Row reducing matrices to obtain the eigenvector
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
asked Nov 30 '18 at 1:55
qbufferqbuffer
13
add a comment |
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
asked Nov 30 '18 at 1:55
qbufferqbuffer
13
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
add a comment |
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
asked Nov 30 '18 at 1:55
qbufferqbuffer
13
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
asked Nov 30 '18 at 1:55
qbufferqbuffer
13
asked Nov 30 '18 at 1:55
qbufferqbuffer
13
asked Nov 30 '18 at 1:55
qbufferqbuffer
13
asked Nov 30 '18 at 1:55
qbufferqbuffer
13
asked Nov 30 '18 at 1:55
qbufferqbuffer
13
13
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
add a comment |
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
add a comment |
1 Answer
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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.
answered Nov 30 '18 at 2:10
Siong Thye GohSiong Thye Goh
99.6k1464117
add a comment |
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1 Answer
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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.
answered Nov 30 '18 at 2:10
Siong Thye GohSiong Thye Goh
99.6k1464117
add a comment |
$$(lambda - lambda_1)(lambda - lambda_2) = lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=lambda^2-trace(A)lambda +det(A)$$
$$trace(A)=-16+10=-6$$
$$det(A)=-160+170=10$$
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.
answered Nov 30 '18 at 2:10
Siong Thye GohSiong Thye Goh
99.6k1464117
add a comment |
$$(lambda - lambda_1)(lambda - lambda_2) = lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=lambda^2-trace(A)lambda +det(A)$$
$$trace(A)=-16+10=-6$$
$$det(A)=-160+170=10$$
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.
answered Nov 30 '18 at 2:10
Siong Thye GohSiong Thye Goh
99.6k1464117
$$(lambda - lambda_1)(lambda - lambda_2) = lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=lambda^2-trace(A)lambda +det(A)$$
$$trace(A)=-16+10=-6$$
$$det(A)=-160+170=10$$
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.
answered Nov 30 '18 at 2:10
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Nov 30 '18 at 2:10
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Nov 30 '18 at 2:10
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Nov 30 '18 at 2:10
Siong Thye GohSiong Thye Goh
99.6k1464117
99.6k1464117
add a comment |
add a comment |
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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