Find all of the eigenvalues and eigenvectors of the 2x2 matrix
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
I got $λ^2 - 3λ - 4 + 6i^2 = 0$
-I isoloated down to $i^2 = -1/6 (λ -4)(λ+1)$ and wrote how λ must be -1 < λ < 4. I don't know what to do next
linear-algebra eigenvalues-eigenvectors
edited Aug 9 at 19:30
mrtaurho
696219
asked Aug 9 at 19:25
Kayy Wang
63
add a comment |Â
up vote
1
down vote
favorite
$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
I got $λ^2 - 3λ - 4 + 6i^2 = 0$
-I isoloated down to $i^2 = -1/6 (λ -4)(λ+1)$ and wrote how λ must be -1 < λ < 4. I don't know what to do next
linear-algebra eigenvalues-eigenvectors
edited Aug 9 at 19:30
mrtaurho
696219
asked Aug 9 at 19:25
Kayy Wang
63
1
Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
– cdipaolo
Aug 9 at 19:30
1
For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
– JMoravitz
Aug 9 at 19:30
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
I got $λ^2 - 3λ - 4 + 6i^2 = 0$
-I isoloated down to $i^2 = -1/6 (λ -4)(λ+1)$ and wrote how λ must be -1 < λ < 4. I don't know what to do next
linear-algebra eigenvalues-eigenvectors
edited Aug 9 at 19:30
mrtaurho
696219
asked Aug 9 at 19:25
Kayy Wang
63
$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
I got $λ^2 - 3λ - 4 + 6i^2 = 0$
-I isoloated down to $i^2 = -1/6 (λ -4)(λ+1)$ and wrote how λ must be -1 < λ < 4. I don't know what to do next
linear-algebra eigenvalues-eigenvectors
edited Aug 9 at 19:30
mrtaurho
696219
asked Aug 9 at 19:25
Kayy Wang
63
edited Aug 9 at 19:30
mrtaurho
696219
edited Aug 9 at 19:30
mrtaurho
696219
edited Aug 9 at 19:30
mrtaurho
696219
696219
asked Aug 9 at 19:25
Kayy Wang
63
asked Aug 9 at 19:25
Kayy Wang
63
asked Aug 9 at 19:25
Kayy Wang
63
63
1
Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
– cdipaolo
Aug 9 at 19:30
1
For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
– JMoravitz
Aug 9 at 19:30
add a comment |Â
1
Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
– cdipaolo
Aug 9 at 19:30
1
For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
– JMoravitz
Aug 9 at 19:30
1
1
Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
– cdipaolo
Aug 9 at 19:30
Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
– cdipaolo
Aug 9 at 19:30
1
1
For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
– JMoravitz
Aug 9 at 19:30
For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
– JMoravitz
Aug 9 at 19:30
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
Just use the fact that $i^2=-1$ within your determinate of the matrix. Furthermore it should be $+3lambda$ instead of $-3lambda$. Therefore you got $lambda^2+3lambda-4+6(-1)=0$. Can you proceed from here?
answered Aug 9 at 19:29
mrtaurho
696219
Thank you very much
– Kayy Wang
Aug 9 at 19:35
You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
– mrtaurho
Aug 9 at 19:37
add a comment |Â
up vote
1
down vote
$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
Now $|A-lambda I|=0$
$$beginvmatrix 1-lambda & 2i \ -3i & -4-lambda endvmatrix=0$$
$$(1-lambda)(-4-lambda)+6i^2=0$$
$$-4-lambda+4lambda+lambda^2-6=0$$
$$lambda^2+3lambda-10=0$$
$$(lambda+5)(lambda-2)$$
$$lambda=-5,2$$
To find the eigenvector you need to find the null space using the above eigen values
answered Aug 9 at 19:34
Key Flex
4,863628
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Just use the fact that $i^2=-1$ within your determinate of the matrix. Furthermore it should be $+3lambda$ instead of $-3lambda$. Therefore you got $lambda^2+3lambda-4+6(-1)=0$. Can you proceed from here?
answered Aug 9 at 19:29
mrtaurho
696219
Thank you very much
– Kayy Wang
Aug 9 at 19:35
You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
– mrtaurho
Aug 9 at 19:37
add a comment |Â
up vote
1
down vote
Just use the fact that $i^2=-1$ within your determinate of the matrix. Furthermore it should be $+3lambda$ instead of $-3lambda$. Therefore you got $lambda^2+3lambda-4+6(-1)=0$. Can you proceed from here?
answered Aug 9 at 19:29
mrtaurho
696219
Thank you very much
– Kayy Wang
Aug 9 at 19:35
You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
– mrtaurho
Aug 9 at 19:37
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Just use the fact that $i^2=-1$ within your determinate of the matrix. Furthermore it should be $+3lambda$ instead of $-3lambda$. Therefore you got $lambda^2+3lambda-4+6(-1)=0$. Can you proceed from here?
answered Aug 9 at 19:29
mrtaurho
696219
Just use the fact that $i^2=-1$ within your determinate of the matrix. Furthermore it should be $+3lambda$ instead of $-3lambda$. Therefore you got $lambda^2+3lambda-4+6(-1)=0$. Can you proceed from here?
answered Aug 9 at 19:29
mrtaurho
696219
answered Aug 9 at 19:29
mrtaurho
696219
answered Aug 9 at 19:29
mrtaurho
696219
answered Aug 9 at 19:29
mrtaurho
696219
696219
Thank you very much
– Kayy Wang
Aug 9 at 19:35
You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
– mrtaurho
Aug 9 at 19:37
add a comment |Â
Thank you very much
– Kayy Wang
Aug 9 at 19:35
You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
– mrtaurho
Aug 9 at 19:37
Thank you very much
– Kayy Wang
Aug 9 at 19:35
Thank you very much
– Kayy Wang
Aug 9 at 19:35
You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
– mrtaurho
Aug 9 at 19:37
You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
– mrtaurho
Aug 9 at 19:37
add a comment |Â
up vote
1
down vote
$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
Now $|A-lambda I|=0$
$$beginvmatrix 1-lambda & 2i \ -3i & -4-lambda endvmatrix=0$$
$$(1-lambda)(-4-lambda)+6i^2=0$$
$$-4-lambda+4lambda+lambda^2-6=0$$
$$lambda^2+3lambda-10=0$$
$$(lambda+5)(lambda-2)$$
$$lambda=-5,2$$
To find the eigenvector you need to find the null space using the above eigen values
answered Aug 9 at 19:34
Key Flex
4,863628
add a comment |Â
up vote
1
down vote
$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
Now $|A-lambda I|=0$
$$beginvmatrix 1-lambda & 2i \ -3i & -4-lambda endvmatrix=0$$
$$(1-lambda)(-4-lambda)+6i^2=0$$
$$-4-lambda+4lambda+lambda^2-6=0$$
$$lambda^2+3lambda-10=0$$
$$(lambda+5)(lambda-2)$$
$$lambda=-5,2$$
To find the eigenvector you need to find the null space using the above eigen values
answered Aug 9 at 19:34
Key Flex
4,863628
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
Now $|A-lambda I|=0$
$$beginvmatrix 1-lambda & 2i \ -3i & -4-lambda endvmatrix=0$$
$$(1-lambda)(-4-lambda)+6i^2=0$$
$$-4-lambda+4lambda+lambda^2-6=0$$
$$lambda^2+3lambda-10=0$$
$$(lambda+5)(lambda-2)$$
$$lambda=-5,2$$
To find the eigenvector you need to find the null space using the above eigen values
answered Aug 9 at 19:34
Key Flex
4,863628
$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
Now $|A-lambda I|=0$
$$beginvmatrix 1-lambda & 2i \ -3i & -4-lambda endvmatrix=0$$
$$(1-lambda)(-4-lambda)+6i^2=0$$
$$-4-lambda+4lambda+lambda^2-6=0$$
$$lambda^2+3lambda-10=0$$
$$(lambda+5)(lambda-2)$$
$$lambda=-5,2$$
To find the eigenvector you need to find the null space using the above eigen values
answered Aug 9 at 19:34
Key Flex
4,863628
answered Aug 9 at 19:34
Key Flex
4,863628
answered Aug 9 at 19:34
Key Flex
4,863628
answered Aug 9 at 19:34
Key Flex
4,863628
4,863628
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2877608%2ffind-all-of-the-eigenvalues-and-eigenvectors-of-the-2x2-matrix%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
– cdipaolo
Aug 9 at 19:30
1
For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
– JMoravitz
Aug 9 at 19:30