Mixing
Could the multiplication of matrix X (with dimensions [d+1 x N]) and its transpose simplify to a matrix with...
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In a machine learning course I'm taking, one of the lectures deals with matrix multiplication.
Could anyone explain why the dot product of these two matrices would "shrink" to [d+1 x d+1] dimensions instead of being [N x N] dimensions? Thank you.
matrices products
asked Sep 11 '14 at 18:01
LucasLucas
64
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add a comment |
$begingroup$
In a machine learning course I'm taking, one of the lectures deals with matrix multiplication.
Could anyone explain why the dot product of these two matrices would "shrink" to [d+1 x d+1] dimensions instead of being [N x N] dimensions? Thank you.
matrices products
asked Sep 11 '14 at 18:01
LucasLucas
64
$endgroup$
add a comment |
$begingroup$
In a machine learning course I'm taking, one of the lectures deals with matrix multiplication.
Could anyone explain why the dot product of these two matrices would "shrink" to [d+1 x d+1] dimensions instead of being [N x N] dimensions? Thank you.
matrices products
asked Sep 11 '14 at 18:01
LucasLucas
64
$endgroup$
In a machine learning course I'm taking, one of the lectures deals with matrix multiplication.
Could anyone explain why the dot product of these two matrices would "shrink" to [d+1 x d+1] dimensions instead of being [N x N] dimensions? Thank you.
matrices products
matrices products
asked Sep 11 '14 at 18:01
LucasLucas
64
asked Sep 11 '14 at 18:01
LucasLucas
64
asked Sep 11 '14 at 18:01
LucasLucas
64
asked Sep 11 '14 at 18:01
LucasLucas
64
asked Sep 11 '14 at 18:01
LucasLucas
64
64
add a comment |
add a comment |
1 Answer
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When you perform matrix multiplication, order matters. Here, when you multiply these two matrices, the final matrix will have rows corresponding to the first matrix (so in your case, $d + 1$) and columns corresponding to the second matrix $(d + 1)$. Now, if it were instead $(mathbf{X}^T)(mathbf{X})$, then the opposite would be true: since $mathbf{X}^T$ has $N$ rows and $mathbf{X}$ has $N$ columns, the matrix multiplication would lead to a matrix with $N times N$ dimensions.
edited Feb 2 at 19:54
dantopa
6,68442245
answered Feb 2 at 19:21
Mark TorresMark Torres
1
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1 Answer
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$begingroup$
When you perform matrix multiplication, order matters. Here, when you multiply these two matrices, the final matrix will have rows corresponding to the first matrix (so in your case, $d + 1$) and columns corresponding to the second matrix $(d + 1)$. Now, if it were instead $(mathbf{X}^T)(mathbf{X})$, then the opposite would be true: since $mathbf{X}^T$ has $N$ rows and $mathbf{X}$ has $N$ columns, the matrix multiplication would lead to a matrix with $N times N$ dimensions.
edited Feb 2 at 19:54
dantopa
6,68442245
answered Feb 2 at 19:21
Mark TorresMark Torres
1
$endgroup$
add a comment |
$begingroup$
When you perform matrix multiplication, order matters. Here, when you multiply these two matrices, the final matrix will have rows corresponding to the first matrix (so in your case, $d + 1$) and columns corresponding to the second matrix $(d + 1)$. Now, if it were instead $(mathbf{X}^T)(mathbf{X})$, then the opposite would be true: since $mathbf{X}^T$ has $N$ rows and $mathbf{X}$ has $N$ columns, the matrix multiplication would lead to a matrix with $N times N$ dimensions.
edited Feb 2 at 19:54
dantopa
6,68442245
answered Feb 2 at 19:21
Mark TorresMark Torres
1
$endgroup$
add a comment |
$begingroup$
When you perform matrix multiplication, order matters. Here, when you multiply these two matrices, the final matrix will have rows corresponding to the first matrix (so in your case, $d + 1$) and columns corresponding to the second matrix $(d + 1)$. Now, if it were instead $(mathbf{X}^T)(mathbf{X})$, then the opposite would be true: since $mathbf{X}^T$ has $N$ rows and $mathbf{X}$ has $N$ columns, the matrix multiplication would lead to a matrix with $N times N$ dimensions.
edited Feb 2 at 19:54
dantopa
6,68442245
answered Feb 2 at 19:21
Mark TorresMark Torres
1
$endgroup$
When you perform matrix multiplication, order matters. Here, when you multiply these two matrices, the final matrix will have rows corresponding to the first matrix (so in your case, $d + 1$) and columns corresponding to the second matrix $(d + 1)$. Now, if it were instead $(mathbf{X}^T)(mathbf{X})$, then the opposite would be true: since $mathbf{X}^T$ has $N$ rows and $mathbf{X}$ has $N$ columns, the matrix multiplication would lead to a matrix with $N times N$ dimensions.
edited Feb 2 at 19:54
dantopa
6,68442245
answered Feb 2 at 19:21
Mark TorresMark Torres
1
edited Feb 2 at 19:54
dantopa
6,68442245
edited Feb 2 at 19:54
dantopa
6,68442245
edited Feb 2 at 19:54
dantopa
6,68442245
6,68442245
answered Feb 2 at 19:21
Mark TorresMark Torres
1
answered Feb 2 at 19:21
Mark TorresMark Torres
1
answered Feb 2 at 19:21
Mark TorresMark Torres
1
1
add a comment |
add a comment |
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