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
= \begin{pmatrix} a & b & \cdots & b \\b& a & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a \end{pmatrix} ;
where and
.
Suppose B=A+ where
is nx1 vector.
Show that,
(a) B is non-singular .
(b)
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,
.
which reduces, .
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
,
we have , when
.
for, this particular problem , =
=
,
, which is non-singular, and clearly
,
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, =
=
=
as and
, 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 why ??? , so,
................(*)
So, ....using(*)
now left multiplying A, to the above matrix equation, we have
. hence , we are done !!
Suppose, it is given that , 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, which one may find for a square matrix C, such that, for a non-null
,
, for a matrix order n, one will find n such scalars or eigenvalues, say
,
then , 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
= \begin{pmatrix} a & b & \cdots & b \\b& a & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a \end{pmatrix} ;
where and
.
Suppose B=A+ where
is nx1 vector.
Show that,
(a) B is non-singular .
(b)
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,
.
which reduces, .
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
,
we have , when
.
for, this particular problem , =
=
,
, which is non-singular, and clearly
,
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, =
=
=
as and
, 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 why ??? , so,
................(*)
So, ....using(*)
now left multiplying A, to the above matrix equation, we have
. hence , we are done !!
Suppose, it is given that , 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, which one may find for a square matrix C, such that, for a non-null
,
, for a matrix order n, one will find n such scalars or eigenvalues, say
,
then , and , for symmetric matrices, all eigenvalues are real, so you don't need to worry much but you can obviously verify it !! ]