Similar Matrices Confer Equivalent Linear Transformations












1












$begingroup$


Apologies - I know there are some similar questions on this site, but I couldn't quite apply what they are saying here.



Let's say I have the following matrices:
$$A=begin{bmatrix}-1&6\-2&6end{bmatrix}; B=begin{bmatrix}1&2\-1&4end{bmatrix}$$
I conclude that they are similar (without finding $P$ such that $PAP^{-1} = B$) because both have eigenvalues 2,3 and thus will be similar to $D=begin{bmatrix}2&0\0&3end{bmatrix}$.



If we let $x=begin{bmatrix}2\3 end{bmatrix}$, we have:
$$Ax=begin{bmatrix}16\14end{bmatrix}; Bx=begin{bmatrix}8\10end{bmatrix}$$



I was told, however, that if two matrices are similar then they induce the same linear transformation. How can they do this if they map the same vector to a different vector? What is meant by 'same' here?



Thanks for help and sorry if an elementary question.






linear-algebra matrices linear-transformations







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Apologies - I know there are some similar questions on this site, but I couldn't quite apply what they are saying here.



    Let's say I have the following matrices:
    $$A=begin{bmatrix}-1&6\-2&6end{bmatrix}; B=begin{bmatrix}1&2\-1&4end{bmatrix}$$
    I conclude that they are similar (without finding $P$ such that $PAP^{-1} = B$) because both have eigenvalues 2,3 and thus will be similar to $D=begin{bmatrix}2&0\0&3end{bmatrix}$.



    If we let $x=begin{bmatrix}2\3 end{bmatrix}$, we have:
    $$Ax=begin{bmatrix}16\14end{bmatrix}; Bx=begin{bmatrix}8\10end{bmatrix}$$



    I was told, however, that if two matrices are similar then they induce the same linear transformation. How can they do this if they map the same vector to a different vector? What is meant by 'same' here?



    Thanks for help and sorry if an elementary question.






    linear-algebra matrices linear-transformations







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Apologies - I know there are some similar questions on this site, but I couldn't quite apply what they are saying here.



      Let's say I have the following matrices:
      $$A=begin{bmatrix}-1&6\-2&6end{bmatrix}; B=begin{bmatrix}1&2\-1&4end{bmatrix}$$
      I conclude that they are similar (without finding $P$ such that $PAP^{-1} = B$) because both have eigenvalues 2,3 and thus will be similar to $D=begin{bmatrix}2&0\0&3end{bmatrix}$.



      If we let $x=begin{bmatrix}2\3 end{bmatrix}$, we have:
      $$Ax=begin{bmatrix}16\14end{bmatrix}; Bx=begin{bmatrix}8\10end{bmatrix}$$



      I was told, however, that if two matrices are similar then they induce the same linear transformation. How can they do this if they map the same vector to a different vector? What is meant by 'same' here?



      Thanks for help and sorry if an elementary question.






      linear-algebra matrices linear-transformations







      share|cite|improve this question









      $endgroup$




      Apologies - I know there are some similar questions on this site, but I couldn't quite apply what they are saying here.



      Let's say I have the following matrices:
      $$A=begin{bmatrix}-1&6\-2&6end{bmatrix}; B=begin{bmatrix}1&2\-1&4end{bmatrix}$$
      I conclude that they are similar (without finding $P$ such that $PAP^{-1} = B$) because both have eigenvalues 2,3 and thus will be similar to $D=begin{bmatrix}2&0\0&3end{bmatrix}$.



      If we let $x=begin{bmatrix}2\3 end{bmatrix}$, we have:
      $$Ax=begin{bmatrix}16\14end{bmatrix}; Bx=begin{bmatrix}8\10end{bmatrix}$$



      I was told, however, that if two matrices are similar then they induce the same linear transformation. How can they do this if they map the same vector to a different vector? What is meant by 'same' here?



      Thanks for help and sorry if an elementary question.






      linear-algebra matrices linear-transformations




      linear-algebra matrices linear-transformations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 5:15









      AndrewAndrew

      346211




      346211






















          1 Answer
          1






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          0












          $begingroup$

          When you write



          $$x = begin{bmatrix} 2\3end{bmatrix}$$



          what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $mathbb{R}^2$. Usually, we assume that



          $$v_1 = begin{bmatrix} 1\0end{bmatrix}qquad v_2 = begin{bmatrix} 0\1end{bmatrix}, $$



          but we don’t need to. Given a vector $x$ and a basis $alpha = {alpha_1, alpha_2}$, if $x = aalpha_1 + balpha_2$, I’ll be explicit and write that



          $$[x]_{alpha} = begin{bmatrix} a\bend{bmatrix}. $$



          Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $alpha$ and $beta$ such that given any $xinmathbb{R}^2$



          $$A[x]_{alpha} = B[x]_{beta}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
            $endgroup$
            – Andrew
            Jan 4 at 9:20










          • $begingroup$
            You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
            $endgroup$
            – Santana Afton
            Jan 4 at 14:41











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          0












          $begingroup$

          When you write



          $$x = begin{bmatrix} 2\3end{bmatrix}$$



          what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $mathbb{R}^2$. Usually, we assume that



          $$v_1 = begin{bmatrix} 1\0end{bmatrix}qquad v_2 = begin{bmatrix} 0\1end{bmatrix}, $$



          but we don’t need to. Given a vector $x$ and a basis $alpha = {alpha_1, alpha_2}$, if $x = aalpha_1 + balpha_2$, I’ll be explicit and write that



          $$[x]_{alpha} = begin{bmatrix} a\bend{bmatrix}. $$



          Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $alpha$ and $beta$ such that given any $xinmathbb{R}^2$



          $$A[x]_{alpha} = B[x]_{beta}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
            $endgroup$
            – Andrew
            Jan 4 at 9:20










          • $begingroup$
            You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
            $endgroup$
            – Santana Afton
            Jan 4 at 14:41
















          0












          $begingroup$

          When you write



          $$x = begin{bmatrix} 2\3end{bmatrix}$$



          what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $mathbb{R}^2$. Usually, we assume that



          $$v_1 = begin{bmatrix} 1\0end{bmatrix}qquad v_2 = begin{bmatrix} 0\1end{bmatrix}, $$



          but we don’t need to. Given a vector $x$ and a basis $alpha = {alpha_1, alpha_2}$, if $x = aalpha_1 + balpha_2$, I’ll be explicit and write that



          $$[x]_{alpha} = begin{bmatrix} a\bend{bmatrix}. $$



          Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $alpha$ and $beta$ such that given any $xinmathbb{R}^2$



          $$A[x]_{alpha} = B[x]_{beta}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
            $endgroup$
            – Andrew
            Jan 4 at 9:20










          • $begingroup$
            You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
            $endgroup$
            – Santana Afton
            Jan 4 at 14:41














          0












          0








          0





          $begingroup$

          When you write



          $$x = begin{bmatrix} 2\3end{bmatrix}$$



          what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $mathbb{R}^2$. Usually, we assume that



          $$v_1 = begin{bmatrix} 1\0end{bmatrix}qquad v_2 = begin{bmatrix} 0\1end{bmatrix}, $$



          but we don’t need to. Given a vector $x$ and a basis $alpha = {alpha_1, alpha_2}$, if $x = aalpha_1 + balpha_2$, I’ll be explicit and write that



          $$[x]_{alpha} = begin{bmatrix} a\bend{bmatrix}. $$



          Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $alpha$ and $beta$ such that given any $xinmathbb{R}^2$



          $$A[x]_{alpha} = B[x]_{beta}.$$






          share|cite|improve this answer











          $endgroup$



          When you write



          $$x = begin{bmatrix} 2\3end{bmatrix}$$



          what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $mathbb{R}^2$. Usually, we assume that



          $$v_1 = begin{bmatrix} 1\0end{bmatrix}qquad v_2 = begin{bmatrix} 0\1end{bmatrix}, $$



          but we don’t need to. Given a vector $x$ and a basis $alpha = {alpha_1, alpha_2}$, if $x = aalpha_1 + balpha_2$, I’ll be explicit and write that



          $$[x]_{alpha} = begin{bmatrix} a\bend{bmatrix}. $$



          Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $alpha$ and $beta$ such that given any $xinmathbb{R}^2$



          $$A[x]_{alpha} = B[x]_{beta}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 4 at 8:24

























          answered Jan 4 at 5:47









          Santana AftonSantana Afton

          2,9132629




          2,9132629












          • $begingroup$
            Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
            $endgroup$
            – Andrew
            Jan 4 at 9:20










          • $begingroup$
            You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
            $endgroup$
            – Santana Afton
            Jan 4 at 14:41


















          • $begingroup$
            Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
            $endgroup$
            – Andrew
            Jan 4 at 9:20










          • $begingroup$
            You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
            $endgroup$
            – Santana Afton
            Jan 4 at 14:41
















          $begingroup$
          Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
          $endgroup$
          – Andrew
          Jan 4 at 9:20




          $begingroup$
          Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
          $endgroup$
          – Andrew
          Jan 4 at 9:20












          $begingroup$
          You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
          $endgroup$
          – Santana Afton
          Jan 4 at 14:41




          $begingroup$
          You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
          $endgroup$
          – Santana Afton
          Jan 4 at 14:41


















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