This is a quite pleasant sample problem from ISI MStat PSB 2011 Problem 1. It is mainly about, patterns in matrices and determinants and using a special kind of determinant decomposition, which is widely used in Statistics . Give it a try !
Let A be a nxn matrix, given as
\(A_{nxn}\) = \begin{pmatrix} a & b & \cdots & b \\b& a & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a \end{pmatrix} ;
where \( a \neq b \) and \( a+(n-1)b= 0 \).
Suppose B=A+ \(\frac{\vec{1} \vec{1'}}{n} \) where \( \vec{1} =(1,1,.....,1)' \) is nx1 vector.
Show that,
(a) B is non-singular .
(b) \(A{B}^{-1} A =A\)
Basic Matrix multiplication
Determinants.
Matrix decomposition .
While attacking a problem related to matrices, the primary approach which I find very helpful is, realizing if there is any subtle pattern hidden. Similarly, here also a pattern is very much prominent exhibited in the matrix A, hence we will break A as,
\(A=(a-b) I_{n} +b \vec{1} \vec{1'} \).
which reduces, \( B = (a-b) I_{n} + (b + \frac{1}{n}) \vec{1}\vec{1'} \).
now, we can find Determinant of B, by using a known decomposition , which follows from determinants of partitioned matrices.
we have the decomposition as for some non-singular M and column vectors \(\vec{u}\) \(\vec{v}\) ,
we have \( |M+\vec{u}\vec{v'}|=|M|(1+ \vec{v'} M^{-1}\vec{u}) \), when \( \vec{v'}M^{-1}\vec{u} \neq -1\).
for, this particular problem , \( \vec{v}\)=\(\vec{u}\)= \( \sqrt{(b+ \frac{1}{n})}\vec{1} \) , \(M= (a-b)I_n \) , which is non-singular, and clearly \( nb \neq 0\),
If you are not familiar with these, then become friendly with this decomposition, as it is has very important and frequent applications in Statistics. Best to derive it !
So, \( |B|= |(a-b)I_{n}+ (b + \frac{1}{n}) \vec{1}\vec{1'}|\)= \( (a-b)^{n} ( 1+\vec{1'} \frac{I_n}{(a-b)}\vec{1}) \) =\( (a-b)^{n-1}( a+(n-1)b+1 ) \) =\( (a-b)^{n-1} \neq 0\)
as \( a+(n-1)b=0 \) and \( a \neq b \), So, hence B is non-singular. Also, B is invertible, we will need this to do the next part
for the second part (b), observe that \(A\vec{1}=\vec{0}\) why ??? , so, \( B\vec{1}=\vec{1} \Rightarrow \vec{1}= B^{-1}\vec{1} \) ................(*)
So, \( B=A+ \frac{\vec{1} \vec{1'}}{n} \Rightarrow I_n= B^{-1}A+ \frac{1}{n} B^{-1}\vec{1}\vec{1'}=B^{-1}A+ \frac{1}{n}\vec{1}\vec{1'} \) ....using(*)
now left multiplying A, to the above matrix equation, we have
\(A=AB^{-1}A+\frac{1}{n}A\vec{1}\vec{1'}=AB^{-1}A \). hence , we are done !!
Suppose, it is given that \(Trace(A)=Trace(A^2)=n \) , Can You show that all the eigenvalues of A are equal to 1 ? Is it true for any symmetric matrix,following the given condition? Give it a thought !
[ In case, you don't know what eigenvalues are, its a scalar, \(\lambda\) which one may find for a square matrix C, such that, for a non-null \(\vec{x}\),
\( C\vec{x}= \lambda\vec{x}\), for a matrix order n, one will find n such scalars or eigenvalues, say \( \lambda_1,.....,\lambda_1\) ,
then \( \lambda_1+ .....+ \lambda_n= Trace(C) \), and , for symmetric matrices, all eigenvalues are real, so you don't need to worry much but you can obviously verify it !! ]
This is a quite pleasant sample problem from ISI MStat PSB 2011 Problem 1. It is mainly about, patterns in matrices and determinants and using a special kind of determinant decomposition, which is widely used in Statistics . Give it a try !
Let A be a nxn matrix, given as
\(A_{nxn}\) = \begin{pmatrix} a & b & \cdots & b \\b& a & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a \end{pmatrix} ;
where \( a \neq b \) and \( a+(n-1)b= 0 \).
Suppose B=A+ \(\frac{\vec{1} \vec{1'}}{n} \) where \( \vec{1} =(1,1,.....,1)' \) is nx1 vector.
Show that,
(a) B is non-singular .
(b) \(A{B}^{-1} A =A\)
Basic Matrix multiplication
Determinants.
Matrix decomposition .
While attacking a problem related to matrices, the primary approach which I find very helpful is, realizing if there is any subtle pattern hidden. Similarly, here also a pattern is very much prominent exhibited in the matrix A, hence we will break A as,
\(A=(a-b) I_{n} +b \vec{1} \vec{1'} \).
which reduces, \( B = (a-b) I_{n} + (b + \frac{1}{n}) \vec{1}\vec{1'} \).
now, we can find Determinant of B, by using a known decomposition , which follows from determinants of partitioned matrices.
we have the decomposition as for some non-singular M and column vectors \(\vec{u}\) \(\vec{v}\) ,
we have \( |M+\vec{u}\vec{v'}|=|M|(1+ \vec{v'} M^{-1}\vec{u}) \), when \( \vec{v'}M^{-1}\vec{u} \neq -1\).
for, this particular problem , \( \vec{v}\)=\(\vec{u}\)= \( \sqrt{(b+ \frac{1}{n})}\vec{1} \) , \(M= (a-b)I_n \) , which is non-singular, and clearly \( nb \neq 0\),
If you are not familiar with these, then become friendly with this decomposition, as it is has very important and frequent applications in Statistics. Best to derive it !
So, \( |B|= |(a-b)I_{n}+ (b + \frac{1}{n}) \vec{1}\vec{1'}|\)= \( (a-b)^{n} ( 1+\vec{1'} \frac{I_n}{(a-b)}\vec{1}) \) =\( (a-b)^{n-1}( a+(n-1)b+1 ) \) =\( (a-b)^{n-1} \neq 0\)
as \( a+(n-1)b=0 \) and \( a \neq b \), So, hence B is non-singular. Also, B is invertible, we will need this to do the next part
for the second part (b), observe that \(A\vec{1}=\vec{0}\) why ??? , so, \( B\vec{1}=\vec{1} \Rightarrow \vec{1}= B^{-1}\vec{1} \) ................(*)
So, \( B=A+ \frac{\vec{1} \vec{1'}}{n} \Rightarrow I_n= B^{-1}A+ \frac{1}{n} B^{-1}\vec{1}\vec{1'}=B^{-1}A+ \frac{1}{n}\vec{1}\vec{1'} \) ....using(*)
now left multiplying A, to the above matrix equation, we have
\(A=AB^{-1}A+\frac{1}{n}A\vec{1}\vec{1'}=AB^{-1}A \). hence , we are done !!
Suppose, it is given that \(Trace(A)=Trace(A^2)=n \) , Can You show that all the eigenvalues of A are equal to 1 ? Is it true for any symmetric matrix,following the given condition? Give it a thought !
[ In case, you don't know what eigenvalues are, its a scalar, \(\lambda\) which one may find for a square matrix C, such that, for a non-null \(\vec{x}\),
\( C\vec{x}= \lambda\vec{x}\), for a matrix order n, one will find n such scalars or eigenvalues, say \( \lambda_1,.....,\lambda_1\) ,
then \( \lambda_1+ .....+ \lambda_n= Trace(C) \), and , for symmetric matrices, all eigenvalues are real, so you don't need to worry much but you can obviously verify it !! ]
