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Laurent Duval
Laurent Duval
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You just have to checkofcheck the rank of the matrix.

When you look a discrete signals, it is customary to express their product the following way. If vectors are considered "column-wise", then you typically use the transpose operation:

$$ x[n_1,n_2] = f[n_1]^T g[n_2]$$

This implies that the 2D signal is of rank one or less (rank being the maximal number of linearly independent rows or columns). Conversely, any (infinite) matrix of rank $0$ or $1$ is separable. Note: the separation is not unique in general.

You just have to checkof the rank of the matrix.

When you look a discrete signals, it is customary to express their product the following way. If vectors are considered "column-wise", then you typically use the transpose operation:

$$ x[n_1,n_2] = f[n_1]^T g[n_2]$$

This implies that the 2D signal is of rank one or less (rank being the maximal number of linearly independent rows or columns). Conversely, any (infinite) matrix of rank $0$ or $1$ is separable. Note: the separation is not unique in general.

You just have to check the rank of the matrix.

When you look a discrete signals, it is customary to express their product the following way. If vectors are considered "column-wise", then you typically use the transpose operation:

$$ x[n_1,n_2] = f[n_1]^T g[n_2]$$

This implies that the 2D signal is of rank one or less (rank being the maximal number of linearly independent rows or columns). Conversely, any (infinite) matrix of rank $0$ or $1$ is separable. Note: the separation is not unique in general.

Source Link
Laurent Duval
Laurent Duval
  • 32.3k
  • 3
  • 35
  • 105

You just have to checkof the rank of the matrix.

When you look a discrete signals, it is customary to express their product the following way. If vectors are considered "column-wise", then you typically use the transpose operation:

$$ x[n_1,n_2] = f[n_1]^T g[n_2]$$

This implies that the 2D signal is of rank one or less (rank being the maximal number of linearly independent rows or columns). Conversely, any (infinite) matrix of rank $0$ or $1$ is separable. Note: the separation is not unique in general.