Row reducing matrices to obtain the eigenvector












0














Find all distinct (real or complex) eigenvalues of $A$. Then find the basic eigenvectors of $A$ corresponding to each eigenvalue. For each eigenvalue, specify the number of basic eigenvectors corresponding to that eigenvalue.



$A = begin{bmatrix}-16 & -10\17& 10end{bmatrix}$



So $A -lambda I = begin{bmatrix}-16-lambda & -10\17& 10-lambdaend{bmatrix}$



Which means $det(A-lambda I) = lambda^2 + 6lambda - 10$



Therefore, using the quadratic formula, the eigenvalues should be:



$lambda = -3pm sqrt(19)i$



So now I get the two eigenvectors which are:



$B = begin{bmatrix}-13 - sqrt19i& -10\17& 13 - sqrt19iend{bmatrix}$



$C = begin{bmatrix}-13 + sqrt19i& -10\17& 13 + sqrt19iend{bmatrix}$



Now the part I am struggling is finding out how to row reduce these matrices in order to obtain the answer:



For example, with $B$:



$Bx = 0$



$Bx = begin{bmatrix}-13 - sqrt19i& -10\17& 13 - sqrt19iend{bmatrix} begin{bmatrix}x1\x2end{bmatrix} = begin{bmatrix}0\0end{bmatrix}$



I know I have to do row operations to first get zeroes in the bottom row, and then to get a 1 in the first row/column, but I don't know how. This requires dealing with complex numbers, and I am not sure how to do it in this scenario. Could someone give me a detailed walkthrough on how to do this?






linear-algebra complex-numbers eigenvalues-eigenvectors







share|cite|improve this question






















  • Eigenvectors are vectors, not matrices. Your $B$ and $C$ are the matrices $A-lambda I$, not the eigenvectors of $A$.
    – amd
    Nov 30 '18 at 2:44
















0














Find all distinct (real or complex) eigenvalues of $A$. Then find the basic eigenvectors of $A$ corresponding to each eigenvalue. For each eigenvalue, specify the number of basic eigenvectors corresponding to that eigenvalue.



$A = begin{bmatrix}-16 & -10\17& 10end{bmatrix}$



So $A -lambda I = begin{bmatrix}-16-lambda & -10\17& 10-lambdaend{bmatrix}$



Which means $det(A-lambda I) = lambda^2 + 6lambda - 10$



Therefore, using the quadratic formula, the eigenvalues should be:



$lambda = -3pm sqrt(19)i$



So now I get the two eigenvectors which are:



$B = begin{bmatrix}-13 - sqrt19i& -10\17& 13 - sqrt19iend{bmatrix}$



$C = begin{bmatrix}-13 + sqrt19i& -10\17& 13 + sqrt19iend{bmatrix}$



Now the part I am struggling is finding out how to row reduce these matrices in order to obtain the answer:



For example, with $B$:



$Bx = 0$



$Bx = begin{bmatrix}-13 - sqrt19i& -10\17& 13 - sqrt19iend{bmatrix} begin{bmatrix}x1\x2end{bmatrix} = begin{bmatrix}0\0end{bmatrix}$



I know I have to do row operations to first get zeroes in the bottom row, and then to get a 1 in the first row/column, but I don't know how. This requires dealing with complex numbers, and I am not sure how to do it in this scenario. Could someone give me a detailed walkthrough on how to do this?






linear-algebra complex-numbers eigenvalues-eigenvectors







share|cite|improve this question






















  • Eigenvectors are vectors, not matrices. Your $B$ and $C$ are the matrices $A-lambda I$, not the eigenvectors of $A$.
    – amd
    Nov 30 '18 at 2:44














0












0








0







Find all distinct (real or complex) eigenvalues of $A$. Then find the basic eigenvectors of $A$ corresponding to each eigenvalue. For each eigenvalue, specify the number of basic eigenvectors corresponding to that eigenvalue.



$A = begin{bmatrix}-16 & -10\17& 10end{bmatrix}$



So $A -lambda I = begin{bmatrix}-16-lambda & -10\17& 10-lambdaend{bmatrix}$



Which means $det(A-lambda I) = lambda^2 + 6lambda - 10$



Therefore, using the quadratic formula, the eigenvalues should be:



$lambda = -3pm sqrt(19)i$



So now I get the two eigenvectors which are:



$B = begin{bmatrix}-13 - sqrt19i& -10\17& 13 - sqrt19iend{bmatrix}$



$C = begin{bmatrix}-13 + sqrt19i& -10\17& 13 + sqrt19iend{bmatrix}$



Now the part I am struggling is finding out how to row reduce these matrices in order to obtain the answer:



For example, with $B$:



$Bx = 0$



$Bx = begin{bmatrix}-13 - sqrt19i& -10\17& 13 - sqrt19iend{bmatrix} begin{bmatrix}x1\x2end{bmatrix} = begin{bmatrix}0\0end{bmatrix}$



I know I have to do row operations to first get zeroes in the bottom row, and then to get a 1 in the first row/column, but I don't know how. This requires dealing with complex numbers, and I am not sure how to do it in this scenario. Could someone give me a detailed walkthrough on how to do this?






linear-algebra complex-numbers eigenvalues-eigenvectors







share|cite|improve this question













Find all distinct (real or complex) eigenvalues of $A$. Then find the basic eigenvectors of $A$ corresponding to each eigenvalue. For each eigenvalue, specify the number of basic eigenvectors corresponding to that eigenvalue.



$A = begin{bmatrix}-16 & -10\17& 10end{bmatrix}$



So $A -lambda I = begin{bmatrix}-16-lambda & -10\17& 10-lambdaend{bmatrix}$



Which means $det(A-lambda I) = lambda^2 + 6lambda - 10$



Therefore, using the quadratic formula, the eigenvalues should be:



$lambda = -3pm sqrt(19)i$



So now I get the two eigenvectors which are:



$B = begin{bmatrix}-13 - sqrt19i& -10\17& 13 - sqrt19iend{bmatrix}$



$C = begin{bmatrix}-13 + sqrt19i& -10\17& 13 + sqrt19iend{bmatrix}$



Now the part I am struggling is finding out how to row reduce these matrices in order to obtain the answer:



For example, with $B$:



$Bx = 0$



$Bx = begin{bmatrix}-13 - sqrt19i& -10\17& 13 - sqrt19iend{bmatrix} begin{bmatrix}x1\x2end{bmatrix} = begin{bmatrix}0\0end{bmatrix}$



I know I have to do row operations to first get zeroes in the bottom row, and then to get a 1 in the first row/column, but I don't know how. This requires dealing with complex numbers, and I am not sure how to do it in this scenario. Could someone give me a detailed walkthrough on how to do this?






linear-algebra complex-numbers eigenvalues-eigenvectors




linear-algebra complex-numbers eigenvalues-eigenvectors






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asked Nov 30 '18 at 1:55









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  • Eigenvectors are vectors, not matrices. Your $B$ and $C$ are the matrices $A-lambda I$, not the eigenvectors of $A$.
    – amd
    Nov 30 '18 at 2:44


















  • Eigenvectors are vectors, not matrices. Your $B$ and $C$ are the matrices $A-lambda I$, not the eigenvectors of $A$.
    – amd
    Nov 30 '18 at 2:44
















Eigenvectors are vectors, not matrices. Your $B$ and $C$ are the matrices $A-lambda I$, not the eigenvectors of $A$.
– amd
Nov 30 '18 at 2:44




Eigenvectors are vectors, not matrices. Your $B$ and $C$ are the matrices $A-lambda I$, not the eigenvectors of $A$.
– amd
Nov 30 '18 at 2:44










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$$(lambda - lambda_1)(lambda - lambda_2) = lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=lambda^2-trace(A)lambda +det(A)$$



$$trace(A)=-16+10=-6$$



$$det(A)=-160+170=10$$



The characteristic polynomial is $lambda^2+6lambda color{red}{+} 10$.



$$lambda = frac{-6 pmsqrt{6^2-40}}{2}$$



Upon trying to solve for the equation $$(A-lambda I)v=0$$



If $lambda$ is indeed an eigenvalue, then the matrix is rank $1$ for this $2 times 2$ matrix. You can just focus on a non-zero row and ignroe the other constraint as it doesn't give you extra information.






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    $$(lambda - lambda_1)(lambda - lambda_2) = lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=lambda^2-trace(A)lambda +det(A)$$



    $$trace(A)=-16+10=-6$$



    $$det(A)=-160+170=10$$



    The characteristic polynomial is $lambda^2+6lambda color{red}{+} 10$.



    $$lambda = frac{-6 pmsqrt{6^2-40}}{2}$$



    Upon trying to solve for the equation $$(A-lambda I)v=0$$



    If $lambda$ is indeed an eigenvalue, then the matrix is rank $1$ for this $2 times 2$ matrix. You can just focus on a non-zero row and ignroe the other constraint as it doesn't give you extra information.






    share|cite|improve this answer


























      0














      $$(lambda - lambda_1)(lambda - lambda_2) = lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=lambda^2-trace(A)lambda +det(A)$$



      $$trace(A)=-16+10=-6$$



      $$det(A)=-160+170=10$$



      The characteristic polynomial is $lambda^2+6lambda color{red}{+} 10$.



      $$lambda = frac{-6 pmsqrt{6^2-40}}{2}$$



      Upon trying to solve for the equation $$(A-lambda I)v=0$$



      If $lambda$ is indeed an eigenvalue, then the matrix is rank $1$ for this $2 times 2$ matrix. You can just focus on a non-zero row and ignroe the other constraint as it doesn't give you extra information.






      share|cite|improve this answer
























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        0






        $$(lambda - lambda_1)(lambda - lambda_2) = lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=lambda^2-trace(A)lambda +det(A)$$



        $$trace(A)=-16+10=-6$$



        $$det(A)=-160+170=10$$



        The characteristic polynomial is $lambda^2+6lambda color{red}{+} 10$.



        $$lambda = frac{-6 pmsqrt{6^2-40}}{2}$$



        Upon trying to solve for the equation $$(A-lambda I)v=0$$



        If $lambda$ is indeed an eigenvalue, then the matrix is rank $1$ for this $2 times 2$ matrix. You can just focus on a non-zero row and ignroe the other constraint as it doesn't give you extra information.






        share|cite|improve this answer












        $$(lambda - lambda_1)(lambda - lambda_2) = lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=lambda^2-trace(A)lambda +det(A)$$



        $$trace(A)=-16+10=-6$$



        $$det(A)=-160+170=10$$



        The characteristic polynomial is $lambda^2+6lambda color{red}{+} 10$.



        $$lambda = frac{-6 pmsqrt{6^2-40}}{2}$$



        Upon trying to solve for the equation $$(A-lambda I)v=0$$



        If $lambda$ is indeed an eigenvalue, then the matrix is rank $1$ for this $2 times 2$ matrix. You can just focus on a non-zero row and ignroe the other constraint as it doesn't give you extra information.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 2:10









        Siong Thye GohSiong Thye Goh

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