Solving for Eigenvectors of an Already Diagonalized Matrix
I'm a bit stuck in solving for the eigenvectors of the following equation.
begin{align}
begin{bmatrix}
2m+M-2komega & 0
\
0 & dfrac{1}{12}Ml^2 + dfrac{1}{2}ml^2 - dfrac{k}{2}l^2omega
end{bmatrix}
begin{bmatrix}
a_1 \ a_2
end{bmatrix}
=0
end{align}
I found the eigenvalues using Mathematica, but finding the Eigenvectors was problematic, so I attempted it by hand. The eigenvalues are:
begin{align}
omega_1=dfrac{3m}{2k} , omega_2 = dfrac{6m+M}{6k}
end{align}
In solving for the eigenvector corresponding to $omega_1$, I get to the following equation.
begin{align}
begin{bmatrix}
M-m & 0
\
0 & dfrac{1}{12}Ml^2 - dfrac{1}{6}ml^2
end{bmatrix}
begin{bmatrix}
a_1 \ a_2
end{bmatrix}
=0
end{align}
I'm totally unsure on how to solve this for the eigenvector. Essentially, my confusion lies in both $a_1$ and $a_2$ equal to zero here, since there are no off-diagonal terms.
Any help would be appreciated, thank you.
linear-algebra eigenvalues-eigenvectors classical-mechanics
asked Nov 29 at 8:31
Kosta
4991516
add a comment |
I'm a bit stuck in solving for the eigenvectors of the following equation.
begin{align}
begin{bmatrix}
2m+M-2komega & 0
\
0 & dfrac{1}{12}Ml^2 + dfrac{1}{2}ml^2 - dfrac{k}{2}l^2omega
end{bmatrix}
begin{bmatrix}
a_1 \ a_2
end{bmatrix}
=0
end{align}
I found the eigenvalues using Mathematica, but finding the Eigenvectors was problematic, so I attempted it by hand. The eigenvalues are:
begin{align}
omega_1=dfrac{3m}{2k} , omega_2 = dfrac{6m+M}{6k}
end{align}
In solving for the eigenvector corresponding to $omega_1$, I get to the following equation.
begin{align}
begin{bmatrix}
M-m & 0
\
0 & dfrac{1}{12}Ml^2 - dfrac{1}{6}ml^2
end{bmatrix}
begin{bmatrix}
a_1 \ a_2
end{bmatrix}
=0
end{align}
I'm totally unsure on how to solve this for the eigenvector. Essentially, my confusion lies in both $a_1$ and $a_2$ equal to zero here, since there are no off-diagonal terms.
Any help would be appreciated, thank you.
linear-algebra eigenvalues-eigenvectors classical-mechanics
asked Nov 29 at 8:31
Kosta
4991516
4
If $$ D = begin{bmatrix} a & 0 \ 0 & bend{bmatrix},$$ then clearly $[1,0]^T, [0,1]^T$ are eigenvectors belong to $a,b$ respectively.
– xbh
Nov 29 at 8:36
@xbh Can you explain this in further detail?
– Kosta
Nov 29 at 8:37
That claim is pretty obvious by verifying the definition $Dv = lambda v$.
– xbh
Nov 29 at 8:39
@xbh So, my eigenvectors here would be what, exactly? This isn't obvious to me.
– Kosta
Nov 29 at 8:47
Instead what did you mean by "eigenvectors of the following equation"? Usually we talk about eigenvectors of matrices, operators.
– xbh
Nov 29 at 9:01
add a comment |
I'm a bit stuck in solving for the eigenvectors of the following equation.
begin{align}
begin{bmatrix}
2m+M-2komega & 0
\
0 & dfrac{1}{12}Ml^2 + dfrac{1}{2}ml^2 - dfrac{k}{2}l^2omega
end{bmatrix}
begin{bmatrix}
a_1 \ a_2
end{bmatrix}
=0
end{align}
I found the eigenvalues using Mathematica, but finding the Eigenvectors was problematic, so I attempted it by hand. The eigenvalues are:
begin{align}
omega_1=dfrac{3m}{2k} , omega_2 = dfrac{6m+M}{6k}
end{align}
In solving for the eigenvector corresponding to $omega_1$, I get to the following equation.
begin{align}
begin{bmatrix}
M-m & 0
\
0 & dfrac{1}{12}Ml^2 - dfrac{1}{6}ml^2
end{bmatrix}
begin{bmatrix}
a_1 \ a_2
end{bmatrix}
=0
end{align}
I'm totally unsure on how to solve this for the eigenvector. Essentially, my confusion lies in both $a_1$ and $a_2$ equal to zero here, since there are no off-diagonal terms.
Any help would be appreciated, thank you.
linear-algebra eigenvalues-eigenvectors classical-mechanics
asked Nov 29 at 8:31
Kosta
4991516
I'm a bit stuck in solving for the eigenvectors of the following equation.
begin{align}
begin{bmatrix}
2m+M-2komega & 0
\
0 & dfrac{1}{12}Ml^2 + dfrac{1}{2}ml^2 - dfrac{k}{2}l^2omega
end{bmatrix}
begin{bmatrix}
a_1 \ a_2
end{bmatrix}
=0
end{align}
I found the eigenvalues using Mathematica, but finding the Eigenvectors was problematic, so I attempted it by hand. The eigenvalues are:
begin{align}
omega_1=dfrac{3m}{2k} , omega_2 = dfrac{6m+M}{6k}
end{align}
In solving for the eigenvector corresponding to $omega_1$, I get to the following equation.
begin{align}
begin{bmatrix}
M-m & 0
\
0 & dfrac{1}{12}Ml^2 - dfrac{1}{6}ml^2
end{bmatrix}
begin{bmatrix}
a_1 \ a_2
end{bmatrix}
=0
end{align}
I'm totally unsure on how to solve this for the eigenvector. Essentially, my confusion lies in both $a_1$ and $a_2$ equal to zero here, since there are no off-diagonal terms.
Any help would be appreciated, thank you.
linear-algebra eigenvalues-eigenvectors classical-mechanics
linear-algebra eigenvalues-eigenvectors classical-mechanics
asked Nov 29 at 8:31
Kosta
4991516
asked Nov 29 at 8:31
Kosta
4991516
asked Nov 29 at 8:31
Kosta
4991516
asked Nov 29 at 8:31
Kosta
4991516
asked Nov 29 at 8:31
Kosta
4991516
4991516
4
If $$ D = begin{bmatrix} a & 0 \ 0 & bend{bmatrix},$$ then clearly $[1,0]^T, [0,1]^T$ are eigenvectors belong to $a,b$ respectively.
– xbh
Nov 29 at 8:36
@xbh Can you explain this in further detail?
– Kosta
Nov 29 at 8:37
That claim is pretty obvious by verifying the definition $Dv = lambda v$.
– xbh
Nov 29 at 8:39
@xbh So, my eigenvectors here would be what, exactly? This isn't obvious to me.
– Kosta
Nov 29 at 8:47
Instead what did you mean by "eigenvectors of the following equation"? Usually we talk about eigenvectors of matrices, operators.
– xbh
Nov 29 at 9:01
add a comment |
4
If $$ D = begin{bmatrix} a & 0 \ 0 & bend{bmatrix},$$ then clearly $[1,0]^T, [0,1]^T$ are eigenvectors belong to $a,b$ respectively.
– xbh
Nov 29 at 8:36
@xbh Can you explain this in further detail?
– Kosta
Nov 29 at 8:37
That claim is pretty obvious by verifying the definition $Dv = lambda v$.
– xbh
Nov 29 at 8:39
@xbh So, my eigenvectors here would be what, exactly? This isn't obvious to me.
– Kosta
Nov 29 at 8:47
Instead what did you mean by "eigenvectors of the following equation"? Usually we talk about eigenvectors of matrices, operators.
– xbh
Nov 29 at 9:01
4
4
If $$ D = begin{bmatrix} a & 0 \ 0 & bend{bmatrix},$$ then clearly $[1,0]^T, [0,1]^T$ are eigenvectors belong to $a,b$ respectively.
– xbh
Nov 29 at 8:36
If $$ D = begin{bmatrix} a & 0 \ 0 & bend{bmatrix},$$ then clearly $[1,0]^T, [0,1]^T$ are eigenvectors belong to $a,b$ respectively.
– xbh
Nov 29 at 8:36
@xbh Can you explain this in further detail?
– Kosta
Nov 29 at 8:37
@xbh Can you explain this in further detail?
– Kosta
Nov 29 at 8:37
That claim is pretty obvious by verifying the definition $Dv = lambda v$.
– xbh
Nov 29 at 8:39
That claim is pretty obvious by verifying the definition $Dv = lambda v$.
– xbh
Nov 29 at 8:39
@xbh So, my eigenvectors here would be what, exactly? This isn't obvious to me.
– Kosta
Nov 29 at 8:47
@xbh So, my eigenvectors here would be what, exactly? This isn't obvious to me.
– Kosta
Nov 29 at 8:47
Instead what did you mean by "eigenvectors of the following equation"? Usually we talk about eigenvectors of matrices, operators.
– xbh
Nov 29 at 9:01
Instead what did you mean by "eigenvectors of the following equation"? Usually we talk about eigenvectors of matrices, operators.
– xbh
Nov 29 at 9:01
add a comment |
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4
If $$ D = begin{bmatrix} a & 0 \ 0 & bend{bmatrix},$$ then clearly $[1,0]^T, [0,1]^T$ are eigenvectors belong to $a,b$ respectively.
– xbh
Nov 29 at 8:36
@xbh Can you explain this in further detail?
– Kosta
Nov 29 at 8:37
That claim is pretty obvious by verifying the definition $Dv = lambda v$.
– xbh
Nov 29 at 8:39
@xbh So, my eigenvectors here would be what, exactly? This isn't obvious to me.
– Kosta
Nov 29 at 8:47
Instead what did you mean by "eigenvectors of the following equation"? Usually we talk about eigenvectors of matrices, operators.
– xbh
Nov 29 at 9:01