On $M_{2×2} (mathbb R)$ define an inner product by $langle A,B rangle = operatorname{tr}(A^T B)$ where $A,B...












0












$begingroup$



On $M_{2×2} (mathbb R)$ define an inner product by $langle A,B rangle = operatorname{tr}(A^T B)$, where $A,B in M_{2×2}(mathbb R)$




Consider the set $W={A in M_{2×2} (mathbb R) : A^T=A, operatorname{tr}(A)=0}$.



Prove that W is a subspace and find $W^perp$.
First proof is done.please give me some hint for the second problem..






linear-algebra inner-product-space







share|cite|improve this question











$endgroup$



closed as off-topic by A. Pongrácz, mrtaurho, Saad, Eevee Trainer, KReiser Jan 7 at 3:35


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – A. Pongrácz, mrtaurho, Saad, Eevee Trainer, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    How did you prove the first claim? Then we can help you better with the second one.
    $endgroup$
    – Dietrich Burde
    Jan 6 at 16:29










  • $begingroup$
    The Null matrix belongs to the set W thus W is non empty .now. For A,B matrix belongs to W and lamda in $mathbb R$....$A+lamda B$ in the set W .hence W is a subspace. Please suggest some edit of my question..I am not too much familiar with this.
    $endgroup$
    – Supriyo Banerjee
    Jan 6 at 16:34












  • $begingroup$
    How do you define $W^{T}?$
    $endgroup$
    – Hello_World
    Jan 6 at 17:00










  • $begingroup$
    $W^T$ is the set of all orthogonal elements with the elements of W with respect to the difend inner product.
    $endgroup$
    – Supriyo Banerjee
    Jan 6 at 17:04
















0












$begingroup$



On $M_{2×2} (mathbb R)$ define an inner product by $langle A,B rangle = operatorname{tr}(A^T B)$, where $A,B in M_{2×2}(mathbb R)$




Consider the set $W={A in M_{2×2} (mathbb R) : A^T=A, operatorname{tr}(A)=0}$.



Prove that W is a subspace and find $W^perp$.
First proof is done.please give me some hint for the second problem..






linear-algebra inner-product-space







share|cite|improve this question











$endgroup$



closed as off-topic by A. Pongrácz, mrtaurho, Saad, Eevee Trainer, KReiser Jan 7 at 3:35


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – A. Pongrácz, mrtaurho, Saad, Eevee Trainer, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    How did you prove the first claim? Then we can help you better with the second one.
    $endgroup$
    – Dietrich Burde
    Jan 6 at 16:29










  • $begingroup$
    The Null matrix belongs to the set W thus W is non empty .now. For A,B matrix belongs to W and lamda in $mathbb R$....$A+lamda B$ in the set W .hence W is a subspace. Please suggest some edit of my question..I am not too much familiar with this.
    $endgroup$
    – Supriyo Banerjee
    Jan 6 at 16:34












  • $begingroup$
    How do you define $W^{T}?$
    $endgroup$
    – Hello_World
    Jan 6 at 17:00










  • $begingroup$
    $W^T$ is the set of all orthogonal elements with the elements of W with respect to the difend inner product.
    $endgroup$
    – Supriyo Banerjee
    Jan 6 at 17:04














0












0








0


0



$begingroup$



On $M_{2×2} (mathbb R)$ define an inner product by $langle A,B rangle = operatorname{tr}(A^T B)$, where $A,B in M_{2×2}(mathbb R)$




Consider the set $W={A in M_{2×2} (mathbb R) : A^T=A, operatorname{tr}(A)=0}$.



Prove that W is a subspace and find $W^perp$.
First proof is done.please give me some hint for the second problem..






linear-algebra inner-product-space







share|cite|improve this question











$endgroup$





On $M_{2×2} (mathbb R)$ define an inner product by $langle A,B rangle = operatorname{tr}(A^T B)$, where $A,B in M_{2×2}(mathbb R)$




Consider the set $W={A in M_{2×2} (mathbb R) : A^T=A, operatorname{tr}(A)=0}$.



Prove that W is a subspace and find $W^perp$.
First proof is done.please give me some hint for the second problem..






linear-algebra inner-product-space




linear-algebra inner-product-space






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 17:29









egreg

180k1485202




180k1485202










asked Jan 6 at 16:26









Supriyo BanerjeeSupriyo Banerjee

1186




1186




closed as off-topic by A. Pongrácz, mrtaurho, Saad, Eevee Trainer, KReiser Jan 7 at 3:35


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – A. Pongrácz, mrtaurho, Saad, Eevee Trainer, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by A. Pongrácz, mrtaurho, Saad, Eevee Trainer, KReiser Jan 7 at 3:35


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – A. Pongrácz, mrtaurho, Saad, Eevee Trainer, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    How did you prove the first claim? Then we can help you better with the second one.
    $endgroup$
    – Dietrich Burde
    Jan 6 at 16:29










  • $begingroup$
    The Null matrix belongs to the set W thus W is non empty .now. For A,B matrix belongs to W and lamda in $mathbb R$....$A+lamda B$ in the set W .hence W is a subspace. Please suggest some edit of my question..I am not too much familiar with this.
    $endgroup$
    – Supriyo Banerjee
    Jan 6 at 16:34












  • $begingroup$
    How do you define $W^{T}?$
    $endgroup$
    – Hello_World
    Jan 6 at 17:00










  • $begingroup$
    $W^T$ is the set of all orthogonal elements with the elements of W with respect to the difend inner product.
    $endgroup$
    – Supriyo Banerjee
    Jan 6 at 17:04














  • 2




    $begingroup$
    How did you prove the first claim? Then we can help you better with the second one.
    $endgroup$
    – Dietrich Burde
    Jan 6 at 16:29










  • $begingroup$
    The Null matrix belongs to the set W thus W is non empty .now. For A,B matrix belongs to W and lamda in $mathbb R$....$A+lamda B$ in the set W .hence W is a subspace. Please suggest some edit of my question..I am not too much familiar with this.
    $endgroup$
    – Supriyo Banerjee
    Jan 6 at 16:34












  • $begingroup$
    How do you define $W^{T}?$
    $endgroup$
    – Hello_World
    Jan 6 at 17:00










  • $begingroup$
    $W^T$ is the set of all orthogonal elements with the elements of W with respect to the difend inner product.
    $endgroup$
    – Supriyo Banerjee
    Jan 6 at 17:04








2




2




$begingroup$
How did you prove the first claim? Then we can help you better with the second one.
$endgroup$
– Dietrich Burde
Jan 6 at 16:29




$begingroup$
How did you prove the first claim? Then we can help you better with the second one.
$endgroup$
– Dietrich Burde
Jan 6 at 16:29












$begingroup$
The Null matrix belongs to the set W thus W is non empty .now. For A,B matrix belongs to W and lamda in $mathbb R$....$A+lamda B$ in the set W .hence W is a subspace. Please suggest some edit of my question..I am not too much familiar with this.
$endgroup$
– Supriyo Banerjee
Jan 6 at 16:34






$begingroup$
The Null matrix belongs to the set W thus W is non empty .now. For A,B matrix belongs to W and lamda in $mathbb R$....$A+lamda B$ in the set W .hence W is a subspace. Please suggest some edit of my question..I am not too much familiar with this.
$endgroup$
– Supriyo Banerjee
Jan 6 at 16:34














$begingroup$
How do you define $W^{T}?$
$endgroup$
– Hello_World
Jan 6 at 17:00




$begingroup$
How do you define $W^{T}?$
$endgroup$
– Hello_World
Jan 6 at 17:00












$begingroup$
$W^T$ is the set of all orthogonal elements with the elements of W with respect to the difend inner product.
$endgroup$
– Supriyo Banerjee
Jan 6 at 17:04




$begingroup$
$W^T$ is the set of all orthogonal elements with the elements of W with respect to the difend inner product.
$endgroup$
– Supriyo Banerjee
Jan 6 at 17:04










3 Answers
3






active

oldest

votes


















1












$begingroup$

Notice that $dim W = 2$ so $dim W^perp = 4 - dim W =2$.



For any $A =begin{bmatrix} a & b \ b &-aend{bmatrix} in W$ we have



$$langle I,Arangle =operatorname{Tr}(A^TI) = operatorname{Tr}(A) = 0$$
$$leftlangle begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix},Arightrangle =operatorname{Tr}left(A^Tbegin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}right) = operatorname{Tr}left(Abegin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}right) = operatorname{Tr}left(begin{bmatrix} b & -a \ -a & -bend{bmatrix}right) =0$$



so $I, begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix} in W^perp$. Since they are linearly independent, we conclude they form a basis for $W^perp$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    You space $W$ consists of symmetric traceless matrices, so these are of the form
    $begin{pmatrix} a & b \ b & -a end{pmatrix}$, and are two dimensional, spanned by the matrices $A =begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix}$ and $B =begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}$. Since the complement has to be two dimensional, it suffices you find two matrices $A'$ and $B'$ that are orthogonal to $A$ and $B$ and linearly independent. One of them is the identity matrix, so what is the other? I would think about complex numbers here.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      The map
      $$
      A=begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix}
      mapsto
      f(A)=begin{bmatrix} a_{11} \ a_{12} \ a_{21} \ a_{22} end{bmatrix}
      $$
      has the property that
      $$
      operatorname{tr}(A^TB)=f(A)^TB
      $$
      so the inner product is transformed into the standard inner product in $mathbb{R}^4$. A basis of $W$ is given by
      $$
      left{A_1=begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix},
      A_2=begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}right}
      $$
      so a basis of $f(W)$ is given by
      $$
      left{v_1=begin{bmatrix} 1 \ 0 \ 0 \ -1 end{bmatrix},
      v_2=begin{bmatrix} 0 \ 1 \ 1 \ 0 end{bmatrix},right}
      $$
      Now it's a matter of finding a basis for the orthogonal complement of $operatorname{span}{v_1,v_2}$ and of going back to the elements of $M_{2times2}(mathbb{R})$ its vectors correspond to through $f$.






      share|cite|improve this answer









      $endgroup$




















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Notice that $dim W = 2$ so $dim W^perp = 4 - dim W =2$.



        For any $A =begin{bmatrix} a & b \ b &-aend{bmatrix} in W$ we have



        $$langle I,Arangle =operatorname{Tr}(A^TI) = operatorname{Tr}(A) = 0$$
        $$leftlangle begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix},Arightrangle =operatorname{Tr}left(A^Tbegin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}right) = operatorname{Tr}left(Abegin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}right) = operatorname{Tr}left(begin{bmatrix} b & -a \ -a & -bend{bmatrix}right) =0$$



        so $I, begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix} in W^perp$. Since they are linearly independent, we conclude they form a basis for $W^perp$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Notice that $dim W = 2$ so $dim W^perp = 4 - dim W =2$.



          For any $A =begin{bmatrix} a & b \ b &-aend{bmatrix} in W$ we have



          $$langle I,Arangle =operatorname{Tr}(A^TI) = operatorname{Tr}(A) = 0$$
          $$leftlangle begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix},Arightrangle =operatorname{Tr}left(A^Tbegin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}right) = operatorname{Tr}left(Abegin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}right) = operatorname{Tr}left(begin{bmatrix} b & -a \ -a & -bend{bmatrix}right) =0$$



          so $I, begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix} in W^perp$. Since they are linearly independent, we conclude they form a basis for $W^perp$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Notice that $dim W = 2$ so $dim W^perp = 4 - dim W =2$.



            For any $A =begin{bmatrix} a & b \ b &-aend{bmatrix} in W$ we have



            $$langle I,Arangle =operatorname{Tr}(A^TI) = operatorname{Tr}(A) = 0$$
            $$leftlangle begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix},Arightrangle =operatorname{Tr}left(A^Tbegin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}right) = operatorname{Tr}left(Abegin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}right) = operatorname{Tr}left(begin{bmatrix} b & -a \ -a & -bend{bmatrix}right) =0$$



            so $I, begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix} in W^perp$. Since they are linearly independent, we conclude they form a basis for $W^perp$






            share|cite|improve this answer









            $endgroup$



            Notice that $dim W = 2$ so $dim W^perp = 4 - dim W =2$.



            For any $A =begin{bmatrix} a & b \ b &-aend{bmatrix} in W$ we have



            $$langle I,Arangle =operatorname{Tr}(A^TI) = operatorname{Tr}(A) = 0$$
            $$leftlangle begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix},Arightrangle =operatorname{Tr}left(A^Tbegin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}right) = operatorname{Tr}left(Abegin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}right) = operatorname{Tr}left(begin{bmatrix} b & -a \ -a & -bend{bmatrix}right) =0$$



            so $I, begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix} in W^perp$. Since they are linearly independent, we conclude they form a basis for $W^perp$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 19:12









            mechanodroidmechanodroid

            27.3k62446




            27.3k62446























                1












                $begingroup$

                You space $W$ consists of symmetric traceless matrices, so these are of the form
                $begin{pmatrix} a & b \ b & -a end{pmatrix}$, and are two dimensional, spanned by the matrices $A =begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix}$ and $B =begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}$. Since the complement has to be two dimensional, it suffices you find two matrices $A'$ and $B'$ that are orthogonal to $A$ and $B$ and linearly independent. One of them is the identity matrix, so what is the other? I would think about complex numbers here.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  You space $W$ consists of symmetric traceless matrices, so these are of the form
                  $begin{pmatrix} a & b \ b & -a end{pmatrix}$, and are two dimensional, spanned by the matrices $A =begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix}$ and $B =begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}$. Since the complement has to be two dimensional, it suffices you find two matrices $A'$ and $B'$ that are orthogonal to $A$ and $B$ and linearly independent. One of them is the identity matrix, so what is the other? I would think about complex numbers here.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    You space $W$ consists of symmetric traceless matrices, so these are of the form
                    $begin{pmatrix} a & b \ b & -a end{pmatrix}$, and are two dimensional, spanned by the matrices $A =begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix}$ and $B =begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}$. Since the complement has to be two dimensional, it suffices you find two matrices $A'$ and $B'$ that are orthogonal to $A$ and $B$ and linearly independent. One of them is the identity matrix, so what is the other? I would think about complex numbers here.






                    share|cite|improve this answer









                    $endgroup$



                    You space $W$ consists of symmetric traceless matrices, so these are of the form
                    $begin{pmatrix} a & b \ b & -a end{pmatrix}$, and are two dimensional, spanned by the matrices $A =begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix}$ and $B =begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}$. Since the complement has to be two dimensional, it suffices you find two matrices $A'$ and $B'$ that are orthogonal to $A$ and $B$ and linearly independent. One of them is the identity matrix, so what is the other? I would think about complex numbers here.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 6 at 17:11









                    Pedro Tamaroff♦Pedro Tamaroff

                    96.6k10153297




                    96.6k10153297























                        1












                        $begingroup$

                        The map
                        $$
                        A=begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix}
                        mapsto
                        f(A)=begin{bmatrix} a_{11} \ a_{12} \ a_{21} \ a_{22} end{bmatrix}
                        $$
                        has the property that
                        $$
                        operatorname{tr}(A^TB)=f(A)^TB
                        $$
                        so the inner product is transformed into the standard inner product in $mathbb{R}^4$. A basis of $W$ is given by
                        $$
                        left{A_1=begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix},
                        A_2=begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}right}
                        $$
                        so a basis of $f(W)$ is given by
                        $$
                        left{v_1=begin{bmatrix} 1 \ 0 \ 0 \ -1 end{bmatrix},
                        v_2=begin{bmatrix} 0 \ 1 \ 1 \ 0 end{bmatrix},right}
                        $$
                        Now it's a matter of finding a basis for the orthogonal complement of $operatorname{span}{v_1,v_2}$ and of going back to the elements of $M_{2times2}(mathbb{R})$ its vectors correspond to through $f$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          The map
                          $$
                          A=begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix}
                          mapsto
                          f(A)=begin{bmatrix} a_{11} \ a_{12} \ a_{21} \ a_{22} end{bmatrix}
                          $$
                          has the property that
                          $$
                          operatorname{tr}(A^TB)=f(A)^TB
                          $$
                          so the inner product is transformed into the standard inner product in $mathbb{R}^4$. A basis of $W$ is given by
                          $$
                          left{A_1=begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix},
                          A_2=begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}right}
                          $$
                          so a basis of $f(W)$ is given by
                          $$
                          left{v_1=begin{bmatrix} 1 \ 0 \ 0 \ -1 end{bmatrix},
                          v_2=begin{bmatrix} 0 \ 1 \ 1 \ 0 end{bmatrix},right}
                          $$
                          Now it's a matter of finding a basis for the orthogonal complement of $operatorname{span}{v_1,v_2}$ and of going back to the elements of $M_{2times2}(mathbb{R})$ its vectors correspond to through $f$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            The map
                            $$
                            A=begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix}
                            mapsto
                            f(A)=begin{bmatrix} a_{11} \ a_{12} \ a_{21} \ a_{22} end{bmatrix}
                            $$
                            has the property that
                            $$
                            operatorname{tr}(A^TB)=f(A)^TB
                            $$
                            so the inner product is transformed into the standard inner product in $mathbb{R}^4$. A basis of $W$ is given by
                            $$
                            left{A_1=begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix},
                            A_2=begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}right}
                            $$
                            so a basis of $f(W)$ is given by
                            $$
                            left{v_1=begin{bmatrix} 1 \ 0 \ 0 \ -1 end{bmatrix},
                            v_2=begin{bmatrix} 0 \ 1 \ 1 \ 0 end{bmatrix},right}
                            $$
                            Now it's a matter of finding a basis for the orthogonal complement of $operatorname{span}{v_1,v_2}$ and of going back to the elements of $M_{2times2}(mathbb{R})$ its vectors correspond to through $f$.






                            share|cite|improve this answer









                            $endgroup$



                            The map
                            $$
                            A=begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix}
                            mapsto
                            f(A)=begin{bmatrix} a_{11} \ a_{12} \ a_{21} \ a_{22} end{bmatrix}
                            $$
                            has the property that
                            $$
                            operatorname{tr}(A^TB)=f(A)^TB
                            $$
                            so the inner product is transformed into the standard inner product in $mathbb{R}^4$. A basis of $W$ is given by
                            $$
                            left{A_1=begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix},
                            A_2=begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}right}
                            $$
                            so a basis of $f(W)$ is given by
                            $$
                            left{v_1=begin{bmatrix} 1 \ 0 \ 0 \ -1 end{bmatrix},
                            v_2=begin{bmatrix} 0 \ 1 \ 1 \ 0 end{bmatrix},right}
                            $$
                            Now it's a matter of finding a basis for the orthogonal complement of $operatorname{span}{v_1,v_2}$ and of going back to the elements of $M_{2times2}(mathbb{R})$ its vectors correspond to through $f$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 6 at 18:58









                            egregegreg

                            180k1485202




                            180k1485202















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                                in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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                                1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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