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 ) = − 12. T − 1 = T3. T is linear4. 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

Question

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  )  =  −  12.       T          −      1        =  T3.   T is linear4. 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

sepolturamo · Accepted Answer

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  =  1The 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 · Accepted Answer

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

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

Answered question

Answer & Explanation

New Questions in Linear algebra