Determine whether a matrix is separable
I wonder whether there is a simple way(e.g. if asked in exam) to determine if a matrix is separable. If yes, how can we write the separated parts? For example, Gaussian Blur kernel is separable.
$$frac1{256}cdotbegin{bmatrix}
1&4&6&4&1\
4&16&24&16&4\
6&24&36&24&6\
4&16&24&16&4\
1&4&6&4&1
end{bmatrix}
=
frac1{256}cdotbegin{bmatrix}
1\4\6\4\1
end{bmatrix}cdotbegin{bmatrix}
1&4&6&4&1
end{bmatrix}
$$
I have two matrices asked in old-exams. How to do my wish on the following matrices?
$$A = frac1{4}cdotbegin{bmatrix}
-1&-2&-1\
-2&16&-2\
-1&-2&-1\
end{bmatrix}
$$
$$B = begin{bmatrix}
-1&-3&-1\
0&0&0\
1&3&1\
end{bmatrix}
$$
linear-algebra matrices matrix-equations
edited Nov 18 at 16:24
Ethan Bolker
41.3k547108
asked Nov 18 at 16:19
itsnotmyrealname
15110
add a comment |
I wonder whether there is a simple way(e.g. if asked in exam) to determine if a matrix is separable. If yes, how can we write the separated parts? For example, Gaussian Blur kernel is separable.
$$frac1{256}cdotbegin{bmatrix}
1&4&6&4&1\
4&16&24&16&4\
6&24&36&24&6\
4&16&24&16&4\
1&4&6&4&1
end{bmatrix}
=
frac1{256}cdotbegin{bmatrix}
1\4\6\4\1
end{bmatrix}cdotbegin{bmatrix}
1&4&6&4&1
end{bmatrix}
$$
I have two matrices asked in old-exams. How to do my wish on the following matrices?
$$A = frac1{4}cdotbegin{bmatrix}
-1&-2&-1\
-2&16&-2\
-1&-2&-1\
end{bmatrix}
$$
$$B = begin{bmatrix}
-1&-3&-1\
0&0&0\
1&3&1\
end{bmatrix}
$$
linear-algebra matrices matrix-equations
edited Nov 18 at 16:24
Ethan Bolker
41.3k547108
asked Nov 18 at 16:19
itsnotmyrealname
15110
1
If it's separable, each row is a scalar multiple of the same row. In particular, the rank is $1$
– saulspatz
Nov 18 at 16:28
... and if you are column-oriented, each column is a scalar multiple of a same column...
– Jean Marie
Nov 18 at 16:43
add a comment |
I wonder whether there is a simple way(e.g. if asked in exam) to determine if a matrix is separable. If yes, how can we write the separated parts? For example, Gaussian Blur kernel is separable.
$$frac1{256}cdotbegin{bmatrix}
1&4&6&4&1\
4&16&24&16&4\
6&24&36&24&6\
4&16&24&16&4\
1&4&6&4&1
end{bmatrix}
=
frac1{256}cdotbegin{bmatrix}
1\4\6\4\1
end{bmatrix}cdotbegin{bmatrix}
1&4&6&4&1
end{bmatrix}
$$
I have two matrices asked in old-exams. How to do my wish on the following matrices?
$$A = frac1{4}cdotbegin{bmatrix}
-1&-2&-1\
-2&16&-2\
-1&-2&-1\
end{bmatrix}
$$
$$B = begin{bmatrix}
-1&-3&-1\
0&0&0\
1&3&1\
end{bmatrix}
$$
linear-algebra matrices matrix-equations
edited Nov 18 at 16:24
Ethan Bolker
41.3k547108
asked Nov 18 at 16:19
itsnotmyrealname
15110
I wonder whether there is a simple way(e.g. if asked in exam) to determine if a matrix is separable. If yes, how can we write the separated parts? For example, Gaussian Blur kernel is separable.
$$frac1{256}cdotbegin{bmatrix}
1&4&6&4&1\
4&16&24&16&4\
6&24&36&24&6\
4&16&24&16&4\
1&4&6&4&1
end{bmatrix}
=
frac1{256}cdotbegin{bmatrix}
1\4\6\4\1
end{bmatrix}cdotbegin{bmatrix}
1&4&6&4&1
end{bmatrix}
$$
I have two matrices asked in old-exams. How to do my wish on the following matrices?
$$A = frac1{4}cdotbegin{bmatrix}
-1&-2&-1\
-2&16&-2\
-1&-2&-1\
end{bmatrix}
$$
$$B = begin{bmatrix}
-1&-3&-1\
0&0&0\
1&3&1\
end{bmatrix}
$$
linear-algebra matrices matrix-equations
linear-algebra matrices matrix-equations
edited Nov 18 at 16:24
Ethan Bolker
41.3k547108
asked Nov 18 at 16:19
itsnotmyrealname
15110
edited Nov 18 at 16:24
Ethan Bolker
41.3k547108
asked Nov 18 at 16:19
itsnotmyrealname
15110
edited Nov 18 at 16:24
Ethan Bolker
41.3k547108
edited Nov 18 at 16:24
Ethan Bolker
41.3k547108
edited Nov 18 at 16:24
Ethan Bolker
41.3k547108
41.3k547108
asked Nov 18 at 16:19
itsnotmyrealname
15110
asked Nov 18 at 16:19
itsnotmyrealname
15110
asked Nov 18 at 16:19
itsnotmyrealname
15110
15110
1
If it's separable, each row is a scalar multiple of the same row. In particular, the rank is $1$
– saulspatz
Nov 18 at 16:28
... and if you are column-oriented, each column is a scalar multiple of a same column...
– Jean Marie
Nov 18 at 16:43
add a comment |
1
If it's separable, each row is a scalar multiple of the same row. In particular, the rank is $1$
– saulspatz
Nov 18 at 16:28
... and if you are column-oriented, each column is a scalar multiple of a same column...
– Jean Marie
Nov 18 at 16:43
1
1
If it's separable, each row is a scalar multiple of the same row. In particular, the rank is $1$
– saulspatz
Nov 18 at 16:28
If it's separable, each row is a scalar multiple of the same row. In particular, the rank is $1$
– saulspatz
Nov 18 at 16:28
... and if you are column-oriented, each column is a scalar multiple of a same column...
– Jean Marie
Nov 18 at 16:43
... and if you are column-oriented, each column is a scalar multiple of a same column...
– Jean Marie
Nov 18 at 16:43
add a comment |
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1
If it's separable, each row is a scalar multiple of the same row. In particular, the rank is $1$
– saulspatz
Nov 18 at 16:28
... and if you are column-oriented, each column is a scalar multiple of a same column...
– Jean Marie
Nov 18 at 16:43