Ok, my issue was reading the wrong thing a thousand times.
It says in the instruction, that if X is positive-definite, not positive semi-definite, it is > 0. Which makes sense because then:
X = USU^T - our premise if X is symmetric
u_j^T X u_j = u_j^T U S U^T u_j
we have thanks to orthonormality of U:
u_j^T X u_j = s_j
as X is said to be positive-definite, the left expression is > 0, thus also s_j is >0.
Ok, my issue was reading the wrong thing a thousand times.
It says in the instruction, that if X is positive-definite, not positive semi-definite, it is > 0. Which makes sense because then:
X = USU^T - our premise if X is symmetric
u_j^T X u_j = u_j^T U S U^T u_j
we have thanks to orthonormality of U:
u_j^T X u_j = s_j
as X is said to be positive-definite, the left expression is > 0, thus also s_j is >0.
Last sentence :)
But in fact I think the solution is erroneous.
If you assume X to be psd --> can prove that s_j >= 0
if assume X to be pd --> can assume that s_j > 0.
The problem is that in the task it states that X is psd, but in the solution they use that it is pd to motivate why s_j > 0 ...
I was told however, that positive in french means >=, so this may be where the confusion comes from in the task.
Exercise Problem Set 12
Hello,
For Problem Set 12, Question 2:
I did the following:
if X NxN symmetric:
A = X X^T = B = X^T X, leading to X = USU^T, as the evectors u_j of A and B are the same.
Now I did:
u_j^T X u_j = u_j^T U S U^T u_j = 1 s_j 1
We can recognize that u_j^T X u_j is the quadratic form of X, which is per definition of X being psd, >= 0.
That implies that s_j >= 0.
What I don't understand is how this \(f(x)\)implies that s_j > 0, as is suggested in the solution.
Thanks!
Ok, my issue was reading the wrong thing a thousand times.
It says in the instruction, that if X is positive-definite, not positive semi-definite, it is > 0. Which makes sense because then:
X = USU^T - our premise if X is symmetric
u_j^T X u_j = u_j^T U S U^T u_j
we have thanks to orthonormality of U:
u_j^T X u_j = s_j
as X is said to be positive-definite, the left expression is > 0, thus also s_j is >0.
1
Did you figure it out?
No, still hoping for an answer
Ok, my issue was reading the wrong thing a thousand times.
It says in the instruction, that if X is positive-definite, not positive semi-definite, it is > 0. Which makes sense because then:
X = USU^T - our premise if X is symmetric
u_j^T X u_j = u_j^T U S U^T u_j
we have thanks to orthonormality of U:
u_j^T X u_j = s_j
as X is said to be positive-definite, the left expression is > 0, thus also s_j is >0.
1
That's also how I understood it, thanks!
Hi! Can you tell me where is written that X is positive-definite? I can't find it and I can't figure out why the solution is true
Last sentence :)
But in fact I think the solution is erroneous.
If you assume X to be psd --> can prove that s_j >= 0
if assume X to be pd --> can assume that s_j > 0.
The problem is that in the task it states that X is psd, but in the solution they use that it is pd to motivate why s_j > 0 ...
I was told however, that positive in french means >=, so this may be where the confusion comes from in the task.
Hope that helps
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