Example of diagonalisable matrix with given property












0














Let $Ain M_5(mathbb C)$ satisfying $(A^2-I)^2=0$ and $A$ is not diagonal matrix.



Then I have To find matrix A



But I tried but adding some terms in up to diagonal Nilpotency occur Which prevent form diagonalisable.



Please Help me to find example






linear-algebra examples-counterexamples diagonalization







share|cite|improve this question



























    0














    Let $Ain M_5(mathbb C)$ satisfying $(A^2-I)^2=0$ and $A$ is not diagonal matrix.



    Then I have To find matrix A



    But I tried but adding some terms in up to diagonal Nilpotency occur Which prevent form diagonalisable.



    Please Help me to find example






    linear-algebra examples-counterexamples diagonalization







    share|cite|improve this question

























      0












      0








      0







      Let $Ain M_5(mathbb C)$ satisfying $(A^2-I)^2=0$ and $A$ is not diagonal matrix.



      Then I have To find matrix A



      But I tried but adding some terms in up to diagonal Nilpotency occur Which prevent form diagonalisable.



      Please Help me to find example






      linear-algebra examples-counterexamples diagonalization







      share|cite|improve this question













      Let $Ain M_5(mathbb C)$ satisfying $(A^2-I)^2=0$ and $A$ is not diagonal matrix.



      Then I have To find matrix A



      But I tried but adding some terms in up to diagonal Nilpotency occur Which prevent form diagonalisable.



      Please Help me to find example






      linear-algebra examples-counterexamples diagonalization




      linear-algebra examples-counterexamples diagonalization






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 16:56









      MathLoverMathLover

      47910




      47910






















          3 Answers
          3






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          oldest

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          0














          Write $(A-Id)^2(A+Id)^2$ instead of $(A^2-I)^2$. Consider for instance the $(4, 4)$ matrix with diagonal entries $1,1,-1,-1, 1$, the ligne just over the diagonal $1,0,1,0$ and ll other coefficients $0$.



          You can also find diagonalisable matrix : any symmetry does the job.






          share|cite|improve this answer























          • Sir I want matrix other than diagonal matrix.I had mentioned that in question
            – MathLover
            Jan 5 at 5:19










          • But the line over the digaonal is (1,0,1,0) , please draw the matrix..
            – Thomas
            Jan 5 at 9:33










          • I tried but That is not diagonalisable
            – MathLover
            Jan 5 at 12:01



















          0














          This is an example:
          $$
          A= left[begin{array}{cccc}
          -11 & 6 & 0 & 0 & 0\
          -20 & 11 & 0 & 0 & 0\
          0 & 0 & -11 & 6 & 0\
          0 & 0 & -20 & 11 & 0\
          0 & 0 & 0 & 0 & 1end{array}right]
          $$






          share|cite|improve this answer























          • Sir but above matrix is not diagonalisable. I wanted Diagonalisable
            – MathLover
            Jan 5 at 11:57










          • @MathLover : I misunderstood your problem statement. Take a look at my new example.
            – DisintegratingByParts
            Jan 5 at 21:19



















          0














          If you want that the matrix is diagonalizable, the minimal polynomial must have distinct roots, so the candidate for the minimal polynomial is $x^2-1$. The eigenvalues must be $1$ and $-1$ (there must be at least two eigenvalues, because otherwise the matrix would necessarily be diagonal).



          One of them has multiplicity $3$ and the other one $2$. Let's find a $2times2$ nondiagonal matrix having eigenvalues $1$ and $-1$, for instance
          begin{bmatrix} 1 & 1 \ -2 & -1 end{bmatrix}
          Then your example is
          $$
          begin{bmatrix}
          1 & 1 & 0 & 0 & 0 \
          -2 & -1 & 0 & 0 & 0 \
          0 & 0 & 1 & 0 & 0 \
          0 & 0 & 0 & 1 & 0 \
          0 & 0 & 0 & 0 & -1
          end{bmatrix}
          $$






          share|cite|improve this answer





















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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0














            Write $(A-Id)^2(A+Id)^2$ instead of $(A^2-I)^2$. Consider for instance the $(4, 4)$ matrix with diagonal entries $1,1,-1,-1, 1$, the ligne just over the diagonal $1,0,1,0$ and ll other coefficients $0$.



            You can also find diagonalisable matrix : any symmetry does the job.






            share|cite|improve this answer























            • Sir I want matrix other than diagonal matrix.I had mentioned that in question
              – MathLover
              Jan 5 at 5:19










            • But the line over the digaonal is (1,0,1,0) , please draw the matrix..
              – Thomas
              Jan 5 at 9:33










            • I tried but That is not diagonalisable
              – MathLover
              Jan 5 at 12:01
















            0














            Write $(A-Id)^2(A+Id)^2$ instead of $(A^2-I)^2$. Consider for instance the $(4, 4)$ matrix with diagonal entries $1,1,-1,-1, 1$, the ligne just over the diagonal $1,0,1,0$ and ll other coefficients $0$.



            You can also find diagonalisable matrix : any symmetry does the job.






            share|cite|improve this answer























            • Sir I want matrix other than diagonal matrix.I had mentioned that in question
              – MathLover
              Jan 5 at 5:19










            • But the line over the digaonal is (1,0,1,0) , please draw the matrix..
              – Thomas
              Jan 5 at 9:33










            • I tried but That is not diagonalisable
              – MathLover
              Jan 5 at 12:01














            0












            0








            0






            Write $(A-Id)^2(A+Id)^2$ instead of $(A^2-I)^2$. Consider for instance the $(4, 4)$ matrix with diagonal entries $1,1,-1,-1, 1$, the ligne just over the diagonal $1,0,1,0$ and ll other coefficients $0$.



            You can also find diagonalisable matrix : any symmetry does the job.






            share|cite|improve this answer














            Write $(A-Id)^2(A+Id)^2$ instead of $(A^2-I)^2$. Consider for instance the $(4, 4)$ matrix with diagonal entries $1,1,-1,-1, 1$, the ligne just over the diagonal $1,0,1,0$ and ll other coefficients $0$.



            You can also find diagonalisable matrix : any symmetry does the job.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 5 at 9:33

























            answered Jan 4 at 17:39









            ThomasThomas

            3,874510




            3,874510












            • Sir I want matrix other than diagonal matrix.I had mentioned that in question
              – MathLover
              Jan 5 at 5:19










            • But the line over the digaonal is (1,0,1,0) , please draw the matrix..
              – Thomas
              Jan 5 at 9:33










            • I tried but That is not diagonalisable
              – MathLover
              Jan 5 at 12:01


















            • Sir I want matrix other than diagonal matrix.I had mentioned that in question
              – MathLover
              Jan 5 at 5:19










            • But the line over the digaonal is (1,0,1,0) , please draw the matrix..
              – Thomas
              Jan 5 at 9:33










            • I tried but That is not diagonalisable
              – MathLover
              Jan 5 at 12:01
















            Sir I want matrix other than diagonal matrix.I had mentioned that in question
            – MathLover
            Jan 5 at 5:19




            Sir I want matrix other than diagonal matrix.I had mentioned that in question
            – MathLover
            Jan 5 at 5:19












            But the line over the digaonal is (1,0,1,0) , please draw the matrix..
            – Thomas
            Jan 5 at 9:33




            But the line over the digaonal is (1,0,1,0) , please draw the matrix..
            – Thomas
            Jan 5 at 9:33












            I tried but That is not diagonalisable
            – MathLover
            Jan 5 at 12:01




            I tried but That is not diagonalisable
            – MathLover
            Jan 5 at 12:01











            0














            This is an example:
            $$
            A= left[begin{array}{cccc}
            -11 & 6 & 0 & 0 & 0\
            -20 & 11 & 0 & 0 & 0\
            0 & 0 & -11 & 6 & 0\
            0 & 0 & -20 & 11 & 0\
            0 & 0 & 0 & 0 & 1end{array}right]
            $$






            share|cite|improve this answer























            • Sir but above matrix is not diagonalisable. I wanted Diagonalisable
              – MathLover
              Jan 5 at 11:57










            • @MathLover : I misunderstood your problem statement. Take a look at my new example.
              – DisintegratingByParts
              Jan 5 at 21:19
















            0














            This is an example:
            $$
            A= left[begin{array}{cccc}
            -11 & 6 & 0 & 0 & 0\
            -20 & 11 & 0 & 0 & 0\
            0 & 0 & -11 & 6 & 0\
            0 & 0 & -20 & 11 & 0\
            0 & 0 & 0 & 0 & 1end{array}right]
            $$






            share|cite|improve this answer























            • Sir but above matrix is not diagonalisable. I wanted Diagonalisable
              – MathLover
              Jan 5 at 11:57










            • @MathLover : I misunderstood your problem statement. Take a look at my new example.
              – DisintegratingByParts
              Jan 5 at 21:19














            0












            0








            0






            This is an example:
            $$
            A= left[begin{array}{cccc}
            -11 & 6 & 0 & 0 & 0\
            -20 & 11 & 0 & 0 & 0\
            0 & 0 & -11 & 6 & 0\
            0 & 0 & -20 & 11 & 0\
            0 & 0 & 0 & 0 & 1end{array}right]
            $$






            share|cite|improve this answer














            This is an example:
            $$
            A= left[begin{array}{cccc}
            -11 & 6 & 0 & 0 & 0\
            -20 & 11 & 0 & 0 & 0\
            0 & 0 & -11 & 6 & 0\
            0 & 0 & -20 & 11 & 0\
            0 & 0 & 0 & 0 & 1end{array}right]
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 5 at 21:39

























            answered Jan 5 at 8:35









            DisintegratingByPartsDisintegratingByParts

            58.7k42579




            58.7k42579












            • Sir but above matrix is not diagonalisable. I wanted Diagonalisable
              – MathLover
              Jan 5 at 11:57










            • @MathLover : I misunderstood your problem statement. Take a look at my new example.
              – DisintegratingByParts
              Jan 5 at 21:19


















            • Sir but above matrix is not diagonalisable. I wanted Diagonalisable
              – MathLover
              Jan 5 at 11:57










            • @MathLover : I misunderstood your problem statement. Take a look at my new example.
              – DisintegratingByParts
              Jan 5 at 21:19
















            Sir but above matrix is not diagonalisable. I wanted Diagonalisable
            – MathLover
            Jan 5 at 11:57




            Sir but above matrix is not diagonalisable. I wanted Diagonalisable
            – MathLover
            Jan 5 at 11:57












            @MathLover : I misunderstood your problem statement. Take a look at my new example.
            – DisintegratingByParts
            Jan 5 at 21:19




            @MathLover : I misunderstood your problem statement. Take a look at my new example.
            – DisintegratingByParts
            Jan 5 at 21:19











            0














            If you want that the matrix is diagonalizable, the minimal polynomial must have distinct roots, so the candidate for the minimal polynomial is $x^2-1$. The eigenvalues must be $1$ and $-1$ (there must be at least two eigenvalues, because otherwise the matrix would necessarily be diagonal).



            One of them has multiplicity $3$ and the other one $2$. Let's find a $2times2$ nondiagonal matrix having eigenvalues $1$ and $-1$, for instance
            begin{bmatrix} 1 & 1 \ -2 & -1 end{bmatrix}
            Then your example is
            $$
            begin{bmatrix}
            1 & 1 & 0 & 0 & 0 \
            -2 & -1 & 0 & 0 & 0 \
            0 & 0 & 1 & 0 & 0 \
            0 & 0 & 0 & 1 & 0 \
            0 & 0 & 0 & 0 & -1
            end{bmatrix}
            $$






            share|cite|improve this answer


























              0














              If you want that the matrix is diagonalizable, the minimal polynomial must have distinct roots, so the candidate for the minimal polynomial is $x^2-1$. The eigenvalues must be $1$ and $-1$ (there must be at least two eigenvalues, because otherwise the matrix would necessarily be diagonal).



              One of them has multiplicity $3$ and the other one $2$. Let's find a $2times2$ nondiagonal matrix having eigenvalues $1$ and $-1$, for instance
              begin{bmatrix} 1 & 1 \ -2 & -1 end{bmatrix}
              Then your example is
              $$
              begin{bmatrix}
              1 & 1 & 0 & 0 & 0 \
              -2 & -1 & 0 & 0 & 0 \
              0 & 0 & 1 & 0 & 0 \
              0 & 0 & 0 & 1 & 0 \
              0 & 0 & 0 & 0 & -1
              end{bmatrix}
              $$






              share|cite|improve this answer
























                0












                0








                0






                If you want that the matrix is diagonalizable, the minimal polynomial must have distinct roots, so the candidate for the minimal polynomial is $x^2-1$. The eigenvalues must be $1$ and $-1$ (there must be at least two eigenvalues, because otherwise the matrix would necessarily be diagonal).



                One of them has multiplicity $3$ and the other one $2$. Let's find a $2times2$ nondiagonal matrix having eigenvalues $1$ and $-1$, for instance
                begin{bmatrix} 1 & 1 \ -2 & -1 end{bmatrix}
                Then your example is
                $$
                begin{bmatrix}
                1 & 1 & 0 & 0 & 0 \
                -2 & -1 & 0 & 0 & 0 \
                0 & 0 & 1 & 0 & 0 \
                0 & 0 & 0 & 1 & 0 \
                0 & 0 & 0 & 0 & -1
                end{bmatrix}
                $$






                share|cite|improve this answer












                If you want that the matrix is diagonalizable, the minimal polynomial must have distinct roots, so the candidate for the minimal polynomial is $x^2-1$. The eigenvalues must be $1$ and $-1$ (there must be at least two eigenvalues, because otherwise the matrix would necessarily be diagonal).



                One of them has multiplicity $3$ and the other one $2$. Let's find a $2times2$ nondiagonal matrix having eigenvalues $1$ and $-1$, for instance
                begin{bmatrix} 1 & 1 \ -2 & -1 end{bmatrix}
                Then your example is
                $$
                begin{bmatrix}
                1 & 1 & 0 & 0 & 0 \
                -2 & -1 & 0 & 0 & 0 \
                0 & 0 & 1 & 0 & 0 \
                0 & 0 & 0 & 1 & 0 \
                0 & 0 & 0 & 0 & -1
                end{bmatrix}
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 0:02









                egregegreg

                179k1485202




                179k1485202






























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