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Show the uniqueness of a matrix

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hello!!! :D
I want to show that the lower triangular matrix L,which has the identity $A=LL^{T}$ ,where A is a positive-definite and symmetric matrix,is unique.
That's what I have done so far:
Suppose that there are two matrices,with that identity.
Then, $$A=LL^{T}=MM^{T} \Rightarrow M^{-1}L=M^{T}(L^{T})^{-1}$$
$M^{-1}L$ is a lower triangular matrix and $M^{T}(L^{T})^{-1}$ is an upper triangular matrix,so,because of the fact that they are equal,they must be equal to the diagonal matrix.
So, $$L=MD \text{ and } M^{T}=DL^{T} $$
$$L=MD \Rightarrow L^{T}=D^{T}M^{T}$$
Replacing it at the other relation we have,we get $$M^{T}=DD^{T}M^{T}$$
So,it must be $DD^{T}=I \Rightarrow DD=I \Rightarrow D=I$.
Is this right?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hey evinda!!!

Your argument is correct.
Still, let me point out a couple of unwarranted jumps...

Hello!!! :D
I want to show that the lower triangular matrix L,which has the identity $A=LL^{T}$ ,where A is a positive-definite and symmetric matrix,is unique.
That's what I have done so far:
Suppose that there are two matrices,with that identity.
Then, $$A=LL^{T}=MM^{T} \Rightarrow M^{-1}L=M^{T}(L^{T})^{-1}$$
You are assuming L and M are invertible.
Can you support that?


$M^{-1}L$ is a lower triangular matrix and $M^{T}(L^{T})^{-1}$ is an upper triangular matrix,so,because of the fact that they are equal,they must be equal to the diagonal matrix.
You are assuming that $L^{-1}$ and $M^{-1}$ are upper triangular matrices.
And also that the product of two lower triangular matrices is a lower triangular matrix.
Can you support that?
Are those properties perhaps given?


So, $$L=MD \text{ and } M^{T}=DL^{T} $$
$$L=MD \Rightarrow L^{T}=D^{T}M^{T}$$
Replacing it at the other relation we have,we get $$M^{T}=DD^{T}M^{T}$$
So,it must be $DD^{T}=I \Rightarrow DD=I \Rightarrow D=I$.
Is this right?
From $DD=I$ you cannot conclude that $D=I$.
It is not true if $D$ has $-1$ on its diagonal.
So you need something more here...
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
From $DD=I$ you cannot conclude that $D=I$.
It is not true if $D$ has $-1$ on its diagonal.
So you need something more here...
Since A is positive-definite,the elements of the main diagonal should be greater than 0,or am I wrong? :confused:
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
You are assuming L and M are invertible.
Can you support that?
I thought that det(L) and det(M) are different from 0,because of the fact that A is positive-definite.Or can't I say it like that?


You are assuming that $L^{-1}$ and $M^{-1}$ are upper triangular matrices.
And also that the product of two lower triangular matrices is a lower triangular matrix.
Can you support that?
Are those properties perhaps given?
No,these properties are not given..I just thought that it is like that.. :confused:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Since A is positive-definite,the elements of the main diagonal should be greater than 0,or am I wrong? :confused:
Hmm, I'm not sure, but I don't think that the elements on the main diagonal of A have to be positive.
Perhaps you mean that all eigenvalues are positive?
Or that A is similar to a diagonal matrix with positive entries on its diagonal?

Anyway, it still doesn't tell us anything about the sign of the entries in D...


I thought that det(L) and det(M) are different from 0,because of the fact that A is positive-definite.Or can't I say it like that?
It's true enough, but I wouldn't say it like that, because it contains jumps that are not evidently true.
I'd say something like:

The fact that A is positive-definite implies that its determinant is different from 0.
Since $\det A = \det(LL^T)=\det(L)\det(L^T)=\det(L)^2$, we can tell that $\det(L)$ is also different from 0.
By the same argument $\det M$ is also different from 0.


No,these properties are not given..I just thought that it is like that.. :confused:
Well it is like that... but it's not something you can blindly assume.
Assuming stuff like that leads to grand mistakes. :eek:
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hmm, I'm not sure, but I don't think that the elements on the main diagonal of A have to be positive.
Perhaps you mean that all eigenvalues are positive?
Or that A is similar to a diagonal matrix with positive entries on its diagonal?

Anyway, it still doesn't tell us anything about the sign of the entries in D...




It's true enough, but I wouldn't say it like that, because it contains jumps that are not evidently true.
I'd say something like:

The fact that A is positive-definite implies that its determinant is different from 0.
Since $\det A = \det(LL^T)=\det(L)\det(L^T)=\det(L)^2$, we can tell that $\det(L)$ is also different from 0.
By the same argument $\det M$ is also different from 0.




Well it is like that... but it's not something you can blindly assume.
Assuming stuff like that leads to grand mistakes. :eek:
Is there an other way I could show that the matrix is unique?? :confused:
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
I am looking at the Cholesky decomposition,and I have to show that there is a unique lower triangular matrix L,with positive elements at the main diagonal,so that:
$$A=LL^{T}$$
Using induction on n,I proved that it exists a lower triangular matrix L,with the above identity.But,how can I show that it is unique?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Since A is positive-definite,the elements of the main diagonal should be greater than 0,or am I wrong? :confused:
The usual definition of a positive definite matrix is that it should be symmetric and that all its eigenvalues should be strictly positive. That implies that each element on the main diagonal, and also the determinant, should be strictly positive. If zero eigenvalues are allowed, the matrix is usually called positive semi-definite. In that case, the Cholesky decomposition is not unique. For example, if $A = \begin{bmatrix} 0&0 \\0&1 \end{bmatrix}$ (with eigenvalues $0$ and $1$) and $L_\theta = \begin{bmatrix} 0&0 \\\sin\theta& \cos\theta \end{bmatrix}$ then $A = L_\theta (L_\theta)^{\text{T}}$ for all $\theta$.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
The usual definition of a positive definite matrix is that it should be symmetric and that all its eigenvalues should be strictly positive. That implies that each element on the main diagonal, and also the determinant, should be strictly positive. If zero eigenvalues are allowed, the matrix is usually called positive semi-definite. In that case, the Cholesky decomposition is not unique. For example, if $A = \begin{bmatrix} 0&0 \\0&1 \end{bmatrix}$ (with eigenvalues $0$ and $1$) and $L_\theta = \begin{bmatrix} 0&0 \\\sin\theta& \cos\theta \end{bmatrix}$ then $A = L_\theta (L_\theta)^{\text{T}}$ for all $\theta$.
I understand...and what can I do then to prove that $M$ and $L$ are equal?What do I have to change?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Is there an other way I could show that the matrix is unique?? :confused:
Your approach is fine. I'd stick with it.
Basically it's only the last step ($DD=I \Rightarrow D=I$) that is incorrect and it actually shows that your problem statement is not true (as it is now).

The matrix D will have entries on its main diagonal that are either $+1$ or $-1$.
That means that $L$ and $M$ do not have to be the same, although you would have that $L=MD$.


I am looking at the Cholesky decomposition,and I have to show that there is a unique lower triangular matrix L,with positive elements at the main diagonal,so that:
$$A=LL^{T}$$
Using induction on n,I proved that it exists a lower triangular matrix L,with the above identity.But,how can I show that it is unique?
You have just added the condition that L has must have positive elements at the main diagonal.
Add that condition to your original problem statement and to your proof and you can conclude that $D$ can only have $+1$ on its diagonal.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
You have just added the condition that L has must have positive elements at the main diagonal.
Add that condition to your original problem statement and to your proof and you can conclude that $D$ can only have $+1$ on its diagonal.
So,we conclude that $D=\pm I$ and then ,at the case $D=-I$ ,we have $L=-D$ and because of the fact that both matrices have to have positive elements at the main diagonal,we only can have $D=I$,or do we reject $D=-I$ at once?? :confused:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
So,we conclude that $D=\pm I$ and then ,at the case $D=-I$ ,we have $L=-D$ and because of the fact that both matrices have to have positive elements at the main diagonal,we only can have $D=I$,or do we reject $D=-I$ at once?? :confused:
Not quite. It is also possible that $D$ has both $+1$ and $-1$ on its diagonal, in which case it is neither $+I$ nor $-I$.

Let's assume that $M$ is a solution with positive entries on its diagonal.

Now suppose that $D$ has at least a $-1$ on its diagonal.
Since we have that $L=MD$, it follows that $L$ will get a negative entry on its diagonal.
With the condition that $L$ must have positive entries on its diagonal we have a contradiction.
In other words, we must have that $D=I$.