is random gaussian matrix invertible?
0
$begingroup$
Is Gaussian Random Matrix invertible?
I mean can we invert a Random Gaussian Square Matrix and also what is nature of its determinant, I mean to say whether determinant is zero or non zero?
matrices random-matrices
asked Dec 7 '18 at 16:38
sujit dassujit das
11
$endgroup$
add a comment |
0
$begingroup$
Is Gaussian Random Matrix invertible?
I mean can we invert a Random Gaussian Square Matrix and also what is nature of its determinant, I mean to say whether determinant is zero or non zero?
matrices random-matrices
asked Dec 7 '18 at 16:38
sujit dassujit das
11
$endgroup$
1
$begingroup$
Theoretically the determinant is a random variable, probability that it is exactly 0 is null. Practically, for large size, the matrix may become very badly conditioned, therefore difficult to invert
$endgroup$
– Damien
Dec 7 '18 at 16:47
1
$begingroup$
When I need to deal with a random matrix, not necessarily Gaussian, I prefer performing a SVD. It allows analysing the matrix, performing inverse or pseudo inverse, doing some light modifications of the matrix to avoid problems with the inverse, etc. However, it implies more computations ...
$endgroup$
– Damien
Dec 7 '18 at 17:02
$begingroup$
I plus-oned Damien's comments, but to emphasize that sujit's question needs more structure, I note that if we let $X$ be a Gaussian random variable, the $2times 2$ random matrix $[X : X; X : X]$ is (surely) not invertible.
$endgroup$
– Michael
Dec 7 '18 at 17:05
1
$begingroup$
On the other hand, if we form a random square matrix by independently choosing the columns, the probability that it is invertible is equal to the probability that the first column is not all-zero, multiplied by the probability that the second column is not in the span of the first, multiplied by the probability that the third column is not in the span of the first two, and so on. So it is easy to see that, in most cases with continuous probability distributions, this is a product of 1s (so the probability it is invertible is also 1, which supports Damien's initial comment).
$endgroup$
– Michael
Dec 7 '18 at 17:09
$begingroup$
Another classical possibility is to approximate the inverse by $(A^TA +lambda I)^{-1}A^T$ where $lambda$ is a small stabilization factor
$endgroup$
– Damien
Dec 7 '18 at 17:17
add a comment |
0
0
0
$begingroup$
Is Gaussian Random Matrix invertible?
I mean can we invert a Random Gaussian Square Matrix and also what is nature of its determinant, I mean to say whether determinant is zero or non zero?
matrices random-matrices
asked Dec 7 '18 at 16:38
sujit dassujit das
11
$endgroup$
Is Gaussian Random Matrix invertible?
I mean can we invert a Random Gaussian Square Matrix and also what is nature of its determinant, I mean to say whether determinant is zero or non zero?
matrices random-matrices
matrices random-matrices
asked Dec 7 '18 at 16:38
sujit dassujit das
11
asked Dec 7 '18 at 16:38
sujit dassujit das
11
asked Dec 7 '18 at 16:38
sujit dassujit das
11
asked Dec 7 '18 at 16:38
sujit dassujit das
11
asked Dec 7 '18 at 16:38
sujit dassujit das
11
11
1
$begingroup$
Theoretically the determinant is a random variable, probability that it is exactly 0 is null. Practically, for large size, the matrix may become very badly conditioned, therefore difficult to invert
$endgroup$
– Damien
Dec 7 '18 at 16:47
1
$begingroup$
When I need to deal with a random matrix, not necessarily Gaussian, I prefer performing a SVD. It allows analysing the matrix, performing inverse or pseudo inverse, doing some light modifications of the matrix to avoid problems with the inverse, etc. However, it implies more computations ...
$endgroup$
– Damien
Dec 7 '18 at 17:02
$begingroup$
I plus-oned Damien's comments, but to emphasize that sujit's question needs more structure, I note that if we let $X$ be a Gaussian random variable, the $2times 2$ random matrix $[X : X; X : X]$ is (surely) not invertible.
$endgroup$
– Michael
Dec 7 '18 at 17:05
1
$begingroup$
On the other hand, if we form a random square matrix by independently choosing the columns, the probability that it is invertible is equal to the probability that the first column is not all-zero, multiplied by the probability that the second column is not in the span of the first, multiplied by the probability that the third column is not in the span of the first two, and so on. So it is easy to see that, in most cases with continuous probability distributions, this is a product of 1s (so the probability it is invertible is also 1, which supports Damien's initial comment).
$endgroup$
– Michael
Dec 7 '18 at 17:09
$begingroup$
Another classical possibility is to approximate the inverse by $(A^TA +lambda I)^{-1}A^T$ where $lambda$ is a small stabilization factor
$endgroup$
– Damien
Dec 7 '18 at 17:17
add a comment |
1
$begingroup$
Theoretically the determinant is a random variable, probability that it is exactly 0 is null. Practically, for large size, the matrix may become very badly conditioned, therefore difficult to invert
$endgroup$
– Damien
Dec 7 '18 at 16:47
1
$begingroup$
When I need to deal with a random matrix, not necessarily Gaussian, I prefer performing a SVD. It allows analysing the matrix, performing inverse or pseudo inverse, doing some light modifications of the matrix to avoid problems with the inverse, etc. However, it implies more computations ...
$endgroup$
– Damien
Dec 7 '18 at 17:02
$begingroup$
I plus-oned Damien's comments, but to emphasize that sujit's question needs more structure, I note that if we let $X$ be a Gaussian random variable, the $2times 2$ random matrix $[X : X; X : X]$ is (surely) not invertible.
$endgroup$
– Michael
Dec 7 '18 at 17:05
1
$begingroup$
On the other hand, if we form a random square matrix by independently choosing the columns, the probability that it is invertible is equal to the probability that the first column is not all-zero, multiplied by the probability that the second column is not in the span of the first, multiplied by the probability that the third column is not in the span of the first two, and so on. So it is easy to see that, in most cases with continuous probability distributions, this is a product of 1s (so the probability it is invertible is also 1, which supports Damien's initial comment).
$endgroup$
– Michael
Dec 7 '18 at 17:09
$begingroup$
Another classical possibility is to approximate the inverse by $(A^TA +lambda I)^{-1}A^T$ where $lambda$ is a small stabilization factor
$endgroup$
– Damien
Dec 7 '18 at 17:17
1
1
$begingroup$
Theoretically the determinant is a random variable, probability that it is exactly 0 is null. Practically, for large size, the matrix may become very badly conditioned, therefore difficult to invert
$endgroup$
– Damien
Dec 7 '18 at 16:47
$begingroup$
Theoretically the determinant is a random variable, probability that it is exactly 0 is null. Practically, for large size, the matrix may become very badly conditioned, therefore difficult to invert
$endgroup$
– Damien
Dec 7 '18 at 16:47
1
1
$begingroup$
When I need to deal with a random matrix, not necessarily Gaussian, I prefer performing a SVD. It allows analysing the matrix, performing inverse or pseudo inverse, doing some light modifications of the matrix to avoid problems with the inverse, etc. However, it implies more computations ...
$endgroup$
– Damien
Dec 7 '18 at 17:02
$begingroup$
When I need to deal with a random matrix, not necessarily Gaussian, I prefer performing a SVD. It allows analysing the matrix, performing inverse or pseudo inverse, doing some light modifications of the matrix to avoid problems with the inverse, etc. However, it implies more computations ...
$endgroup$
– Damien
Dec 7 '18 at 17:02
$begingroup$
I plus-oned Damien's comments, but to emphasize that sujit's question needs more structure, I note that if we let $X$ be a Gaussian random variable, the $2times 2$ random matrix $[X : X; X : X]$ is (surely) not invertible.
$endgroup$
– Michael
Dec 7 '18 at 17:05
$begingroup$
I plus-oned Damien's comments, but to emphasize that sujit's question needs more structure, I note that if we let $X$ be a Gaussian random variable, the $2times 2$ random matrix $[X : X; X : X]$ is (surely) not invertible.
$endgroup$
– Michael
Dec 7 '18 at 17:05
1
1
$begingroup$
On the other hand, if we form a random square matrix by independently choosing the columns, the probability that it is invertible is equal to the probability that the first column is not all-zero, multiplied by the probability that the second column is not in the span of the first, multiplied by the probability that the third column is not in the span of the first two, and so on. So it is easy to see that, in most cases with continuous probability distributions, this is a product of 1s (so the probability it is invertible is also 1, which supports Damien's initial comment).
$endgroup$
– Michael
Dec 7 '18 at 17:09
$begingroup$
On the other hand, if we form a random square matrix by independently choosing the columns, the probability that it is invertible is equal to the probability that the first column is not all-zero, multiplied by the probability that the second column is not in the span of the first, multiplied by the probability that the third column is not in the span of the first two, and so on. So it is easy to see that, in most cases with continuous probability distributions, this is a product of 1s (so the probability it is invertible is also 1, which supports Damien's initial comment).
$endgroup$
– Michael
Dec 7 '18 at 17:09
$begingroup$
Another classical possibility is to approximate the inverse by $(A^TA +lambda I)^{-1}A^T$ where $lambda$ is a small stabilization factor
$endgroup$
– Damien
Dec 7 '18 at 17:17
$begingroup$
Another classical possibility is to approximate the inverse by $(A^TA +lambda I)^{-1}A^T$ where $lambda$ is a small stabilization factor
$endgroup$
– Damien
Dec 7 '18 at 17:17
add a comment |
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1
$begingroup$
Theoretically the determinant is a random variable, probability that it is exactly 0 is null. Practically, for large size, the matrix may become very badly conditioned, therefore difficult to invert
$endgroup$
– Damien
Dec 7 '18 at 16:47
1
$begingroup$
When I need to deal with a random matrix, not necessarily Gaussian, I prefer performing a SVD. It allows analysing the matrix, performing inverse or pseudo inverse, doing some light modifications of the matrix to avoid problems with the inverse, etc. However, it implies more computations ...
$endgroup$
– Damien
Dec 7 '18 at 17:02
$begingroup$
I plus-oned Damien's comments, but to emphasize that sujit's question needs more structure, I note that if we let $X$ be a Gaussian random variable, the $2times 2$ random matrix $[X : X; X : X]$ is (surely) not invertible.
$endgroup$
– Michael
Dec 7 '18 at 17:05
1
$begingroup$
On the other hand, if we form a random square matrix by independently choosing the columns, the probability that it is invertible is equal to the probability that the first column is not all-zero, multiplied by the probability that the second column is not in the span of the first, multiplied by the probability that the third column is not in the span of the first two, and so on. So it is easy to see that, in most cases with continuous probability distributions, this is a product of 1s (so the probability it is invertible is also 1, which supports Damien's initial comment).
$endgroup$
– Michael
Dec 7 '18 at 17:09
$begingroup$
Another classical possibility is to approximate the inverse by $(A^TA +lambda I)^{-1}A^T$ where $lambda$ is a small stabilization factor
$endgroup$
– Damien
Dec 7 '18 at 17:17