Find a matrix $P$ that diagonalizes the matrix $A$, and determine $P^{-1}AP$












2












$begingroup$


$A =begin{bmatrix} -14 & 12 \ -20 & 17 \ end{bmatrix} $



I did this by first calculating the eigenvalues which turn out to be $λ = 2$ and $λ = 1$



Then I calculated the eigenvectors by first substituting $λ = 2$ and then $λ = 1$ in the $(Iλ - A)$ matrix and reducing both matrices before calculating the eigenvectors.



E.g for $λ = 2$



$begin{bmatrix} 1 & frac{-3}{4} \ 0 & 0 \ end{bmatrix}begin{bmatrix} x_1 \ x_2end{bmatrix} =begin{bmatrix} 0 \ 0end{bmatrix} $



In both cases I made $x_2 = t$



which resulted in the first eigenvector being $t begin{bmatrix} frac{3}{4} \ 1end{bmatrix}$ and the second one $t begin{bmatrix} frac{4}{5} \ 1end{bmatrix}$



So I made $t = 4$ in the first one and $t = 5$ in the second one to get $ begin{bmatrix} 3 \ 4end{bmatrix}$ and $begin{bmatrix} 4 \ 5end{bmatrix}$ which represent the $P_1$ and $P_2$ columns of the matrix $P = begin{bmatrix} 3 & 4 \ 4 & 5 \ end{bmatrix}$ which gives: $$P^{-1} AP = begin{bmatrix} 2 & 0 \ 0 & 1 \ end{bmatrix}$$



In my textbook solution they made $x_1 = t$ for both cases, instead of $x_2$ like I did. Which makes $P_1 = begin{bmatrix} 4 \ 5end{bmatrix}$ and $P_2 = begin{bmatrix} 3 \ 4end{bmatrix}$, so same $P$ but in different order $P = begin{bmatrix} 4 & 5 \ 3 & 4 \ end{bmatrix}$. When doing $P^{-1} AP$ with this new matrix $P$ the diagonal matrix turns out to be: $$P^{-1} AP = begin{bmatrix} 1 & 0 \ 0 & 2 \ end{bmatrix}$$



Would this be the same transformation just that in a different basis (represented by the eigenvectors in different order) and is equally valid as a diagonalized matrix $A$?






linear-algebra eigenvalues-eigenvectors linear-transformations diagonalization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It is formally not the same transformation due to the permutation of the basic vectors, but it is equally valid as diagonalization of $A$.
    $endgroup$
    – A.Γ.
    Jan 2 at 14:27










  • $begingroup$
    Just to be specific, you mean it's not even the same transformation represented in different bases? So which one is the valid one equal to A in a different basis (applied to vectors in the same different basis)?
    $endgroup$
    – Mango164
    Jan 2 at 15:06






  • 1




    $begingroup$
    Ok, by transformation I meant the change of basis $P$. If you mean the transformation to be the linear mapping defined by $A$ then yes, it is the same transformation in different bases.
    $endgroup$
    – A.Γ.
    Jan 2 at 15:09












  • $begingroup$
    Yeah that's what I meant, thanks for clarifying.
    $endgroup$
    – Mango164
    Jan 2 at 15:13
















2












$begingroup$


$A =begin{bmatrix} -14 & 12 \ -20 & 17 \ end{bmatrix} $



I did this by first calculating the eigenvalues which turn out to be $λ = 2$ and $λ = 1$



Then I calculated the eigenvectors by first substituting $λ = 2$ and then $λ = 1$ in the $(Iλ - A)$ matrix and reducing both matrices before calculating the eigenvectors.



E.g for $λ = 2$



$begin{bmatrix} 1 & frac{-3}{4} \ 0 & 0 \ end{bmatrix}begin{bmatrix} x_1 \ x_2end{bmatrix} =begin{bmatrix} 0 \ 0end{bmatrix} $



In both cases I made $x_2 = t$



which resulted in the first eigenvector being $t begin{bmatrix} frac{3}{4} \ 1end{bmatrix}$ and the second one $t begin{bmatrix} frac{4}{5} \ 1end{bmatrix}$



So I made $t = 4$ in the first one and $t = 5$ in the second one to get $ begin{bmatrix} 3 \ 4end{bmatrix}$ and $begin{bmatrix} 4 \ 5end{bmatrix}$ which represent the $P_1$ and $P_2$ columns of the matrix $P = begin{bmatrix} 3 & 4 \ 4 & 5 \ end{bmatrix}$ which gives: $$P^{-1} AP = begin{bmatrix} 2 & 0 \ 0 & 1 \ end{bmatrix}$$



In my textbook solution they made $x_1 = t$ for both cases, instead of $x_2$ like I did. Which makes $P_1 = begin{bmatrix} 4 \ 5end{bmatrix}$ and $P_2 = begin{bmatrix} 3 \ 4end{bmatrix}$, so same $P$ but in different order $P = begin{bmatrix} 4 & 5 \ 3 & 4 \ end{bmatrix}$. When doing $P^{-1} AP$ with this new matrix $P$ the diagonal matrix turns out to be: $$P^{-1} AP = begin{bmatrix} 1 & 0 \ 0 & 2 \ end{bmatrix}$$



Would this be the same transformation just that in a different basis (represented by the eigenvectors in different order) and is equally valid as a diagonalized matrix $A$?






linear-algebra eigenvalues-eigenvectors linear-transformations diagonalization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It is formally not the same transformation due to the permutation of the basic vectors, but it is equally valid as diagonalization of $A$.
    $endgroup$
    – A.Γ.
    Jan 2 at 14:27










  • $begingroup$
    Just to be specific, you mean it's not even the same transformation represented in different bases? So which one is the valid one equal to A in a different basis (applied to vectors in the same different basis)?
    $endgroup$
    – Mango164
    Jan 2 at 15:06






  • 1




    $begingroup$
    Ok, by transformation I meant the change of basis $P$. If you mean the transformation to be the linear mapping defined by $A$ then yes, it is the same transformation in different bases.
    $endgroup$
    – A.Γ.
    Jan 2 at 15:09












  • $begingroup$
    Yeah that's what I meant, thanks for clarifying.
    $endgroup$
    – Mango164
    Jan 2 at 15:13














2












2








2





$begingroup$


$A =begin{bmatrix} -14 & 12 \ -20 & 17 \ end{bmatrix} $



I did this by first calculating the eigenvalues which turn out to be $λ = 2$ and $λ = 1$



Then I calculated the eigenvectors by first substituting $λ = 2$ and then $λ = 1$ in the $(Iλ - A)$ matrix and reducing both matrices before calculating the eigenvectors.



E.g for $λ = 2$



$begin{bmatrix} 1 & frac{-3}{4} \ 0 & 0 \ end{bmatrix}begin{bmatrix} x_1 \ x_2end{bmatrix} =begin{bmatrix} 0 \ 0end{bmatrix} $



In both cases I made $x_2 = t$



which resulted in the first eigenvector being $t begin{bmatrix} frac{3}{4} \ 1end{bmatrix}$ and the second one $t begin{bmatrix} frac{4}{5} \ 1end{bmatrix}$



So I made $t = 4$ in the first one and $t = 5$ in the second one to get $ begin{bmatrix} 3 \ 4end{bmatrix}$ and $begin{bmatrix} 4 \ 5end{bmatrix}$ which represent the $P_1$ and $P_2$ columns of the matrix $P = begin{bmatrix} 3 & 4 \ 4 & 5 \ end{bmatrix}$ which gives: $$P^{-1} AP = begin{bmatrix} 2 & 0 \ 0 & 1 \ end{bmatrix}$$



In my textbook solution they made $x_1 = t$ for both cases, instead of $x_2$ like I did. Which makes $P_1 = begin{bmatrix} 4 \ 5end{bmatrix}$ and $P_2 = begin{bmatrix} 3 \ 4end{bmatrix}$, so same $P$ but in different order $P = begin{bmatrix} 4 & 5 \ 3 & 4 \ end{bmatrix}$. When doing $P^{-1} AP$ with this new matrix $P$ the diagonal matrix turns out to be: $$P^{-1} AP = begin{bmatrix} 1 & 0 \ 0 & 2 \ end{bmatrix}$$



Would this be the same transformation just that in a different basis (represented by the eigenvectors in different order) and is equally valid as a diagonalized matrix $A$?






linear-algebra eigenvalues-eigenvectors linear-transformations diagonalization







share|cite|improve this question











$endgroup$




$A =begin{bmatrix} -14 & 12 \ -20 & 17 \ end{bmatrix} $



I did this by first calculating the eigenvalues which turn out to be $λ = 2$ and $λ = 1$



Then I calculated the eigenvectors by first substituting $λ = 2$ and then $λ = 1$ in the $(Iλ - A)$ matrix and reducing both matrices before calculating the eigenvectors.



E.g for $λ = 2$



$begin{bmatrix} 1 & frac{-3}{4} \ 0 & 0 \ end{bmatrix}begin{bmatrix} x_1 \ x_2end{bmatrix} =begin{bmatrix} 0 \ 0end{bmatrix} $



In both cases I made $x_2 = t$



which resulted in the first eigenvector being $t begin{bmatrix} frac{3}{4} \ 1end{bmatrix}$ and the second one $t begin{bmatrix} frac{4}{5} \ 1end{bmatrix}$



So I made $t = 4$ in the first one and $t = 5$ in the second one to get $ begin{bmatrix} 3 \ 4end{bmatrix}$ and $begin{bmatrix} 4 \ 5end{bmatrix}$ which represent the $P_1$ and $P_2$ columns of the matrix $P = begin{bmatrix} 3 & 4 \ 4 & 5 \ end{bmatrix}$ which gives: $$P^{-1} AP = begin{bmatrix} 2 & 0 \ 0 & 1 \ end{bmatrix}$$



In my textbook solution they made $x_1 = t$ for both cases, instead of $x_2$ like I did. Which makes $P_1 = begin{bmatrix} 4 \ 5end{bmatrix}$ and $P_2 = begin{bmatrix} 3 \ 4end{bmatrix}$, so same $P$ but in different order $P = begin{bmatrix} 4 & 5 \ 3 & 4 \ end{bmatrix}$. When doing $P^{-1} AP$ with this new matrix $P$ the diagonal matrix turns out to be: $$P^{-1} AP = begin{bmatrix} 1 & 0 \ 0 & 2 \ end{bmatrix}$$



Would this be the same transformation just that in a different basis (represented by the eigenvectors in different order) and is equally valid as a diagonalized matrix $A$?






linear-algebra eigenvalues-eigenvectors linear-transformations diagonalization




linear-algebra eigenvalues-eigenvectors linear-transformations diagonalization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 14:24







Mango164

















asked Jan 2 at 14:18









Mango164Mango164

504




504








  • 1




    $begingroup$
    It is formally not the same transformation due to the permutation of the basic vectors, but it is equally valid as diagonalization of $A$.
    $endgroup$
    – A.Γ.
    Jan 2 at 14:27










  • $begingroup$
    Just to be specific, you mean it's not even the same transformation represented in different bases? So which one is the valid one equal to A in a different basis (applied to vectors in the same different basis)?
    $endgroup$
    – Mango164
    Jan 2 at 15:06






  • 1




    $begingroup$
    Ok, by transformation I meant the change of basis $P$. If you mean the transformation to be the linear mapping defined by $A$ then yes, it is the same transformation in different bases.
    $endgroup$
    – A.Γ.
    Jan 2 at 15:09












  • $begingroup$
    Yeah that's what I meant, thanks for clarifying.
    $endgroup$
    – Mango164
    Jan 2 at 15:13














  • 1




    $begingroup$
    It is formally not the same transformation due to the permutation of the basic vectors, but it is equally valid as diagonalization of $A$.
    $endgroup$
    – A.Γ.
    Jan 2 at 14:27










  • $begingroup$
    Just to be specific, you mean it's not even the same transformation represented in different bases? So which one is the valid one equal to A in a different basis (applied to vectors in the same different basis)?
    $endgroup$
    – Mango164
    Jan 2 at 15:06






  • 1




    $begingroup$
    Ok, by transformation I meant the change of basis $P$. If you mean the transformation to be the linear mapping defined by $A$ then yes, it is the same transformation in different bases.
    $endgroup$
    – A.Γ.
    Jan 2 at 15:09












  • $begingroup$
    Yeah that's what I meant, thanks for clarifying.
    $endgroup$
    – Mango164
    Jan 2 at 15:13








1




1




$begingroup$
It is formally not the same transformation due to the permutation of the basic vectors, but it is equally valid as diagonalization of $A$.
$endgroup$
– A.Γ.
Jan 2 at 14:27




$begingroup$
It is formally not the same transformation due to the permutation of the basic vectors, but it is equally valid as diagonalization of $A$.
$endgroup$
– A.Γ.
Jan 2 at 14:27












$begingroup$
Just to be specific, you mean it's not even the same transformation represented in different bases? So which one is the valid one equal to A in a different basis (applied to vectors in the same different basis)?
$endgroup$
– Mango164
Jan 2 at 15:06




$begingroup$
Just to be specific, you mean it's not even the same transformation represented in different bases? So which one is the valid one equal to A in a different basis (applied to vectors in the same different basis)?
$endgroup$
– Mango164
Jan 2 at 15:06




1




1




$begingroup$
Ok, by transformation I meant the change of basis $P$. If you mean the transformation to be the linear mapping defined by $A$ then yes, it is the same transformation in different bases.
$endgroup$
– A.Γ.
Jan 2 at 15:09






$begingroup$
Ok, by transformation I meant the change of basis $P$. If you mean the transformation to be the linear mapping defined by $A$ then yes, it is the same transformation in different bases.
$endgroup$
– A.Γ.
Jan 2 at 15:09














$begingroup$
Yeah that's what I meant, thanks for clarifying.
$endgroup$
– Mango164
Jan 2 at 15:13




$begingroup$
Yeah that's what I meant, thanks for clarifying.
$endgroup$
– Mango164
Jan 2 at 15:13










1 Answer
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$begingroup$

It is all okay. You just calculated the eigenvalues base in other order. In fact, those two matrices are equivalent.



It is (and will be for you) a well known fact that if the eigenvalues are real and all distinct, then your matrix is diagonalizable and the diagonal matrix is the diagonal matrix with the eigenvalues in the diagonal. (edit: in the ground field, but I said that because I assumed that you were working over the reals)






share|cite|improve this answer











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    $begingroup$

    It is all okay. You just calculated the eigenvalues base in other order. In fact, those two matrices are equivalent.



    It is (and will be for you) a well known fact that if the eigenvalues are real and all distinct, then your matrix is diagonalizable and the diagonal matrix is the diagonal matrix with the eigenvalues in the diagonal. (edit: in the ground field, but I said that because I assumed that you were working over the reals)






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      It is all okay. You just calculated the eigenvalues base in other order. In fact, those two matrices are equivalent.



      It is (and will be for you) a well known fact that if the eigenvalues are real and all distinct, then your matrix is diagonalizable and the diagonal matrix is the diagonal matrix with the eigenvalues in the diagonal. (edit: in the ground field, but I said that because I assumed that you were working over the reals)






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        It is all okay. You just calculated the eigenvalues base in other order. In fact, those two matrices are equivalent.



        It is (and will be for you) a well known fact that if the eigenvalues are real and all distinct, then your matrix is diagonalizable and the diagonal matrix is the diagonal matrix with the eigenvalues in the diagonal. (edit: in the ground field, but I said that because I assumed that you were working over the reals)






        share|cite|improve this answer











        $endgroup$



        It is all okay. You just calculated the eigenvalues base in other order. In fact, those two matrices are equivalent.



        It is (and will be for you) a well known fact that if the eigenvalues are real and all distinct, then your matrix is diagonalizable and the diagonal matrix is the diagonal matrix with the eigenvalues in the diagonal. (edit: in the ground field, but I said that because I assumed that you were working over the reals)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 15:27

























        answered Jan 2 at 14:39









        José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

        802110




        802110






























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