This is a beautiful problem from ISI MStat 2017 PSB Problem 1 based on matrices. We provide a detailed solution with the prerequisites mentioned explicitly.
Let and
be real numbers. Show that there exists a unique
real symmetric matrix
with
and
if and only if
.
Let , be a unique
real symmetric matrix , with x,y,z belongs to real .
Given , trace(A)= ----(1)and
Det(A)= ----(2)
1st solution : from (1) putting this in (2) we get
Now using , Sridhar Acharya Formula we get ,
Now as given , A is unique so y can't take two different values hence this part must be zero
i.e
Again , as A is unique matrix z can't take two different values . Hence z must be equal to zero
i.e ( Hence proved)
2nd solution : See if we interchange x by y then all the properties of A remains same i.e trace and determinant . But it can't be possible as we assume that A is unique hence x and y must be equal i.e .
From (1) we get
Again if we interchange z by -z then all the properties of A remains same i.e trace and determinant . But it can't be possible as we assume that A is unique hence z and -z must be equal i.e .
From (2) we get (Hence proved )
Now , we will assume that , where a and b are real and show that the matrix is unique .
Let , be a
real symmetric matrix with x,y,z belongs to real
Given trace(A)=---(3)
and Det(A)= ---(4)
Another thing is given that
So using (3) and (4) in (5) we get ,
i.e sum of two squares is equal to 0 which implies individual squares are equal to zero .
Hence , and
and
which give and
Hence , is unique ( proved )
we have proved both if and only if part . Hence we are done!
This is a beautiful problem from ISI MStat 2017 PSB Problem 1 based on matrices. We provide a detailed solution with the prerequisites mentioned explicitly.
Let and
be real numbers. Show that there exists a unique
real symmetric matrix
with
and
if and only if
.
Let , be a unique
real symmetric matrix , with x,y,z belongs to real .
Given , trace(A)= ----(1)and
Det(A)= ----(2)
1st solution : from (1) putting this in (2) we get
Now using , Sridhar Acharya Formula we get ,
Now as given , A is unique so y can't take two different values hence this part must be zero
i.e
Again , as A is unique matrix z can't take two different values . Hence z must be equal to zero
i.e ( Hence proved)
2nd solution : See if we interchange x by y then all the properties of A remains same i.e trace and determinant . But it can't be possible as we assume that A is unique hence x and y must be equal i.e .
From (1) we get
Again if we interchange z by -z then all the properties of A remains same i.e trace and determinant . But it can't be possible as we assume that A is unique hence z and -z must be equal i.e .
From (2) we get (Hence proved )
Now , we will assume that , where a and b are real and show that the matrix is unique .
Let , be a
real symmetric matrix with x,y,z belongs to real
Given trace(A)=---(3)
and Det(A)= ---(4)
Another thing is given that
So using (3) and (4) in (5) we get ,
i.e sum of two squares is equal to 0 which implies individual squares are equal to zero .
Hence , and
and
which give and
Hence , is unique ( proved )
we have proved both if and only if part . Hence we are done!