Mixing
Showing that two matrices are similar by showing they are similar to the same diagonal matrix
$begingroup$
The exercise gives two matrices $A$ and $B$ and asks you to show they are similar by showing that they are similar to the same diagonal matrix, and then after that find and invertible matrix $P$ such that $P^{-1}AP=B$.
$$
A = begin{bmatrix}
1 & 0 & 2\[0.3em]
1 & -1 & 1\[0.3em]
2 & 0 & 1\[0.3em]
end{bmatrix},
%
B = begin{bmatrix}
-3 & -2 & 0\[0.3em]
6 & 5 & 0\[0.3em]
4 & 4 & -1\[0.3em]
end{bmatrix}
$$
I started with with calculating the eigenvalues of matrix $A$ like so:
$$
det(A) = det(begin{bmatrix}
1-lambda & 0 & 2\[0.3em]
1 & -1-lambda & 1\[0.3em]
2 & 0 & 1-lambda\[0.3em]
end{bmatrix})= lambda^3-lambda^2-5lambda-3 = (lambda-3)(lambda+1)^2
$$
Which gives $lambda = -1, 3$. Taking $lambda = -1$ gives $$[A--1 cdot I|0] =
begin{bmatrix}
2 & 0 & 2 & 0\[0.3em]
1 & 0 & 1 & 0\[0.3em]
2 & 0 & 2 & 0\[0.3em]
end{bmatrix}=
%
begin{bmatrix}
1 & 0 & 1 & 0\[0.3em]
0 & 0 & 0 & 0\[0.3em]
0 & 0 & 0 & 0\[0.3em]
end{bmatrix}
$$
But this is where I get stuck, because $x_2$ is a free variable, and so this gives
$$
E_{-1}=Bigg{s
begin{bmatrix}
0\[0.3em]
1\[0.3em]
0\[0.3em]
end{bmatrix}+
%
tbegin{bmatrix}
-1\[0.3em]
0\[0.3em]
1\[0.3em]
end{bmatrix}Bigg} = span
%
Bigg(begin{bmatrix}
0\[0.3em]
1\[0.3em]
0\[0.3em]
end{bmatrix},
%
begin{bmatrix}
-1\[0.3em]
0\[0.3em]
1\[0.3em]
end{bmatrix}
Bigg)
$$
right? I thought $P ={E_{-1}, E_{-1}, E_{3}}$ for example, so how does this work? I might be completely missing the mark, but I thought you had to calculate the eigenvalues of matrix $A$ and $B$, and if they are the same, then they are similar, then after this, calculate the eigenvectors of matrix $A$ to obtain matrix $P$, and then plug these matrices into the formula given above ($P^{-1}AP=B$) to see if it's true or not?
I uploaded pictures of how I've done it here, is that the correct way?
linear-algebra matrices diagonalization
asked Mar 9 '16 at 20:30
EsoemahEsoemah
132211
$endgroup$
add a comment |
$begingroup$
The exercise gives two matrices $A$ and $B$ and asks you to show they are similar by showing that they are similar to the same diagonal matrix, and then after that find and invertible matrix $P$ such that $P^{-1}AP=B$.
$$
A = begin{bmatrix}
1 & 0 & 2\[0.3em]
1 & -1 & 1\[0.3em]
2 & 0 & 1\[0.3em]
end{bmatrix},
%
B = begin{bmatrix}
-3 & -2 & 0\[0.3em]
6 & 5 & 0\[0.3em]
4 & 4 & -1\[0.3em]
end{bmatrix}
$$
I started with with calculating the eigenvalues of matrix $A$ like so:
$$
det(A) = det(begin{bmatrix}
1-lambda & 0 & 2\[0.3em]
1 & -1-lambda & 1\[0.3em]
2 & 0 & 1-lambda\[0.3em]
end{bmatrix})= lambda^3-lambda^2-5lambda-3 = (lambda-3)(lambda+1)^2
$$
Which gives $lambda = -1, 3$. Taking $lambda = -1$ gives $$[A--1 cdot I|0] =
begin{bmatrix}
2 & 0 & 2 & 0\[0.3em]
1 & 0 & 1 & 0\[0.3em]
2 & 0 & 2 & 0\[0.3em]
end{bmatrix}=
%
begin{bmatrix}
1 & 0 & 1 & 0\[0.3em]
0 & 0 & 0 & 0\[0.3em]
0 & 0 & 0 & 0\[0.3em]
end{bmatrix}
$$
But this is where I get stuck, because $x_2$ is a free variable, and so this gives
$$
E_{-1}=Bigg{s
begin{bmatrix}
0\[0.3em]
1\[0.3em]
0\[0.3em]
end{bmatrix}+
%
tbegin{bmatrix}
-1\[0.3em]
0\[0.3em]
1\[0.3em]
end{bmatrix}Bigg} = span
%
Bigg(begin{bmatrix}
0\[0.3em]
1\[0.3em]
0\[0.3em]
end{bmatrix},
%
begin{bmatrix}
-1\[0.3em]
0\[0.3em]
1\[0.3em]
end{bmatrix}
Bigg)
$$
right? I thought $P ={E_{-1}, E_{-1}, E_{3}}$ for example, so how does this work? I might be completely missing the mark, but I thought you had to calculate the eigenvalues of matrix $A$ and $B$, and if they are the same, then they are similar, then after this, calculate the eigenvectors of matrix $A$ to obtain matrix $P$, and then plug these matrices into the formula given above ($P^{-1}AP=B$) to see if it's true or not?
I uploaded pictures of how I've done it here, is that the correct way?
linear-algebra matrices diagonalization
asked Mar 9 '16 at 20:30
EsoemahEsoemah
132211
$endgroup$
$begingroup$
Why are you stuck? You seem to have just stopped rather than continuing doing precisely what you say you think you should be doing.
$endgroup$
– Tobias Kildetoft
Mar 9 '16 at 20:32
$begingroup$
@TobiasKildetoft I thought $E_{-1}$ was supposed to be 1 vector, so then I must have done something wrong. Should I just continue with $lambda = 3$ and then use that together with $E_{-1}$ to obtain matrix $P$?
$endgroup$
– Esoemah
Mar 9 '16 at 20:34
$begingroup$
Why should it be just one vector? You only have two eigenvalues, so you need two eigenvectors from one of them to be able to diagonalize.
$endgroup$
– Tobias Kildetoft
Mar 9 '16 at 20:36
$begingroup$
@TobiasKildetoft Okay thanks, I must have been confused for some reason. Is my reasoning at the bottom correct and should I just continue with that?
$endgroup$
– Esoemah
Mar 9 '16 at 20:38
$begingroup$
@TobiasKildetoft I edited in pictures of my work, is this correct?
$endgroup$
– Esoemah
Mar 9 '16 at 22:26
add a comment |
$begingroup$
The exercise gives two matrices $A$ and $B$ and asks you to show they are similar by showing that they are similar to the same diagonal matrix, and then after that find and invertible matrix $P$ such that $P^{-1}AP=B$.
$$
A = begin{bmatrix}
1 & 0 & 2\[0.3em]
1 & -1 & 1\[0.3em]
2 & 0 & 1\[0.3em]
end{bmatrix},
%
B = begin{bmatrix}
-3 & -2 & 0\[0.3em]
6 & 5 & 0\[0.3em]
4 & 4 & -1\[0.3em]
end{bmatrix}
$$
I started with with calculating the eigenvalues of matrix $A$ like so:
$$
det(A) = det(begin{bmatrix}
1-lambda & 0 & 2\[0.3em]
1 & -1-lambda & 1\[0.3em]
2 & 0 & 1-lambda\[0.3em]
end{bmatrix})= lambda^3-lambda^2-5lambda-3 = (lambda-3)(lambda+1)^2
$$
Which gives $lambda = -1, 3$. Taking $lambda = -1$ gives $$[A--1 cdot I|0] =
begin{bmatrix}
2 & 0 & 2 & 0\[0.3em]
1 & 0 & 1 & 0\[0.3em]
2 & 0 & 2 & 0\[0.3em]
end{bmatrix}=
%
begin{bmatrix}
1 & 0 & 1 & 0\[0.3em]
0 & 0 & 0 & 0\[0.3em]
0 & 0 & 0 & 0\[0.3em]
end{bmatrix}
$$
But this is where I get stuck, because $x_2$ is a free variable, and so this gives
$$
E_{-1}=Bigg{s
begin{bmatrix}
0\[0.3em]
1\[0.3em]
0\[0.3em]
end{bmatrix}+
%
tbegin{bmatrix}
-1\[0.3em]
0\[0.3em]
1\[0.3em]
end{bmatrix}Bigg} = span
%
Bigg(begin{bmatrix}
0\[0.3em]
1\[0.3em]
0\[0.3em]
end{bmatrix},
%
begin{bmatrix}
-1\[0.3em]
0\[0.3em]
1\[0.3em]
end{bmatrix}
Bigg)
$$
right? I thought $P ={E_{-1}, E_{-1}, E_{3}}$ for example, so how does this work? I might be completely missing the mark, but I thought you had to calculate the eigenvalues of matrix $A$ and $B$, and if they are the same, then they are similar, then after this, calculate the eigenvectors of matrix $A$ to obtain matrix $P$, and then plug these matrices into the formula given above ($P^{-1}AP=B$) to see if it's true or not?
I uploaded pictures of how I've done it here, is that the correct way?
linear-algebra matrices diagonalization
asked Mar 9 '16 at 20:30
EsoemahEsoemah
132211
$endgroup$
The exercise gives two matrices $A$ and $B$ and asks you to show they are similar by showing that they are similar to the same diagonal matrix, and then after that find and invertible matrix $P$ such that $P^{-1}AP=B$.
$$
A = begin{bmatrix}
1 & 0 & 2\[0.3em]
1 & -1 & 1\[0.3em]
2 & 0 & 1\[0.3em]
end{bmatrix},
%
B = begin{bmatrix}
-3 & -2 & 0\[0.3em]
6 & 5 & 0\[0.3em]
4 & 4 & -1\[0.3em]
end{bmatrix}
$$
I started with with calculating the eigenvalues of matrix $A$ like so:
$$
det(A) = det(begin{bmatrix}
1-lambda & 0 & 2\[0.3em]
1 & -1-lambda & 1\[0.3em]
2 & 0 & 1-lambda\[0.3em]
end{bmatrix})= lambda^3-lambda^2-5lambda-3 = (lambda-3)(lambda+1)^2
$$
Which gives $lambda = -1, 3$. Taking $lambda = -1$ gives $$[A--1 cdot I|0] =
begin{bmatrix}
2 & 0 & 2 & 0\[0.3em]
1 & 0 & 1 & 0\[0.3em]
2 & 0 & 2 & 0\[0.3em]
end{bmatrix}=
%
begin{bmatrix}
1 & 0 & 1 & 0\[0.3em]
0 & 0 & 0 & 0\[0.3em]
0 & 0 & 0 & 0\[0.3em]
end{bmatrix}
$$
But this is where I get stuck, because $x_2$ is a free variable, and so this gives
$$
E_{-1}=Bigg{s
begin{bmatrix}
0\[0.3em]
1\[0.3em]
0\[0.3em]
end{bmatrix}+
%
tbegin{bmatrix}
-1\[0.3em]
0\[0.3em]
1\[0.3em]
end{bmatrix}Bigg} = span
%
Bigg(begin{bmatrix}
0\[0.3em]
1\[0.3em]
0\[0.3em]
end{bmatrix},
%
begin{bmatrix}
-1\[0.3em]
0\[0.3em]
1\[0.3em]
end{bmatrix}
Bigg)
$$
right? I thought $P ={E_{-1}, E_{-1}, E_{3}}$ for example, so how does this work? I might be completely missing the mark, but I thought you had to calculate the eigenvalues of matrix $A$ and $B$, and if they are the same, then they are similar, then after this, calculate the eigenvectors of matrix $A$ to obtain matrix $P$, and then plug these matrices into the formula given above ($P^{-1}AP=B$) to see if it's true or not?
I uploaded pictures of how I've done it here, is that the correct way?
linear-algebra matrices diagonalization
linear-algebra matrices diagonalization
asked Mar 9 '16 at 20:30
EsoemahEsoemah
132211
asked Mar 9 '16 at 20:30
EsoemahEsoemah
132211
edited Mar 9 '16 at 22:26
Esoemah
asked Mar 9 '16 at 20:30
EsoemahEsoemah
132211
asked Mar 9 '16 at 20:30
EsoemahEsoemah
132211
asked Mar 9 '16 at 20:30
EsoemahEsoemah
132211
132211
$begingroup$
Why are you stuck? You seem to have just stopped rather than continuing doing precisely what you say you think you should be doing.
$endgroup$
– Tobias Kildetoft
Mar 9 '16 at 20:32
$begingroup$
@TobiasKildetoft I thought $E_{-1}$ was supposed to be 1 vector, so then I must have done something wrong. Should I just continue with $lambda = 3$ and then use that together with $E_{-1}$ to obtain matrix $P$?
$endgroup$
– Esoemah
Mar 9 '16 at 20:34
$begingroup$
Why should it be just one vector? You only have two eigenvalues, so you need two eigenvectors from one of them to be able to diagonalize.
$endgroup$
– Tobias Kildetoft
Mar 9 '16 at 20:36
$begingroup$
@TobiasKildetoft Okay thanks, I must have been confused for some reason. Is my reasoning at the bottom correct and should I just continue with that?
$endgroup$
– Esoemah
Mar 9 '16 at 20:38
$begingroup$
@TobiasKildetoft I edited in pictures of my work, is this correct?
$endgroup$
– Esoemah
Mar 9 '16 at 22:26
add a comment |
$begingroup$
Why are you stuck? You seem to have just stopped rather than continuing doing precisely what you say you think you should be doing.
$endgroup$
– Tobias Kildetoft
Mar 9 '16 at 20:32
$begingroup$
@TobiasKildetoft I thought $E_{-1}$ was supposed to be 1 vector, so then I must have done something wrong. Should I just continue with $lambda = 3$ and then use that together with $E_{-1}$ to obtain matrix $P$?
$endgroup$
– Esoemah
Mar 9 '16 at 20:34
$begingroup$
Why should it be just one vector? You only have two eigenvalues, so you need two eigenvectors from one of them to be able to diagonalize.
$endgroup$
– Tobias Kildetoft
Mar 9 '16 at 20:36
$begingroup$
@TobiasKildetoft Okay thanks, I must have been confused for some reason. Is my reasoning at the bottom correct and should I just continue with that?
$endgroup$
– Esoemah
Mar 9 '16 at 20:38
$begingroup$
@TobiasKildetoft I edited in pictures of my work, is this correct?
$endgroup$
– Esoemah
Mar 9 '16 at 22:26
$begingroup$
Why are you stuck? You seem to have just stopped rather than continuing doing precisely what you say you think you should be doing.
$endgroup$
– Tobias Kildetoft
Mar 9 '16 at 20:32
$begingroup$
Why are you stuck? You seem to have just stopped rather than continuing doing precisely what you say you think you should be doing.
$endgroup$
– Tobias Kildetoft
Mar 9 '16 at 20:32
$begingroup$
@TobiasKildetoft I thought $E_{-1}$ was supposed to be 1 vector, so then I must have done something wrong. Should I just continue with $lambda = 3$ and then use that together with $E_{-1}$ to obtain matrix $P$?
$endgroup$
– Esoemah
Mar 9 '16 at 20:34
$begingroup$
@TobiasKildetoft I thought $E_{-1}$ was supposed to be 1 vector, so then I must have done something wrong. Should I just continue with $lambda = 3$ and then use that together with $E_{-1}$ to obtain matrix $P$?
$endgroup$
– Esoemah
Mar 9 '16 at 20:34
$begingroup$
Why should it be just one vector? You only have two eigenvalues, so you need two eigenvectors from one of them to be able to diagonalize.
$endgroup$
– Tobias Kildetoft
Mar 9 '16 at 20:36
$begingroup$
Why should it be just one vector? You only have two eigenvalues, so you need two eigenvectors from one of them to be able to diagonalize.
$endgroup$
– Tobias Kildetoft
Mar 9 '16 at 20:36
$begingroup$
@TobiasKildetoft Okay thanks, I must have been confused for some reason. Is my reasoning at the bottom correct and should I just continue with that?
$endgroup$
– Esoemah
Mar 9 '16 at 20:38
$begingroup$
@TobiasKildetoft Okay thanks, I must have been confused for some reason. Is my reasoning at the bottom correct and should I just continue with that?
$endgroup$
– Esoemah
Mar 9 '16 at 20:38
$begingroup$
@TobiasKildetoft I edited in pictures of my work, is this correct?
$endgroup$
– Esoemah
Mar 9 '16 at 22:26
$begingroup$
@TobiasKildetoft I edited in pictures of my work, is this correct?
$endgroup$
– Esoemah
Mar 9 '16 at 22:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$det A=3neq27=det Bimplies A,,B;;text{cannot be similar}$$
answered Mar 9 '16 at 20:40
DonAntonioDonAntonio
179k1494230
$endgroup$
$begingroup$
But they have the same eigenvalues. I don't really think this is a helpful answer.
$endgroup$
– Esoemah
Mar 9 '16 at 21:10
$begingroup$
@Esoemah Even if that was true, and I'm not sure it is, similar matrices have the same trace and the same determinant. As these two don't they can't be similar. Check the determinants, perhaps I made I mistake, or perhaps you miscopied. For example, if that first entry in $;B;$ was $;3;$ and not $;-3;$ , as written there, then they'd have the same determinant.
$endgroup$
– DonAntonio
Mar 9 '16 at 21:13
$begingroup$
I miscopied, the 2 should be a -2, and then the determinants are the same. I'm sorry about that.
$endgroup$
– Esoemah
Mar 9 '16 at 21:17
$begingroup$
I edited in pictures of my work, is this correct?
$endgroup$
– Esoemah
Mar 9 '16 at 22:27
add a comment |
$begingroup$
If you have an eigenvalue of multplicity, you will get multiple eigenvalues associated with that eigenvalue.
$begin{bmatrix}
1 & 0 \
1 & -1\
2 & 0\
end{bmatrix}begin{bmatrix}
0&-1 \
1&0\
0&1\
end{bmatrix} = begin{bmatrix}
0&1 \
-1&0\
0&-1\
end{bmatrix}= begin{bmatrix}0&-1 \
1&0\
0&1\
end{bmatrix}(-I)$
Find the 3rd eigenvetor and you will have
$AP = Pbegin{bmatrix} -1&\&-1\&&3end{bmatrix}\
P^{-1}AP = begin{bmatrix} -1&\&-1\&&3end{bmatrix}$
answered Jan 18 at 19:19
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
$$det A=3neq27=det Bimplies A,,B;;text{cannot be similar}$$
answered Mar 9 '16 at 20:40
DonAntonioDonAntonio
179k1494230
$endgroup$
$begingroup$
But they have the same eigenvalues. I don't really think this is a helpful answer.
$endgroup$
– Esoemah
Mar 9 '16 at 21:10
$begingroup$
@Esoemah Even if that was true, and I'm not sure it is, similar matrices have the same trace and the same determinant. As these two don't they can't be similar. Check the determinants, perhaps I made I mistake, or perhaps you miscopied. For example, if that first entry in $;B;$ was $;3;$ and not $;-3;$ , as written there, then they'd have the same determinant.
$endgroup$
– DonAntonio
Mar 9 '16 at 21:13
$begingroup$
I miscopied, the 2 should be a -2, and then the determinants are the same. I'm sorry about that.
$endgroup$
– Esoemah
Mar 9 '16 at 21:17
$begingroup$
I edited in pictures of my work, is this correct?
$endgroup$
– Esoemah
Mar 9 '16 at 22:27
add a comment |
$begingroup$
$$det A=3neq27=det Bimplies A,,B;;text{cannot be similar}$$
answered Mar 9 '16 at 20:40
DonAntonioDonAntonio
179k1494230
$endgroup$
$begingroup$
But they have the same eigenvalues. I don't really think this is a helpful answer.
$endgroup$
– Esoemah
Mar 9 '16 at 21:10
$begingroup$
@Esoemah Even if that was true, and I'm not sure it is, similar matrices have the same trace and the same determinant. As these two don't they can't be similar. Check the determinants, perhaps I made I mistake, or perhaps you miscopied. For example, if that first entry in $;B;$ was $;3;$ and not $;-3;$ , as written there, then they'd have the same determinant.
$endgroup$
– DonAntonio
Mar 9 '16 at 21:13
$begingroup$
I miscopied, the 2 should be a -2, and then the determinants are the same. I'm sorry about that.
$endgroup$
– Esoemah
Mar 9 '16 at 21:17
$begingroup$
I edited in pictures of my work, is this correct?
$endgroup$
– Esoemah
Mar 9 '16 at 22:27
add a comment |
$begingroup$
$$det A=3neq27=det Bimplies A,,B;;text{cannot be similar}$$
answered Mar 9 '16 at 20:40
DonAntonioDonAntonio
179k1494230
$endgroup$
$$det A=3neq27=det Bimplies A,,B;;text{cannot be similar}$$
answered Mar 9 '16 at 20:40
DonAntonioDonAntonio
179k1494230
answered Mar 9 '16 at 20:40
DonAntonioDonAntonio
179k1494230
answered Mar 9 '16 at 20:40
DonAntonioDonAntonio
179k1494230
answered Mar 9 '16 at 20:40
DonAntonioDonAntonio
179k1494230
179k1494230
$begingroup$
But they have the same eigenvalues. I don't really think this is a helpful answer.
$endgroup$
– Esoemah
Mar 9 '16 at 21:10
$begingroup$
@Esoemah Even if that was true, and I'm not sure it is, similar matrices have the same trace and the same determinant. As these two don't they can't be similar. Check the determinants, perhaps I made I mistake, or perhaps you miscopied. For example, if that first entry in $;B;$ was $;3;$ and not $;-3;$ , as written there, then they'd have the same determinant.
$endgroup$
– DonAntonio
Mar 9 '16 at 21:13
$begingroup$
I miscopied, the 2 should be a -2, and then the determinants are the same. I'm sorry about that.
$endgroup$
– Esoemah
Mar 9 '16 at 21:17
$begingroup$
I edited in pictures of my work, is this correct?
$endgroup$
– Esoemah
Mar 9 '16 at 22:27
add a comment |
$begingroup$
But they have the same eigenvalues. I don't really think this is a helpful answer.
$endgroup$
– Esoemah
Mar 9 '16 at 21:10
$begingroup$
@Esoemah Even if that was true, and I'm not sure it is, similar matrices have the same trace and the same determinant. As these two don't they can't be similar. Check the determinants, perhaps I made I mistake, or perhaps you miscopied. For example, if that first entry in $;B;$ was $;3;$ and not $;-3;$ , as written there, then they'd have the same determinant.
$endgroup$
– DonAntonio
Mar 9 '16 at 21:13
$begingroup$
I miscopied, the 2 should be a -2, and then the determinants are the same. I'm sorry about that.
$endgroup$
– Esoemah
Mar 9 '16 at 21:17
$begingroup$
I edited in pictures of my work, is this correct?
$endgroup$
– Esoemah
Mar 9 '16 at 22:27
$begingroup$
But they have the same eigenvalues. I don't really think this is a helpful answer.
$endgroup$
– Esoemah
Mar 9 '16 at 21:10
$begingroup$
But they have the same eigenvalues. I don't really think this is a helpful answer.
$endgroup$
– Esoemah
Mar 9 '16 at 21:10
$begingroup$
@Esoemah Even if that was true, and I'm not sure it is, similar matrices have the same trace and the same determinant. As these two don't they can't be similar. Check the determinants, perhaps I made I mistake, or perhaps you miscopied. For example, if that first entry in $;B;$ was $;3;$ and not $;-3;$ , as written there, then they'd have the same determinant.
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– DonAntonio
Mar 9 '16 at 21:13
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@Esoemah Even if that was true, and I'm not sure it is, similar matrices have the same trace and the same determinant. As these two don't they can't be similar. Check the determinants, perhaps I made I mistake, or perhaps you miscopied. For example, if that first entry in $;B;$ was $;3;$ and not $;-3;$ , as written there, then they'd have the same determinant.
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– DonAntonio
Mar 9 '16 at 21:13
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I miscopied, the 2 should be a -2, and then the determinants are the same. I'm sorry about that.
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– Esoemah
Mar 9 '16 at 21:17
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I miscopied, the 2 should be a -2, and then the determinants are the same. I'm sorry about that.
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– Esoemah
Mar 9 '16 at 21:17
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I edited in pictures of my work, is this correct?
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– Esoemah
Mar 9 '16 at 22:27
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I edited in pictures of my work, is this correct?
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– Esoemah
Mar 9 '16 at 22:27
add a comment |
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If you have an eigenvalue of multplicity, you will get multiple eigenvalues associated with that eigenvalue.
$begin{bmatrix}
1 & 0 \
1 & -1\
2 & 0\
end{bmatrix}begin{bmatrix}
0&-1 \
1&0\
0&1\
end{bmatrix} = begin{bmatrix}
0&1 \
-1&0\
0&-1\
end{bmatrix}= begin{bmatrix}0&-1 \
1&0\
0&1\
end{bmatrix}(-I)$
Find the 3rd eigenvetor and you will have
$AP = Pbegin{bmatrix} -1&\&-1\&&3end{bmatrix}\
P^{-1}AP = begin{bmatrix} -1&\&-1\&&3end{bmatrix}$
answered Jan 18 at 19:19
Doug MDoug M
45.3k31954
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add a comment |
$begingroup$
If you have an eigenvalue of multplicity, you will get multiple eigenvalues associated with that eigenvalue.
$begin{bmatrix}
1 & 0 \
1 & -1\
2 & 0\
end{bmatrix}begin{bmatrix}
0&-1 \
1&0\
0&1\
end{bmatrix} = begin{bmatrix}
0&1 \
-1&0\
0&-1\
end{bmatrix}= begin{bmatrix}0&-1 \
1&0\
0&1\
end{bmatrix}(-I)$
Find the 3rd eigenvetor and you will have
$AP = Pbegin{bmatrix} -1&\&-1\&&3end{bmatrix}\
P^{-1}AP = begin{bmatrix} -1&\&-1\&&3end{bmatrix}$
answered Jan 18 at 19:19
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
If you have an eigenvalue of multplicity, you will get multiple eigenvalues associated with that eigenvalue.
$begin{bmatrix}
1 & 0 \
1 & -1\
2 & 0\
end{bmatrix}begin{bmatrix}
0&-1 \
1&0\
0&1\
end{bmatrix} = begin{bmatrix}
0&1 \
-1&0\
0&-1\
end{bmatrix}= begin{bmatrix}0&-1 \
1&0\
0&1\
end{bmatrix}(-I)$
Find the 3rd eigenvetor and you will have
$AP = Pbegin{bmatrix} -1&\&-1\&&3end{bmatrix}\
P^{-1}AP = begin{bmatrix} -1&\&-1\&&3end{bmatrix}$
answered Jan 18 at 19:19
Doug MDoug M
45.3k31954
$endgroup$
If you have an eigenvalue of multplicity, you will get multiple eigenvalues associated with that eigenvalue.
$begin{bmatrix}
1 & 0 \
1 & -1\
2 & 0\
end{bmatrix}begin{bmatrix}
0&-1 \
1&0\
0&1\
end{bmatrix} = begin{bmatrix}
0&1 \
-1&0\
0&-1\
end{bmatrix}= begin{bmatrix}0&-1 \
1&0\
0&1\
end{bmatrix}(-I)$
Find the 3rd eigenvetor and you will have
$AP = Pbegin{bmatrix} -1&\&-1\&&3end{bmatrix}\
P^{-1}AP = begin{bmatrix} -1&\&-1\&&3end{bmatrix}$
answered Jan 18 at 19:19
Doug MDoug M
45.3k31954
answered Jan 18 at 19:19
Doug MDoug M
45.3k31954
answered Jan 18 at 19:19
Doug MDoug M
45.3k31954
answered Jan 18 at 19:19
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
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Why are you stuck? You seem to have just stopped rather than continuing doing precisely what you say you think you should be doing.
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– Tobias Kildetoft
Mar 9 '16 at 20:32
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@TobiasKildetoft I thought $E_{-1}$ was supposed to be 1 vector, so then I must have done something wrong. Should I just continue with $lambda = 3$ and then use that together with $E_{-1}$ to obtain matrix $P$?
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– Esoemah
Mar 9 '16 at 20:34
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Why should it be just one vector? You only have two eigenvalues, so you need two eigenvectors from one of them to be able to diagonalize.
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– Tobias Kildetoft
Mar 9 '16 at 20:36
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@TobiasKildetoft Okay thanks, I must have been confused for some reason. Is my reasoning at the bottom correct and should I just continue with that?
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– Esoemah
Mar 9 '16 at 20:38
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@TobiasKildetoft I edited in pictures of my work, is this correct?
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– Esoemah
Mar 9 '16 at 22:26