Let T be the transformation of 2 by 2 real symmetric matrices defined by: [ <mtable

kunyatia33

kunyatia33

Answered question

2022-05-26

Let T be the transformation of 2 by 2 real symmetric matrices defined by:
[ a b b c ]
[ c b b a ]
then which of the following statements is NOT true?
1. det ( T ) = 1
2. T 1 = T
3. T is linear
4. the space of 2 by 2 real symmetric matricies with only zeros in the main diagonal is an eigenspace of T.
5. λ = 2 is an eigen value of T

Answer & Explanation

sepolturamo

sepolturamo

Beginner2022-05-27Added 14 answers

A basis of the vector space of symmetric 2 × 2 matrices is
B = { A 1 = [ 1 0 0 0 ] ; A 2 = [ 0 1 1 0 ] ; A 3 = [ 0 0 0 1 ] }
If T is linear, then its matrix with respect to B is given by
X = [ C B ( T ( A 1 ) ) C B ( T ( A 2 ) ) C B ( T ( A 3 ) ) ]
where C B ( A ) is the coordinate vector of A with respect to B. Then, since T ( A 1 ) = A 3 , T ( A 2 ) = A 2 and T ( A 3 ) = A 1 ,
X = [ 0 0 1 0 1 0 1 0 0 ]
Now, for A = [ a b b c ] , we have
C B ( A ) = [ a b c ]
and
X [ a b c ] = [ 0 0 1 0 1 0 1 0 0 ] [ a b c ] = [ c b a ] = C B ( T ( A ) )
Therefore T is linear. Note that this could be proved directly, by simple computations, but T is useful for the other questions.
The determinant of T is the same as the determinant of any associated matrix, so
det T = det X = 1
The eigenvalues of T are the same as the eigenvalues of X. Since
det ( X λ I ) = det [ λ 0 1 0 1 λ 0 1 0 λ ] = ( 1 λ ) 2 ( 1 λ )
the eigenvalues are 1 and -1.
Is the subspace of symmetric matrices with 0 on the diagonal an eigenspace? This is easier without X: the eigenspace relative to -1 is the set of matrices A = [ a b b c ] such that T ( A ) = A or
[ c b b a ] = [ a b b c ]
So the condition is a=c and this is not having 0 on the diagonal.
The eigenspace relative to 1 is the space of matrices A such that T ( A ) = A and the condition reads
[ c b b a ] = [ a b b c ]
which gives a=c and b=0.
The fact that T = T 1 is obvious: just observe that T ( T ( A ) ) = A for any (symmetric) matrix A This already tells you that 2 is not an eigenvalue. Why?
Brooke Ayala

Brooke Ayala

Beginner2022-05-28Added 4 answers

Your answer is very detailed, and you explain the answer well. thank you!

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