Mixing
Let $alpha, beta,gamma,delta $ be the eigenvalues of the matrix find $alpha ^2+beta^2+gamma^2+delta^2 $
$begingroup$
Let $alpha, beta,gamma,delta $ be the eigen values of the matrix
$$ A=begin{pmatrix} 0 & 0 & 0& 0\ 1 & 0 & 0 &-2 \ 0 & 1 & 0 & 1
\ 0 & 0 &1 & 2 end{pmatrix} $$
then $alpha ^2+beta^2+gamma^2+delta^2$ is ?
Matrix has determinant $|A|=0$
Thus one eigen value is zero. I assumed $delta=0$.
$trace(A)=alpha +beta+gamma=2$
$alpha ^2+beta^2+gamma^2=(alpha +beta+gamma)^2-2(alpha beta+beta gamma+alpha gamma)=4-2(alpha beta+beta gamma+alpha gamma)$
I am not sure how to figure out $2(alpha beta+beta gamma+alpha gamma)$ term.I am not going for long method of finding eigen values by characteristic polynomial since this question came in $2$ marks so I am trying to find out quick way to solve this problem.
linear-algebra eigenvalues-eigenvectors
asked Jan 25 at 5:10
Daman deepDaman deep
756419
$endgroup$
add a comment |
$begingroup$
Let $alpha, beta,gamma,delta $ be the eigen values of the matrix
$$ A=begin{pmatrix} 0 & 0 & 0& 0\ 1 & 0 & 0 &-2 \ 0 & 1 & 0 & 1
\ 0 & 0 &1 & 2 end{pmatrix} $$
then $alpha ^2+beta^2+gamma^2+delta^2$ is ?
Matrix has determinant $|A|=0$
Thus one eigen value is zero. I assumed $delta=0$.
$trace(A)=alpha +beta+gamma=2$
$alpha ^2+beta^2+gamma^2=(alpha +beta+gamma)^2-2(alpha beta+beta gamma+alpha gamma)=4-2(alpha beta+beta gamma+alpha gamma)$
I am not sure how to figure out $2(alpha beta+beta gamma+alpha gamma)$ term.I am not going for long method of finding eigen values by characteristic polynomial since this question came in $2$ marks so I am trying to find out quick way to solve this problem.
linear-algebra eigenvalues-eigenvectors
asked Jan 25 at 5:10
Daman deepDaman deep
756419
$endgroup$
5
$begingroup$
Why you do not compute $A^2$ and taking its trace?
$endgroup$
– Chinnapparaj R
Jan 25 at 5:21
$begingroup$
@ChinnapparajR oh yessss how in the world I forgot that thank you sir .
$endgroup$
– Daman deep
Jan 25 at 5:23
add a comment |
$begingroup$
Let $alpha, beta,gamma,delta $ be the eigen values of the matrix
$$ A=begin{pmatrix} 0 & 0 & 0& 0\ 1 & 0 & 0 &-2 \ 0 & 1 & 0 & 1
\ 0 & 0 &1 & 2 end{pmatrix} $$
then $alpha ^2+beta^2+gamma^2+delta^2$ is ?
Matrix has determinant $|A|=0$
Thus one eigen value is zero. I assumed $delta=0$.
$trace(A)=alpha +beta+gamma=2$
$alpha ^2+beta^2+gamma^2=(alpha +beta+gamma)^2-2(alpha beta+beta gamma+alpha gamma)=4-2(alpha beta+beta gamma+alpha gamma)$
I am not sure how to figure out $2(alpha beta+beta gamma+alpha gamma)$ term.I am not going for long method of finding eigen values by characteristic polynomial since this question came in $2$ marks so I am trying to find out quick way to solve this problem.
linear-algebra eigenvalues-eigenvectors
asked Jan 25 at 5:10
Daman deepDaman deep
756419
$endgroup$
Let $alpha, beta,gamma,delta $ be the eigen values of the matrix
$$ A=begin{pmatrix} 0 & 0 & 0& 0\ 1 & 0 & 0 &-2 \ 0 & 1 & 0 & 1
\ 0 & 0 &1 & 2 end{pmatrix} $$
then $alpha ^2+beta^2+gamma^2+delta^2$ is ?
Matrix has determinant $|A|=0$
Thus one eigen value is zero. I assumed $delta=0$.
$trace(A)=alpha +beta+gamma=2$
$alpha ^2+beta^2+gamma^2=(alpha +beta+gamma)^2-2(alpha beta+beta gamma+alpha gamma)=4-2(alpha beta+beta gamma+alpha gamma)$
I am not sure how to figure out $2(alpha beta+beta gamma+alpha gamma)$ term.I am not going for long method of finding eigen values by characteristic polynomial since this question came in $2$ marks so I am trying to find out quick way to solve this problem.
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Jan 25 at 5:10
Daman deepDaman deep
756419
asked Jan 25 at 5:10
Daman deepDaman deep
756419
asked Jan 25 at 5:10
Daman deepDaman deep
756419
asked Jan 25 at 5:10
Daman deepDaman deep
756419
asked Jan 25 at 5:10
Daman deepDaman deep
756419
756419
5
$begingroup$
Why you do not compute $A^2$ and taking its trace?
$endgroup$
– Chinnapparaj R
Jan 25 at 5:21
$begingroup$
@ChinnapparajR oh yessss how in the world I forgot that thank you sir .
$endgroup$
– Daman deep
Jan 25 at 5:23
add a comment |
5
$begingroup$
Why you do not compute $A^2$ and taking its trace?
$endgroup$
– Chinnapparaj R
Jan 25 at 5:21
$begingroup$
@ChinnapparajR oh yessss how in the world I forgot that thank you sir .
$endgroup$
– Daman deep
Jan 25 at 5:23
5
5
$begingroup$
Why you do not compute $A^2$ and taking its trace?
$endgroup$
– Chinnapparaj R
Jan 25 at 5:21
$begingroup$
Why you do not compute $A^2$ and taking its trace?
$endgroup$
– Chinnapparaj R
Jan 25 at 5:21
$begingroup$
@ChinnapparajR oh yessss how in the world I forgot that thank you sir .
$endgroup$
– Daman deep
Jan 25 at 5:23
$begingroup$
@ChinnapparajR oh yessss how in the world I forgot that thank you sir .
$endgroup$
– Daman deep
Jan 25 at 5:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You did most of the work!
Let $P(lambda)$ be $A$'s characteristic polynomial. Then $alpha beta+beta gamma+alpha gamma$ is the coefficient of $lambda^2$. To see this, write $$P(lambda)=lambda(lambda-alpha)(lambda-beta)(lambda-gamma)$$
If I'm correct, $P(lambda)=lambda^4-2lambda^3-lambda^2+2lambda$. So $alpha beta+beta gamma+alpha gamma=-1$.
So $$alpha ^2+beta^2+gamma^2+delta^2=6$$
answered Jan 25 at 5:32
Stefan LafonStefan Lafon
2,96019
$endgroup$
$begingroup$
How did you calculate it $-1$? Can you add more steps ?
$endgroup$
– Daman deep
Jan 25 at 5:34
$begingroup$
Look at the expression for $P(lambda)$, and take the coefficient in front of $lambda^2$. It is $-1$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:35
$begingroup$
Can you tell me whole expression in general of characteristic polynomial of 4th degree?
$endgroup$
– Daman deep
Jan 25 at 5:37
$begingroup$
I computed it from matrix $A$ via $P(lambda)=det(A-lambda I)$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:41
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You did most of the work!
Let $P(lambda)$ be $A$'s characteristic polynomial. Then $alpha beta+beta gamma+alpha gamma$ is the coefficient of $lambda^2$. To see this, write $$P(lambda)=lambda(lambda-alpha)(lambda-beta)(lambda-gamma)$$
If I'm correct, $P(lambda)=lambda^4-2lambda^3-lambda^2+2lambda$. So $alpha beta+beta gamma+alpha gamma=-1$.
So $$alpha ^2+beta^2+gamma^2+delta^2=6$$
answered Jan 25 at 5:32
Stefan LafonStefan Lafon
2,96019
$endgroup$
$begingroup$
How did you calculate it $-1$? Can you add more steps ?
$endgroup$
– Daman deep
Jan 25 at 5:34
$begingroup$
Look at the expression for $P(lambda)$, and take the coefficient in front of $lambda^2$. It is $-1$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:35
$begingroup$
Can you tell me whole expression in general of characteristic polynomial of 4th degree?
$endgroup$
– Daman deep
Jan 25 at 5:37
$begingroup$
I computed it from matrix $A$ via $P(lambda)=det(A-lambda I)$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:41
add a comment |
$begingroup$
You did most of the work!
Let $P(lambda)$ be $A$'s characteristic polynomial. Then $alpha beta+beta gamma+alpha gamma$ is the coefficient of $lambda^2$. To see this, write $$P(lambda)=lambda(lambda-alpha)(lambda-beta)(lambda-gamma)$$
If I'm correct, $P(lambda)=lambda^4-2lambda^3-lambda^2+2lambda$. So $alpha beta+beta gamma+alpha gamma=-1$.
So $$alpha ^2+beta^2+gamma^2+delta^2=6$$
answered Jan 25 at 5:32
Stefan LafonStefan Lafon
2,96019
$endgroup$
$begingroup$
How did you calculate it $-1$? Can you add more steps ?
$endgroup$
– Daman deep
Jan 25 at 5:34
$begingroup$
Look at the expression for $P(lambda)$, and take the coefficient in front of $lambda^2$. It is $-1$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:35
$begingroup$
Can you tell me whole expression in general of characteristic polynomial of 4th degree?
$endgroup$
– Daman deep
Jan 25 at 5:37
$begingroup$
I computed it from matrix $A$ via $P(lambda)=det(A-lambda I)$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:41
add a comment |
$begingroup$
You did most of the work!
Let $P(lambda)$ be $A$'s characteristic polynomial. Then $alpha beta+beta gamma+alpha gamma$ is the coefficient of $lambda^2$. To see this, write $$P(lambda)=lambda(lambda-alpha)(lambda-beta)(lambda-gamma)$$
If I'm correct, $P(lambda)=lambda^4-2lambda^3-lambda^2+2lambda$. So $alpha beta+beta gamma+alpha gamma=-1$.
So $$alpha ^2+beta^2+gamma^2+delta^2=6$$
answered Jan 25 at 5:32
Stefan LafonStefan Lafon
2,96019
$endgroup$
You did most of the work!
Let $P(lambda)$ be $A$'s characteristic polynomial. Then $alpha beta+beta gamma+alpha gamma$ is the coefficient of $lambda^2$. To see this, write $$P(lambda)=lambda(lambda-alpha)(lambda-beta)(lambda-gamma)$$
If I'm correct, $P(lambda)=lambda^4-2lambda^3-lambda^2+2lambda$. So $alpha beta+beta gamma+alpha gamma=-1$.
So $$alpha ^2+beta^2+gamma^2+delta^2=6$$
answered Jan 25 at 5:32
Stefan LafonStefan Lafon
2,96019
answered Jan 25 at 5:32
Stefan LafonStefan Lafon
2,96019
answered Jan 25 at 5:32
Stefan LafonStefan Lafon
2,96019
answered Jan 25 at 5:32
Stefan LafonStefan Lafon
2,96019
2,96019
$begingroup$
How did you calculate it $-1$? Can you add more steps ?
$endgroup$
– Daman deep
Jan 25 at 5:34
$begingroup$
Look at the expression for $P(lambda)$, and take the coefficient in front of $lambda^2$. It is $-1$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:35
$begingroup$
Can you tell me whole expression in general of characteristic polynomial of 4th degree?
$endgroup$
– Daman deep
Jan 25 at 5:37
$begingroup$
I computed it from matrix $A$ via $P(lambda)=det(A-lambda I)$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:41
add a comment |
$begingroup$
How did you calculate it $-1$? Can you add more steps ?
$endgroup$
– Daman deep
Jan 25 at 5:34
$begingroup$
Look at the expression for $P(lambda)$, and take the coefficient in front of $lambda^2$. It is $-1$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:35
$begingroup$
Can you tell me whole expression in general of characteristic polynomial of 4th degree?
$endgroup$
– Daman deep
Jan 25 at 5:37
$begingroup$
I computed it from matrix $A$ via $P(lambda)=det(A-lambda I)$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:41
$begingroup$
How did you calculate it $-1$? Can you add more steps ?
$endgroup$
– Daman deep
Jan 25 at 5:34
$begingroup$
How did you calculate it $-1$? Can you add more steps ?
$endgroup$
– Daman deep
Jan 25 at 5:34
$begingroup$
Look at the expression for $P(lambda)$, and take the coefficient in front of $lambda^2$. It is $-1$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:35
$begingroup$
Look at the expression for $P(lambda)$, and take the coefficient in front of $lambda^2$. It is $-1$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:35
$begingroup$
Can you tell me whole expression in general of characteristic polynomial of 4th degree?
$endgroup$
– Daman deep
Jan 25 at 5:37
$begingroup$
Can you tell me whole expression in general of characteristic polynomial of 4th degree?
$endgroup$
– Daman deep
Jan 25 at 5:37
$begingroup$
I computed it from matrix $A$ via $P(lambda)=det(A-lambda I)$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:41
$begingroup$
I computed it from matrix $A$ via $P(lambda)=det(A-lambda I)$.
$endgroup$
– Stefan Lafon
Jan 25 at 5:41
add a comment |
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5
$begingroup$
Why you do not compute $A^2$ and taking its trace?
$endgroup$
– Chinnapparaj R
Jan 25 at 5:21
$begingroup$
@ChinnapparajR oh yessss how in the world I forgot that thank you sir .
$endgroup$
– Daman deep
Jan 25 at 5:23