Find Lipschtiz constant for a function in matrix
I have the following function in $X in R^{n times k}$
$$f(X) = -4A XLambda_1 + 4(XLambda_1 X^T XLambda_1) - 4A^TXYLambda_2Y^T + 4XYLambda_2Y^TX^TXYLambda_2Y^T$$
where $A in R^{n times n}$, $Y in R^{k times k1}$, $Lambda_1 in R^{k times k}$ and $Lambda_2 in R^{k1 times k1}$, and $Lambda_1 , Lambda_2$ are diagonal matrices. How can I find Lipschitz constant of the above function with respect to frobenious norm? Would it be possible to find local Lipschitz constant I know ||X||? The f function is gradient with respect to X and I am trying to do a gradient descent, so, I will have bounds on ||X||.
matrices analysis lipschitz-functions
asked Nov 19 '18 at 0:25
Dushyant Sahoo
528
|
show 2 more comments
I have the following function in $X in R^{n times k}$
$$f(X) = -4A XLambda_1 + 4(XLambda_1 X^T XLambda_1) - 4A^TXYLambda_2Y^T + 4XYLambda_2Y^TX^TXYLambda_2Y^T$$
where $A in R^{n times n}$, $Y in R^{k times k1}$, $Lambda_1 in R^{k times k}$ and $Lambda_2 in R^{k1 times k1}$, and $Lambda_1 , Lambda_2$ are diagonal matrices. How can I find Lipschitz constant of the above function with respect to frobenious norm? Would it be possible to find local Lipschitz constant I know ||X||? The f function is gradient with respect to X and I am trying to do a gradient descent, so, I will have bounds on ||X||.
matrices analysis lipschitz-functions
asked Nov 19 '18 at 0:25
Dushyant Sahoo
528
1
There is no global Lipschitz constant for $f$.
– copper.hat
Nov 19 '18 at 0:28
@copper.hat Why is there no global Lipschitz?
– Dushyant Sahoo
Nov 19 '18 at 1:05
Well, it depends on the actual values of the matrices above, but consider the formula for scalars, there is an $X^3$ term which is not Lipschitz.
– copper.hat
Nov 19 '18 at 1:11
@copper.hat What if I know ||X||? The $f$ function is gradient with respect to $X$ and I am trying to do a gradient descent, so, I will have bounds on ||X||.
– Dushyant Sahoo
Nov 19 '18 at 1:31
1
I don't know what to say. In general, the function is not globally Lipschitz. On any bounded set you can find a local Lipschitz constant. One way is to find the derivative of $f$ with respect to $X$ and bound the derivative.
– copper.hat
Nov 19 '18 at 1:55
|
show 2 more comments
I have the following function in $X in R^{n times k}$
$$f(X) = -4A XLambda_1 + 4(XLambda_1 X^T XLambda_1) - 4A^TXYLambda_2Y^T + 4XYLambda_2Y^TX^TXYLambda_2Y^T$$
where $A in R^{n times n}$, $Y in R^{k times k1}$, $Lambda_1 in R^{k times k}$ and $Lambda_2 in R^{k1 times k1}$, and $Lambda_1 , Lambda_2$ are diagonal matrices. How can I find Lipschitz constant of the above function with respect to frobenious norm? Would it be possible to find local Lipschitz constant I know ||X||? The f function is gradient with respect to X and I am trying to do a gradient descent, so, I will have bounds on ||X||.
matrices analysis lipschitz-functions
asked Nov 19 '18 at 0:25
Dushyant Sahoo
528
I have the following function in $X in R^{n times k}$
$$f(X) = -4A XLambda_1 + 4(XLambda_1 X^T XLambda_1) - 4A^TXYLambda_2Y^T + 4XYLambda_2Y^TX^TXYLambda_2Y^T$$
where $A in R^{n times n}$, $Y in R^{k times k1}$, $Lambda_1 in R^{k times k}$ and $Lambda_2 in R^{k1 times k1}$, and $Lambda_1 , Lambda_2$ are diagonal matrices. How can I find Lipschitz constant of the above function with respect to frobenious norm? Would it be possible to find local Lipschitz constant I know ||X||? The f function is gradient with respect to X and I am trying to do a gradient descent, so, I will have bounds on ||X||.
matrices analysis lipschitz-functions
matrices analysis lipschitz-functions
asked Nov 19 '18 at 0:25
Dushyant Sahoo
528
asked Nov 19 '18 at 0:25
Dushyant Sahoo
528
edited Nov 19 '18 at 2:04
asked Nov 19 '18 at 0:25
Dushyant Sahoo
528
asked Nov 19 '18 at 0:25
Dushyant Sahoo
528
asked Nov 19 '18 at 0:25
Dushyant Sahoo
528
528
1
There is no global Lipschitz constant for $f$.
– copper.hat
Nov 19 '18 at 0:28
@copper.hat Why is there no global Lipschitz?
– Dushyant Sahoo
Nov 19 '18 at 1:05
Well, it depends on the actual values of the matrices above, but consider the formula for scalars, there is an $X^3$ term which is not Lipschitz.
– copper.hat
Nov 19 '18 at 1:11
@copper.hat What if I know ||X||? The $f$ function is gradient with respect to $X$ and I am trying to do a gradient descent, so, I will have bounds on ||X||.
– Dushyant Sahoo
Nov 19 '18 at 1:31
1
I don't know what to say. In general, the function is not globally Lipschitz. On any bounded set you can find a local Lipschitz constant. One way is to find the derivative of $f$ with respect to $X$ and bound the derivative.
– copper.hat
Nov 19 '18 at 1:55
|
show 2 more comments
1
There is no global Lipschitz constant for $f$.
– copper.hat
Nov 19 '18 at 0:28
@copper.hat Why is there no global Lipschitz?
– Dushyant Sahoo
Nov 19 '18 at 1:05
Well, it depends on the actual values of the matrices above, but consider the formula for scalars, there is an $X^3$ term which is not Lipschitz.
– copper.hat
Nov 19 '18 at 1:11
@copper.hat What if I know ||X||? The $f$ function is gradient with respect to $X$ and I am trying to do a gradient descent, so, I will have bounds on ||X||.
– Dushyant Sahoo
Nov 19 '18 at 1:31
1
I don't know what to say. In general, the function is not globally Lipschitz. On any bounded set you can find a local Lipschitz constant. One way is to find the derivative of $f$ with respect to $X$ and bound the derivative.
– copper.hat
Nov 19 '18 at 1:55
1
1
There is no global Lipschitz constant for $f$.
– copper.hat
Nov 19 '18 at 0:28
There is no global Lipschitz constant for $f$.
– copper.hat
Nov 19 '18 at 0:28
@copper.hat Why is there no global Lipschitz?
– Dushyant Sahoo
Nov 19 '18 at 1:05
@copper.hat Why is there no global Lipschitz?
– Dushyant Sahoo
Nov 19 '18 at 1:05
Well, it depends on the actual values of the matrices above, but consider the formula for scalars, there is an $X^3$ term which is not Lipschitz.
– copper.hat
Nov 19 '18 at 1:11
Well, it depends on the actual values of the matrices above, but consider the formula for scalars, there is an $X^3$ term which is not Lipschitz.
– copper.hat
Nov 19 '18 at 1:11
@copper.hat What if I know ||X||? The $f$ function is gradient with respect to $X$ and I am trying to do a gradient descent, so, I will have bounds on ||X||.
– Dushyant Sahoo
Nov 19 '18 at 1:31
@copper.hat What if I know ||X||? The $f$ function is gradient with respect to $X$ and I am trying to do a gradient descent, so, I will have bounds on ||X||.
– Dushyant Sahoo
Nov 19 '18 at 1:31
1
1
I don't know what to say. In general, the function is not globally Lipschitz. On any bounded set you can find a local Lipschitz constant. One way is to find the derivative of $f$ with respect to $X$ and bound the derivative.
– copper.hat
Nov 19 '18 at 1:55
I don't know what to say. In general, the function is not globally Lipschitz. On any bounded set you can find a local Lipschitz constant. One way is to find the derivative of $f$ with respect to $X$ and bound the derivative.
– copper.hat
Nov 19 '18 at 1:55
|
show 2 more comments
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1
There is no global Lipschitz constant for $f$.
– copper.hat
Nov 19 '18 at 0:28
@copper.hat Why is there no global Lipschitz?
– Dushyant Sahoo
Nov 19 '18 at 1:05
Well, it depends on the actual values of the matrices above, but consider the formula for scalars, there is an $X^3$ term which is not Lipschitz.
– copper.hat
Nov 19 '18 at 1:11
@copper.hat What if I know ||X||? The $f$ function is gradient with respect to $X$ and I am trying to do a gradient descent, so, I will have bounds on ||X||.
– Dushyant Sahoo
Nov 19 '18 at 1:31
1
I don't know what to say. In general, the function is not globally Lipschitz. On any bounded set you can find a local Lipschitz constant. One way is to find the derivative of $f$ with respect to $X$ and bound the derivative.
– copper.hat
Nov 19 '18 at 1:55