Let
Which is Hermitian Matrix.
Again
Which is skew-Hermitian matrix.
Now,
Matrix has been expressed as the sum of Hermitian and skew-Hermitian matrix.
Appeared in Year: 2007
Let be the vector space of all polynomial with real coefficient of degree less than or equal to two considered over the real field , such that and . Determine a basis for and hence its dimension. (CS main Paper 1)
Let
be the given vector space of all polynomials with real coefficient of degree less than or equal to two considered over the real field , such that and .
To determine a basis for .
Given that
And
the given vector space becomes
To find a basis for a finite subset of
s. t
Let,
Then where .
From eq. , we have .
i. e. spans .
Let s. t
is L. I subse
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Appeared in Year: 2013
Let where is a cube root of unity. If denote the eigen values of show that| . (CS main Paper 1)
Given
is cube root of unity.
and and .
Eigenvalue of are
Now
Appeared in Year: 2011
Let be the eigen value of a square matrix with corresponding eigen vector . if is Matrix Similar to Show that the eigen value of are same as that of and eigen vectors of A. (CS main Paper 1)
is Matrix Similar to (Given)
So, inverible Matrix such that
Matrices and have the same characteristics polynomial and hence the same set of given values.
Now, Let is given value of Matrix
Then
with
is eigen vector of corresponding to eigen value . This is relation b/w eigenvector of two similar matrices.
Appeared in Year: 2013
Let denote the vector space of all polynomials of degree at most and be a linear transformation given by
Find the matrix of with respect to the bases and of respectively. Also, find the null space of T. (CS main Paper 1)
Given be a linear transformation defined by
And basis of respectively.
Now,
If is matrix of w. r. t. Basis of .
Now to find Null Space of T
Let (Null Space of )
Appeared in Year: 2014
Verify Cayley-Hamilton Theorem for the Matrix and hence find its inverse.
Also find the Matrix represented by (CS main Paper 1)
Given
The characteristic equation of is
We have to prove that satisfies the equation
Now,
=
=
So, Cayley – Hamilton Theorem is Satisfied.
Now,
It is Inverse of Matrix
Now, We find the Matrix represented by
We have and
=
=
So, =
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