Constructing a basis for the row space and column space and their orthogonal complements The 2019 Stack Overflow Developer Survey Results Are InColumn Vectors orthogonal implies Row Vectors also orthogonal?Understanding how to find a basis for the row space/column space of some matrix A.Finding the basis, difference between row space and column spaceUnderstanding the significance of row space and column space basisVisulizing column/row space and null/left null space, A and xSVD, the connection between the column space and the row space?What is the difference between orthogonal subspaces and orthogonal complements?Basis and dimension of row/column spaceWhy is the basis for the column space of a matrix $A$ merely the columns that which have pivots in $rref(A)$?Basis for the row space, column space and null space of a matrix
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Constructing a basis for the row space and column space and their orthogonal complements
The 2019 Stack Overflow Developer Survey Results Are InColumn Vectors orthogonal implies Row Vectors also orthogonal?Understanding how to find a basis for the row space/column space of some matrix A.Finding the basis, difference between row space and column spaceUnderstanding the significance of row space and column space basisVisulizing column/row space and null/left null space, A and xSVD, the connection between the column space and the row space?What is the difference between orthogonal subspaces and orthogonal complements?Basis and dimension of row/column spaceWhy is the basis for the column space of a matrix $A$ merely the columns that which have pivots in $rref(A)$?Basis for the row space, column space and null space of a matrix
$begingroup$
I just want to make sure my way of solving this right.
(a)The row space is the number of non - zero rows so the row space here is the first two rows.
(b)The orthogonal complement is the null space of the matrix and vector I got is (-5,-1,3,1).
(c)The column space is the pivot columns so it is only the 1st and 3rd columns.
(d) The column space is the null space of A transpose and the vector I have is, however it seems that only the zero vector works here , would that still be valid as the orthogonal complement?
linear-algebra
asked Mar 30 at 21:42
Samurai BaleSamurai Bale
746
$endgroup$
add a comment |
$begingroup$
I just want to make sure my way of solving this right.
(a)The row space is the number of non - zero rows so the row space here is the first two rows.
(b)The orthogonal complement is the null space of the matrix and vector I got is (-5,-1,3,1).
(c)The column space is the pivot columns so it is only the 1st and 3rd columns.
(d) The column space is the null space of A transpose and the vector I have is, however it seems that only the zero vector works here , would that still be valid as the orthogonal complement?
linear-algebra
asked Mar 30 at 21:42
Samurai BaleSamurai Bale
746
$endgroup$
1
$begingroup$
Your terminology confuses spaces, their dimension, sets of vectors and bases of vector spaces. For instance, the row space is not “the first two rows.” It is a vector space spanned by those rows. They happen to be linearly independent, so they are also a basis for this space.
$endgroup$
– amd
Mar 30 at 21:58
add a comment |
$begingroup$
I just want to make sure my way of solving this right.
(a)The row space is the number of non - zero rows so the row space here is the first two rows.
(b)The orthogonal complement is the null space of the matrix and vector I got is (-5,-1,3,1).
(c)The column space is the pivot columns so it is only the 1st and 3rd columns.
(d) The column space is the null space of A transpose and the vector I have is, however it seems that only the zero vector works here , would that still be valid as the orthogonal complement?
linear-algebra
asked Mar 30 at 21:42
Samurai BaleSamurai Bale
746
$endgroup$
I just want to make sure my way of solving this right.
(a)The row space is the number of non - zero rows so the row space here is the first two rows.
(b)The orthogonal complement is the null space of the matrix and vector I got is (-5,-1,3,1).
(c)The column space is the pivot columns so it is only the 1st and 3rd columns.
(d) The column space is the null space of A transpose and the vector I have is, however it seems that only the zero vector works here , would that still be valid as the orthogonal complement?
linear-algebra
linear-algebra
asked Mar 30 at 21:42
Samurai BaleSamurai Bale
746
asked Mar 30 at 21:42
Samurai BaleSamurai Bale
746
asked Mar 30 at 21:42
Samurai BaleSamurai Bale
746
asked Mar 30 at 21:42
Samurai BaleSamurai Bale
746
asked Mar 30 at 21:42
Samurai BaleSamurai Bale
746
746
1
$begingroup$
Your terminology confuses spaces, their dimension, sets of vectors and bases of vector spaces. For instance, the row space is not “the first two rows.” It is a vector space spanned by those rows. They happen to be linearly independent, so they are also a basis for this space.
$endgroup$
– amd
Mar 30 at 21:58
add a comment |
1
$begingroup$
Your terminology confuses spaces, their dimension, sets of vectors and bases of vector spaces. For instance, the row space is not “the first two rows.” It is a vector space spanned by those rows. They happen to be linearly independent, so they are also a basis for this space.
$endgroup$
– amd
Mar 30 at 21:58
1
1
$begingroup$
Your terminology confuses spaces, their dimension, sets of vectors and bases of vector spaces. For instance, the row space is not “the first two rows.” It is a vector space spanned by those rows. They happen to be linearly independent, so they are also a basis for this space.
$endgroup$
– amd
Mar 30 at 21:58
$begingroup$
Your terminology confuses spaces, their dimension, sets of vectors and bases of vector spaces. For instance, the row space is not “the first two rows.” It is a vector space spanned by those rows. They happen to be linearly independent, so they are also a basis for this space.
$endgroup$
– amd
Mar 30 at 21:58
add a comment |
0
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$begingroup$
Your terminology confuses spaces, their dimension, sets of vectors and bases of vector spaces. For instance, the row space is not “the first two rows.” It is a vector space spanned by those rows. They happen to be linearly independent, so they are also a basis for this space.
$endgroup$
– amd
Mar 30 at 21:58