Eigenvalues of Matrix with variables
$begingroup$
Im supposed to find eigenvalues of following matrix:
$begin{bmatrix}1 & -x^2+x-2 & 0\1& 2x-1 & x-1\ 0& 4 & 1end{bmatrix}$
My attempt was to solve it the "normal way" to solve det(A-$lambda$I) = 0
But since there is a variable in the matrix, this leads to a big cubic equation which i cannot solve easily, which looks like:
$$lambda^3 - lambda ^2 -2xlambda + 5lambda + x^2 lambda -xlambda +3x - x^2 -5 =0$$
I know that this equation will add up for $lambda =1 $ but other than that i have no idea
am I missing something? is there a different way to attempt this?
Thanks
linear-algebra eigenvalues-eigenvectors
asked Jan 5 at 22:50
Nicola ZauggNicola Zaugg
91
$endgroup$
add a comment |
$begingroup$
Im supposed to find eigenvalues of following matrix:
$begin{bmatrix}1 & -x^2+x-2 & 0\1& 2x-1 & x-1\ 0& 4 & 1end{bmatrix}$
My attempt was to solve it the "normal way" to solve det(A-$lambda$I) = 0
But since there is a variable in the matrix, this leads to a big cubic equation which i cannot solve easily, which looks like:
$$lambda^3 - lambda ^2 -2xlambda + 5lambda + x^2 lambda -xlambda +3x - x^2 -5 =0$$
I know that this equation will add up for $lambda =1 $ but other than that i have no idea
am I missing something? is there a different way to attempt this?
Thanks
linear-algebra eigenvalues-eigenvectors
asked Jan 5 at 22:50
Nicola ZauggNicola Zaugg
91
$endgroup$
add a comment |
$begingroup$
Im supposed to find eigenvalues of following matrix:
$begin{bmatrix}1 & -x^2+x-2 & 0\1& 2x-1 & x-1\ 0& 4 & 1end{bmatrix}$
My attempt was to solve it the "normal way" to solve det(A-$lambda$I) = 0
But since there is a variable in the matrix, this leads to a big cubic equation which i cannot solve easily, which looks like:
$$lambda^3 - lambda ^2 -2xlambda + 5lambda + x^2 lambda -xlambda +3x - x^2 -5 =0$$
I know that this equation will add up for $lambda =1 $ but other than that i have no idea
am I missing something? is there a different way to attempt this?
Thanks
linear-algebra eigenvalues-eigenvectors
asked Jan 5 at 22:50
Nicola ZauggNicola Zaugg
91
$endgroup$
Im supposed to find eigenvalues of following matrix:
$begin{bmatrix}1 & -x^2+x-2 & 0\1& 2x-1 & x-1\ 0& 4 & 1end{bmatrix}$
My attempt was to solve it the "normal way" to solve det(A-$lambda$I) = 0
But since there is a variable in the matrix, this leads to a big cubic equation which i cannot solve easily, which looks like:
$$lambda^3 - lambda ^2 -2xlambda + 5lambda + x^2 lambda -xlambda +3x - x^2 -5 =0$$
I know that this equation will add up for $lambda =1 $ but other than that i have no idea
am I missing something? is there a different way to attempt this?
Thanks
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Jan 5 at 22:50
Nicola ZauggNicola Zaugg
91
asked Jan 5 at 22:50
Nicola ZauggNicola Zaugg
91
asked Jan 5 at 22:50
Nicola ZauggNicola Zaugg
91
asked Jan 5 at 22:50
Nicola ZauggNicola Zaugg
91
asked Jan 5 at 22:50
Nicola ZauggNicola Zaugg
91
91
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since you know that $ lambda = 1$ works, note that your equation factors as
$$(lambda - 1) (x^2 - 3x + 5 + lambda^2) = 0.$$
This gives that
$$lambda = 1, lambda = pm sqrt{3x - x^2 - 5}.$$
answered Jan 5 at 22:58
Klint QinamiKlint Qinami
1,137510
$endgroup$
add a comment |
$begingroup$
Since you know that $1$ is a root of your polynomial, divide it by $lambda-1$, getting$$lambda^3-lambda^2-2xlambda+5lambda+x^2lambda-xlambda+3x-x^2-5=(lambda-1)(lambda^2+x^2-3x+5).$$
answered Jan 5 at 22:59
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since you know that $ lambda = 1$ works, note that your equation factors as
$$(lambda - 1) (x^2 - 3x + 5 + lambda^2) = 0.$$
This gives that
$$lambda = 1, lambda = pm sqrt{3x - x^2 - 5}.$$
answered Jan 5 at 22:58
Klint QinamiKlint Qinami
1,137510
$endgroup$
add a comment |
$begingroup$
Since you know that $ lambda = 1$ works, note that your equation factors as
$$(lambda - 1) (x^2 - 3x + 5 + lambda^2) = 0.$$
This gives that
$$lambda = 1, lambda = pm sqrt{3x - x^2 - 5}.$$
answered Jan 5 at 22:58
Klint QinamiKlint Qinami
1,137510
$endgroup$
add a comment |
$begingroup$
Since you know that $ lambda = 1$ works, note that your equation factors as
$$(lambda - 1) (x^2 - 3x + 5 + lambda^2) = 0.$$
This gives that
$$lambda = 1, lambda = pm sqrt{3x - x^2 - 5}.$$
answered Jan 5 at 22:58
Klint QinamiKlint Qinami
1,137510
$endgroup$
Since you know that $ lambda = 1$ works, note that your equation factors as
$$(lambda - 1) (x^2 - 3x + 5 + lambda^2) = 0.$$
This gives that
$$lambda = 1, lambda = pm sqrt{3x - x^2 - 5}.$$
answered Jan 5 at 22:58
Klint QinamiKlint Qinami
1,137510
answered Jan 5 at 22:58
Klint QinamiKlint Qinami
1,137510
answered Jan 5 at 22:58
Klint QinamiKlint Qinami
1,137510
answered Jan 5 at 22:58
Klint QinamiKlint Qinami
1,137510
1,137510
add a comment |
add a comment |
$begingroup$
Since you know that $1$ is a root of your polynomial, divide it by $lambda-1$, getting$$lambda^3-lambda^2-2xlambda+5lambda+x^2lambda-xlambda+3x-x^2-5=(lambda-1)(lambda^2+x^2-3x+5).$$
answered Jan 5 at 22:59
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
Since you know that $1$ is a root of your polynomial, divide it by $lambda-1$, getting$$lambda^3-lambda^2-2xlambda+5lambda+x^2lambda-xlambda+3x-x^2-5=(lambda-1)(lambda^2+x^2-3x+5).$$
answered Jan 5 at 22:59
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
Since you know that $1$ is a root of your polynomial, divide it by $lambda-1$, getting$$lambda^3-lambda^2-2xlambda+5lambda+x^2lambda-xlambda+3x-x^2-5=(lambda-1)(lambda^2+x^2-3x+5).$$
answered Jan 5 at 22:59
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
Since you know that $1$ is a root of your polynomial, divide it by $lambda-1$, getting$$lambda^3-lambda^2-2xlambda+5lambda+x^2lambda-xlambda+3x-x^2-5=(lambda-1)(lambda^2+x^2-3x+5).$$
answered Jan 5 at 22:59
José Carlos SantosJosé Carlos Santos
173k23133241
answered Jan 5 at 22:59
José Carlos SantosJosé Carlos Santos
173k23133241
answered Jan 5 at 22:59
José Carlos SantosJosé Carlos Santos
173k23133241
answered Jan 5 at 22:59
José Carlos SantosJosé Carlos Santos
173k23133241
173k23133241
add a comment |
add a comment |
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