a comment |
$begingroup$
Let $W= operatorname{span}{(1,-2,5,-3), (2,3,1,-4), (3,8,-3,-5)}$. In order to find the linearly independent vectors, I wrote the vectors column wise and reduced the matrix to Echelon form and found that rank is 2. Since leading elements were found only in columns 1 and 2, I concluded that only the first two vectors are linearly independent and hence these two vectors form basis and therefore dim(W)= 2. Am I right?
What if I had written the vectors row wise? Will it still be the same?
linear-algebra
edited Dec 2 '18 at 11:46
gimusi
1
asked Dec 2 '18 at 11:42
user46697user46697
191111
$endgroup$
add a comment |
$begingroup$
Let $W= operatorname{span}{(1,-2,5,-3), (2,3,1,-4), (3,8,-3,-5)}$. In order to find the linearly independent vectors, I wrote the vectors column wise and reduced the matrix to Echelon form and found that rank is 2. Since leading elements were found only in columns 1 and 2, I concluded that only the first two vectors are linearly independent and hence these two vectors form basis and therefore dim(W)= 2. Am I right?
What if I had written the vectors row wise? Will it still be the same?
linear-algebra
edited Dec 2 '18 at 11:46
gimusi
1
asked Dec 2 '18 at 11:42
user46697user46697
191111
$endgroup$
Let $W= operatorname{span}{(1,-2,5,-3), (2,3,1,-4), (3,8,-3,-5)}$. In order to find the linearly independent vectors, I wrote the vectors column wise and reduced the matrix to Echelon form and found that rank is 2. Since leading elements were found only in columns 1 and 2, I concluded that only the first two vectors are linearly independent and hence these two vectors form basis and therefore dim(W)= 2. Am I right?
What if I had written the vectors row wise? Will it still be the same?
linear-algebra
linear-algebra
edited Dec 2 '18 at 11:46
gimusi
1
asked Dec 2 '18 at 11:42
user46697user46697
191111
edited Dec 2 '18 at 11:46
gimusi
1
asked Dec 2 '18 at 11:42
user46697user46697
191111
edited Dec 2 '18 at 11:46
gimusi
1
edited Dec 2 '18 at 11:46
gimusi
1
edited Dec 2 '18 at 11:46
gimusi
1
1
asked Dec 2 '18 at 11:42
user46697user46697
191111
asked Dec 2 '18 at 11:42
user46697user46697
191111
asked Dec 2 '18 at 11:42
user46697user46697
191111
191111
add a comment |
add a comment |
1 Answer
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Yes your procedure is fine and you can also work by rows. Since row operations preserve the row space, the advantage in that latter case is that at the end also the pivot row vectors in the RREF are basis vectors for the subspace spanned by the two basis vectors whereas by working by columns we need to refer to the original vectors.
answered Dec 2 '18 at 11:46
gimusigimusi
1
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1 Answer
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1 Answer
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$begingroup$
Yes your procedure is fine and you can also work by rows. Since row operations preserve the row space, the advantage in that latter case is that at the end also the pivot row vectors in the RREF are basis vectors for the subspace spanned by the two basis vectors whereas by working by columns we need to refer to the original vectors.
answered Dec 2 '18 at 11:46
gimusigimusi
1
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add a comment |
$begingroup$
Yes your procedure is fine and you can also work by rows. Since row operations preserve the row space, the advantage in that latter case is that at the end also the pivot row vectors in the RREF are basis vectors for the subspace spanned by the two basis vectors whereas by working by columns we need to refer to the original vectors.
answered Dec 2 '18 at 11:46
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
Yes your procedure is fine and you can also work by rows. Since row operations preserve the row space, the advantage in that latter case is that at the end also the pivot row vectors in the RREF are basis vectors for the subspace spanned by the two basis vectors whereas by working by columns we need to refer to the original vectors.
answered Dec 2 '18 at 11:46
gimusigimusi
1
$endgroup$
Yes your procedure is fine and you can also work by rows. Since row operations preserve the row space, the advantage in that latter case is that at the end also the pivot row vectors in the RREF are basis vectors for the subspace spanned by the two basis vectors whereas by working by columns we need to refer to the original vectors.
answered Dec 2 '18 at 11:46
gimusigimusi
1
answered Dec 2 '18 at 11:46
gimusigimusi
1
answered Dec 2 '18 at 11:46
gimusigimusi
1
answered Dec 2 '18 at 11:46
gimusigimusi
1
1
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