What does this linear transformation do?

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The following is given:



  • $F_4$ is the vector space of all symmetrical matrices

  • $F = operatornameSpanleftbeginbmatrix 1 & 0 \ 0 & 0 endbmatrix; beginbmatrix 0 & 1 \ 1 & 0 endbmatrix ; beginbmatrix 0 & 0 \ 0 & 1 endbmatrix right$

  • $ L:F rightarrow F : A rightarrow Abeginbmatrix 0 & 0 \ 0 & 1 endbmatrix + beginbmatrix 0 & 0 \ 0 & 1 endbmatrixA$

The following is asked:



  • What does this linear transformation do ?

  • Give an exact description of the kernel and range of the transformation.

  • Give the matrix representation M of this transformation

  • Eigenvalues and Eigenvectors of the transformation

  • What is the dimension of the Eigenspace ?

The kernel, range, matrix representation, eigenvalues and eigenvectors are pretty straightforward. I am struggling with the explanation of what this transformation does and the dimension of the Eigenspace. Can someone explain to me what this transformation does and hint me on the eigenspace question ?







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  • Have you written down the 5th point correctly? There are three eigenspaces ...
    – ancientmathematician
    Jul 30 at 8:59














up vote
2
down vote

favorite












The following is given:



  • $F_4$ is the vector space of all symmetrical matrices

  • $F = operatornameSpanleftbeginbmatrix 1 & 0 \ 0 & 0 endbmatrix; beginbmatrix 0 & 1 \ 1 & 0 endbmatrix ; beginbmatrix 0 & 0 \ 0 & 1 endbmatrix right$

  • $ L:F rightarrow F : A rightarrow Abeginbmatrix 0 & 0 \ 0 & 1 endbmatrix + beginbmatrix 0 & 0 \ 0 & 1 endbmatrixA$

The following is asked:



  • What does this linear transformation do ?

  • Give an exact description of the kernel and range of the transformation.

  • Give the matrix representation M of this transformation

  • Eigenvalues and Eigenvectors of the transformation

  • What is the dimension of the Eigenspace ?

The kernel, range, matrix representation, eigenvalues and eigenvectors are pretty straightforward. I am struggling with the explanation of what this transformation does and the dimension of the Eigenspace. Can someone explain to me what this transformation does and hint me on the eigenspace question ?







share|cite|improve this question





















  • Have you written down the 5th point correctly? There are three eigenspaces ...
    – ancientmathematician
    Jul 30 at 8:59












up vote
2
down vote

favorite









up vote
2
down vote

favorite











The following is given:



  • $F_4$ is the vector space of all symmetrical matrices

  • $F = operatornameSpanleftbeginbmatrix 1 & 0 \ 0 & 0 endbmatrix; beginbmatrix 0 & 1 \ 1 & 0 endbmatrix ; beginbmatrix 0 & 0 \ 0 & 1 endbmatrix right$

  • $ L:F rightarrow F : A rightarrow Abeginbmatrix 0 & 0 \ 0 & 1 endbmatrix + beginbmatrix 0 & 0 \ 0 & 1 endbmatrixA$

The following is asked:



  • What does this linear transformation do ?

  • Give an exact description of the kernel and range of the transformation.

  • Give the matrix representation M of this transformation

  • Eigenvalues and Eigenvectors of the transformation

  • What is the dimension of the Eigenspace ?

The kernel, range, matrix representation, eigenvalues and eigenvectors are pretty straightforward. I am struggling with the explanation of what this transformation does and the dimension of the Eigenspace. Can someone explain to me what this transformation does and hint me on the eigenspace question ?







share|cite|improve this question













The following is given:



  • $F_4$ is the vector space of all symmetrical matrices

  • $F = operatornameSpanleftbeginbmatrix 1 & 0 \ 0 & 0 endbmatrix; beginbmatrix 0 & 1 \ 1 & 0 endbmatrix ; beginbmatrix 0 & 0 \ 0 & 1 endbmatrix right$

  • $ L:F rightarrow F : A rightarrow Abeginbmatrix 0 & 0 \ 0 & 1 endbmatrix + beginbmatrix 0 & 0 \ 0 & 1 endbmatrixA$

The following is asked:



  • What does this linear transformation do ?

  • Give an exact description of the kernel and range of the transformation.

  • Give the matrix representation M of this transformation

  • Eigenvalues and Eigenvectors of the transformation

  • What is the dimension of the Eigenspace ?

The kernel, range, matrix representation, eigenvalues and eigenvectors are pretty straightforward. I am struggling with the explanation of what this transformation does and the dimension of the Eigenspace. Can someone explain to me what this transformation does and hint me on the eigenspace question ?









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edited Jul 30 at 1:04









Michael Hardy

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asked Jul 30 at 1:01









mbouchi

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  • Have you written down the 5th point correctly? There are three eigenspaces ...
    – ancientmathematician
    Jul 30 at 8:59
















  • Have you written down the 5th point correctly? There are three eigenspaces ...
    – ancientmathematician
    Jul 30 at 8:59















Have you written down the 5th point correctly? There are three eigenspaces ...
– ancientmathematician
Jul 30 at 8:59




Have you written down the 5th point correctly? There are three eigenspaces ...
– ancientmathematician
Jul 30 at 8:59










1 Answer
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Hint Write $A= beginbmatrix a & b \ c &d endbmatrix$. Then
$$L(A)= beginbmatrix a & b \ c &d endbmatrixrightarrow Abeginbmatrix 0 & 0 \ 0 & 1 endbmatrix + beginbmatrix 0 & 0 \ 0 & 1 endbmatrixbeginbmatrix a & b \ c &d endbmatrix= beginbmatrix ?? & ?? \ ?? &??endbmatrix$$



-Kernel: Solve $L(A)=0$.



-Range: describe all values of $L(A)$.



-Matrix Representation: calculate $L$ of the cannonical basis.



-Eigenvalues/Eigenvectors: For which $A$ do you get $L(A)=lambda A$?






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    1 Answer
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    1 Answer
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    Hint Write $A= beginbmatrix a & b \ c &d endbmatrix$. Then
    $$L(A)= beginbmatrix a & b \ c &d endbmatrixrightarrow Abeginbmatrix 0 & 0 \ 0 & 1 endbmatrix + beginbmatrix 0 & 0 \ 0 & 1 endbmatrixbeginbmatrix a & b \ c &d endbmatrix= beginbmatrix ?? & ?? \ ?? &??endbmatrix$$



    -Kernel: Solve $L(A)=0$.



    -Range: describe all values of $L(A)$.



    -Matrix Representation: calculate $L$ of the cannonical basis.



    -Eigenvalues/Eigenvectors: For which $A$ do you get $L(A)=lambda A$?






    share|cite|improve this answer

























      up vote
      1
      down vote













      Hint Write $A= beginbmatrix a & b \ c &d endbmatrix$. Then
      $$L(A)= beginbmatrix a & b \ c &d endbmatrixrightarrow Abeginbmatrix 0 & 0 \ 0 & 1 endbmatrix + beginbmatrix 0 & 0 \ 0 & 1 endbmatrixbeginbmatrix a & b \ c &d endbmatrix= beginbmatrix ?? & ?? \ ?? &??endbmatrix$$



      -Kernel: Solve $L(A)=0$.



      -Range: describe all values of $L(A)$.



      -Matrix Representation: calculate $L$ of the cannonical basis.



      -Eigenvalues/Eigenvectors: For which $A$ do you get $L(A)=lambda A$?






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        Hint Write $A= beginbmatrix a & b \ c &d endbmatrix$. Then
        $$L(A)= beginbmatrix a & b \ c &d endbmatrixrightarrow Abeginbmatrix 0 & 0 \ 0 & 1 endbmatrix + beginbmatrix 0 & 0 \ 0 & 1 endbmatrixbeginbmatrix a & b \ c &d endbmatrix= beginbmatrix ?? & ?? \ ?? &??endbmatrix$$



        -Kernel: Solve $L(A)=0$.



        -Range: describe all values of $L(A)$.



        -Matrix Representation: calculate $L$ of the cannonical basis.



        -Eigenvalues/Eigenvectors: For which $A$ do you get $L(A)=lambda A$?






        share|cite|improve this answer













        Hint Write $A= beginbmatrix a & b \ c &d endbmatrix$. Then
        $$L(A)= beginbmatrix a & b \ c &d endbmatrixrightarrow Abeginbmatrix 0 & 0 \ 0 & 1 endbmatrix + beginbmatrix 0 & 0 \ 0 & 1 endbmatrixbeginbmatrix a & b \ c &d endbmatrix= beginbmatrix ?? & ?? \ ?? &??endbmatrix$$



        -Kernel: Solve $L(A)=0$.



        -Range: describe all values of $L(A)$.



        -Matrix Representation: calculate $L$ of the cannonical basis.



        -Eigenvalues/Eigenvectors: For which $A$ do you get $L(A)=lambda A$?







        share|cite|improve this answer













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        share|cite|improve this answer











        answered Jul 30 at 1:12









        N. S.

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