Orthogonal Diagonalization of a $3$ by $3$ Matrix












0














$M$ $=$ $begin{pmatrix}3&2&2\ 2&3&2\ 2&2&3end{pmatrix}$. Diagonalize $M$ using an orthogonal matrix.



So I got that the eigenvalues for $M$ were $1$ and $7$. For the eigenvalue of $1$, I got the eigenvectors $begin{pmatrix}-1\ 0\ 1end{pmatrix}$ and $begin{pmatrix}-1\ 1\ 0end{pmatrix}$, and for the eigenvalue of $7$, I got the eigenvector $begin{pmatrix}1\ 1\ 1end{pmatrix}$. This gave me the diagonal matrix $begin{pmatrix}1&0&0\ 0&1&0\ 0&0&7end{pmatrix}$ and the orthogonal matrix $begin{pmatrix}-frac{1}{sqrt{2}}&-frac{1}{sqrt{2}}&frac{1}{sqrt{3}}\ 0&frac{1}{sqrt{2}}&frac{1}{sqrt{3}}\ frac{1}{sqrt{2}}&0&frac{1}{sqrt{3}}end{pmatrix}$.



But when I multiply the orthogonal matrix by the diagonal matrix and then its transpose, I get an answer that is slightly off what $M$ is, but I am not sure why.



If anyone knows where I may have gone wrong, I would greatly appreciate you telling me!






linear-algebra eigenvalues-eigenvectors diagonalization orthogonal-matrices







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  • 1




    Both eigenvectors for $;lambda=1;$ are wrong, as you can easily check. The eigenvectors corresponding to different eigenvalues are orthogonal (because the matrix is symmetric), so you must only do GM in each eigenspace...
    – DonAntonio
    Nov 19 '18 at 2:35


















0














$M$ $=$ $begin{pmatrix}3&2&2\ 2&3&2\ 2&2&3end{pmatrix}$. Diagonalize $M$ using an orthogonal matrix.



So I got that the eigenvalues for $M$ were $1$ and $7$. For the eigenvalue of $1$, I got the eigenvectors $begin{pmatrix}-1\ 0\ 1end{pmatrix}$ and $begin{pmatrix}-1\ 1\ 0end{pmatrix}$, and for the eigenvalue of $7$, I got the eigenvector $begin{pmatrix}1\ 1\ 1end{pmatrix}$. This gave me the diagonal matrix $begin{pmatrix}1&0&0\ 0&1&0\ 0&0&7end{pmatrix}$ and the orthogonal matrix $begin{pmatrix}-frac{1}{sqrt{2}}&-frac{1}{sqrt{2}}&frac{1}{sqrt{3}}\ 0&frac{1}{sqrt{2}}&frac{1}{sqrt{3}}\ frac{1}{sqrt{2}}&0&frac{1}{sqrt{3}}end{pmatrix}$.



But when I multiply the orthogonal matrix by the diagonal matrix and then its transpose, I get an answer that is slightly off what $M$ is, but I am not sure why.



If anyone knows where I may have gone wrong, I would greatly appreciate you telling me!






linear-algebra eigenvalues-eigenvectors diagonalization orthogonal-matrices







share|cite|improve this question


















  • 1




    Both eigenvectors for $;lambda=1;$ are wrong, as you can easily check. The eigenvectors corresponding to different eigenvalues are orthogonal (because the matrix is symmetric), so you must only do GM in each eigenspace...
    – DonAntonio
    Nov 19 '18 at 2:35
















0












0








0







$M$ $=$ $begin{pmatrix}3&2&2\ 2&3&2\ 2&2&3end{pmatrix}$. Diagonalize $M$ using an orthogonal matrix.



So I got that the eigenvalues for $M$ were $1$ and $7$. For the eigenvalue of $1$, I got the eigenvectors $begin{pmatrix}-1\ 0\ 1end{pmatrix}$ and $begin{pmatrix}-1\ 1\ 0end{pmatrix}$, and for the eigenvalue of $7$, I got the eigenvector $begin{pmatrix}1\ 1\ 1end{pmatrix}$. This gave me the diagonal matrix $begin{pmatrix}1&0&0\ 0&1&0\ 0&0&7end{pmatrix}$ and the orthogonal matrix $begin{pmatrix}-frac{1}{sqrt{2}}&-frac{1}{sqrt{2}}&frac{1}{sqrt{3}}\ 0&frac{1}{sqrt{2}}&frac{1}{sqrt{3}}\ frac{1}{sqrt{2}}&0&frac{1}{sqrt{3}}end{pmatrix}$.



But when I multiply the orthogonal matrix by the diagonal matrix and then its transpose, I get an answer that is slightly off what $M$ is, but I am not sure why.



If anyone knows where I may have gone wrong, I would greatly appreciate you telling me!






linear-algebra eigenvalues-eigenvectors diagonalization orthogonal-matrices







share|cite|improve this question













$M$ $=$ $begin{pmatrix}3&2&2\ 2&3&2\ 2&2&3end{pmatrix}$. Diagonalize $M$ using an orthogonal matrix.



So I got that the eigenvalues for $M$ were $1$ and $7$. For the eigenvalue of $1$, I got the eigenvectors $begin{pmatrix}-1\ 0\ 1end{pmatrix}$ and $begin{pmatrix}-1\ 1\ 0end{pmatrix}$, and for the eigenvalue of $7$, I got the eigenvector $begin{pmatrix}1\ 1\ 1end{pmatrix}$. This gave me the diagonal matrix $begin{pmatrix}1&0&0\ 0&1&0\ 0&0&7end{pmatrix}$ and the orthogonal matrix $begin{pmatrix}-frac{1}{sqrt{2}}&-frac{1}{sqrt{2}}&frac{1}{sqrt{3}}\ 0&frac{1}{sqrt{2}}&frac{1}{sqrt{3}}\ frac{1}{sqrt{2}}&0&frac{1}{sqrt{3}}end{pmatrix}$.



But when I multiply the orthogonal matrix by the diagonal matrix and then its transpose, I get an answer that is slightly off what $M$ is, but I am not sure why.



If anyone knows where I may have gone wrong, I would greatly appreciate you telling me!






linear-algebra eigenvalues-eigenvectors diagonalization orthogonal-matrices




linear-algebra eigenvalues-eigenvectors diagonalization orthogonal-matrices






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asked Nov 19 '18 at 2:11









Dev SR

859




859








  • 1




    Both eigenvectors for $;lambda=1;$ are wrong, as you can easily check. The eigenvectors corresponding to different eigenvalues are orthogonal (because the matrix is symmetric), so you must only do GM in each eigenspace...
    – DonAntonio
    Nov 19 '18 at 2:35
















  • 1




    Both eigenvectors for $;lambda=1;$ are wrong, as you can easily check. The eigenvectors corresponding to different eigenvalues are orthogonal (because the matrix is symmetric), so you must only do GM in each eigenspace...
    – DonAntonio
    Nov 19 '18 at 2:35










1




1




Both eigenvectors for $;lambda=1;$ are wrong, as you can easily check. The eigenvectors corresponding to different eigenvalues are orthogonal (because the matrix is symmetric), so you must only do GM in each eigenspace...
– DonAntonio
Nov 19 '18 at 2:35






Both eigenvectors for $;lambda=1;$ are wrong, as you can easily check. The eigenvectors corresponding to different eigenvalues are orthogonal (because the matrix is symmetric), so you must only do GM in each eigenspace...
– DonAntonio
Nov 19 '18 at 2:35












2 Answers
2






active

oldest

votes


















1














$$(-1, 0, 1) cdot (-1, 1, 0)=1$$



They are not orthogonal.



Just do a gram-schmidt step to find a set of orthogonal eigenvectors for eigenvalues corresponding to $1$.






share|cite|improve this answer





























    1














    start with



    $$
    left(
    begin{array}{rrr}
    1 & -1 & -1 \
    1 & 1 & -1 \
    1 & 0 & 2 \
    end{array}
    right).
    $$
    and divide the columns by $sqrt 3, sqrt 2, sqrt 6$



    If you had the analogous problem in 4 by 4, you could begin with



    $$
    left(
    begin{array}{rrrr}
    1 & -1 & -1 & -1 \
    1 & 1 & -1 & -1 \
    1 & 0 & 2 & -1 \
    1 & 0 & 0 & 3 \
    end{array}
    right).
    $$
    and divide the columns by $2,sqrt 2, sqrt 6, sqrt {12}$



    for 5 by 5
    $$
    left(
    begin{array}{rrrrr}
    1 & -1 & -1 & -1 & -1 \
    1 & 1 & -1 & -1 & -1 \
    1 & 0 & 2 & -1 & -1 \
    1 & 0 & 0 & 3 & -1 \
    1 & 0 & 0 & 0 & 4 \
    end{array}
    right).
    $$
    $sqrt 5,sqrt 2, sqrt 6, sqrt {12}, sqrt{20}$






    share|cite|improve this answer























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      $$(-1, 0, 1) cdot (-1, 1, 0)=1$$



      They are not orthogonal.



      Just do a gram-schmidt step to find a set of orthogonal eigenvectors for eigenvalues corresponding to $1$.






      share|cite|improve this answer


























        1














        $$(-1, 0, 1) cdot (-1, 1, 0)=1$$



        They are not orthogonal.



        Just do a gram-schmidt step to find a set of orthogonal eigenvectors for eigenvalues corresponding to $1$.






        share|cite|improve this answer
























          1












          1








          1






          $$(-1, 0, 1) cdot (-1, 1, 0)=1$$



          They are not orthogonal.



          Just do a gram-schmidt step to find a set of orthogonal eigenvectors for eigenvalues corresponding to $1$.






          share|cite|improve this answer












          $$(-1, 0, 1) cdot (-1, 1, 0)=1$$



          They are not orthogonal.



          Just do a gram-schmidt step to find a set of orthogonal eigenvectors for eigenvalues corresponding to $1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 '18 at 2:22









          Siong Thye Goh

          99.3k1464117




          99.3k1464117























              1














              start with



              $$
              left(
              begin{array}{rrr}
              1 & -1 & -1 \
              1 & 1 & -1 \
              1 & 0 & 2 \
              end{array}
              right).
              $$
              and divide the columns by $sqrt 3, sqrt 2, sqrt 6$



              If you had the analogous problem in 4 by 4, you could begin with



              $$
              left(
              begin{array}{rrrr}
              1 & -1 & -1 & -1 \
              1 & 1 & -1 & -1 \
              1 & 0 & 2 & -1 \
              1 & 0 & 0 & 3 \
              end{array}
              right).
              $$
              and divide the columns by $2,sqrt 2, sqrt 6, sqrt {12}$



              for 5 by 5
              $$
              left(
              begin{array}{rrrrr}
              1 & -1 & -1 & -1 & -1 \
              1 & 1 & -1 & -1 & -1 \
              1 & 0 & 2 & -1 & -1 \
              1 & 0 & 0 & 3 & -1 \
              1 & 0 & 0 & 0 & 4 \
              end{array}
              right).
              $$
              $sqrt 5,sqrt 2, sqrt 6, sqrt {12}, sqrt{20}$






              share|cite|improve this answer




























                1














                start with



                $$
                left(
                begin{array}{rrr}
                1 & -1 & -1 \
                1 & 1 & -1 \
                1 & 0 & 2 \
                end{array}
                right).
                $$
                and divide the columns by $sqrt 3, sqrt 2, sqrt 6$



                If you had the analogous problem in 4 by 4, you could begin with



                $$
                left(
                begin{array}{rrrr}
                1 & -1 & -1 & -1 \
                1 & 1 & -1 & -1 \
                1 & 0 & 2 & -1 \
                1 & 0 & 0 & 3 \
                end{array}
                right).
                $$
                and divide the columns by $2,sqrt 2, sqrt 6, sqrt {12}$



                for 5 by 5
                $$
                left(
                begin{array}{rrrrr}
                1 & -1 & -1 & -1 & -1 \
                1 & 1 & -1 & -1 & -1 \
                1 & 0 & 2 & -1 & -1 \
                1 & 0 & 0 & 3 & -1 \
                1 & 0 & 0 & 0 & 4 \
                end{array}
                right).
                $$
                $sqrt 5,sqrt 2, sqrt 6, sqrt {12}, sqrt{20}$






                share|cite|improve this answer


























                  1












                  1








                  1






                  start with



                  $$
                  left(
                  begin{array}{rrr}
                  1 & -1 & -1 \
                  1 & 1 & -1 \
                  1 & 0 & 2 \
                  end{array}
                  right).
                  $$
                  and divide the columns by $sqrt 3, sqrt 2, sqrt 6$



                  If you had the analogous problem in 4 by 4, you could begin with



                  $$
                  left(
                  begin{array}{rrrr}
                  1 & -1 & -1 & -1 \
                  1 & 1 & -1 & -1 \
                  1 & 0 & 2 & -1 \
                  1 & 0 & 0 & 3 \
                  end{array}
                  right).
                  $$
                  and divide the columns by $2,sqrt 2, sqrt 6, sqrt {12}$



                  for 5 by 5
                  $$
                  left(
                  begin{array}{rrrrr}
                  1 & -1 & -1 & -1 & -1 \
                  1 & 1 & -1 & -1 & -1 \
                  1 & 0 & 2 & -1 & -1 \
                  1 & 0 & 0 & 3 & -1 \
                  1 & 0 & 0 & 0 & 4 \
                  end{array}
                  right).
                  $$
                  $sqrt 5,sqrt 2, sqrt 6, sqrt {12}, sqrt{20}$






                  share|cite|improve this answer














                  start with



                  $$
                  left(
                  begin{array}{rrr}
                  1 & -1 & -1 \
                  1 & 1 & -1 \
                  1 & 0 & 2 \
                  end{array}
                  right).
                  $$
                  and divide the columns by $sqrt 3, sqrt 2, sqrt 6$



                  If you had the analogous problem in 4 by 4, you could begin with



                  $$
                  left(
                  begin{array}{rrrr}
                  1 & -1 & -1 & -1 \
                  1 & 1 & -1 & -1 \
                  1 & 0 & 2 & -1 \
                  1 & 0 & 0 & 3 \
                  end{array}
                  right).
                  $$
                  and divide the columns by $2,sqrt 2, sqrt 6, sqrt {12}$



                  for 5 by 5
                  $$
                  left(
                  begin{array}{rrrrr}
                  1 & -1 & -1 & -1 & -1 \
                  1 & 1 & -1 & -1 & -1 \
                  1 & 0 & 2 & -1 & -1 \
                  1 & 0 & 0 & 3 & -1 \
                  1 & 0 & 0 & 0 & 4 \
                  end{array}
                  right).
                  $$
                  $sqrt 5,sqrt 2, sqrt 6, sqrt {12}, sqrt{20}$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 19 '18 at 2:49

























                  answered Nov 19 '18 at 2:34









                  Will Jagy

                  101k599199




                  101k599199






























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