Faster way to calculate the inverse of matrice $C $(when diagonisable.. $C-1AC = D$)
$begingroup$
We found the orthonormal basis for the eigen spaces.
We got $C$ to be the matrix
[ 1/squareroot(2) 1/squareroot(6) 1/squareroot(3)
-1/squareroot(2) 1/squareroot(6) 1/squareroot(3)
0 -2/squareroot(6) 1/squareroot(3) ]
And the original matrix $A $ is
[4 2 2
2 4 2
2 2 4]
After finding $C$, my notes jump to:
therefore $C^-1 A C = $
[2 0 0
0 2 0
0 0 8]
They do not show any steps on how to calculate the inverse of $C$. Is there an easy way of calculating it? How would I start off reducing it to RREF? How would I get rid of the square roots? (normally, I'm used to just dealing with regular integers).
Thanks in advance!
matrices diagonalization
edited Dec 21 '18 at 11:32
dmtri
1,7612521
asked Dec 21 '18 at 10:43
Jay PatelJay Patel
656
$endgroup$
add a comment |
$begingroup$
We found the orthonormal basis for the eigen spaces.
We got $C$ to be the matrix
[ 1/squareroot(2) 1/squareroot(6) 1/squareroot(3)
-1/squareroot(2) 1/squareroot(6) 1/squareroot(3)
0 -2/squareroot(6) 1/squareroot(3) ]
And the original matrix $A $ is
[4 2 2
2 4 2
2 2 4]
After finding $C$, my notes jump to:
therefore $C^-1 A C = $
[2 0 0
0 2 0
0 0 8]
They do not show any steps on how to calculate the inverse of $C$. Is there an easy way of calculating it? How would I start off reducing it to RREF? How would I get rid of the square roots? (normally, I'm used to just dealing with regular integers).
Thanks in advance!
matrices diagonalization
edited Dec 21 '18 at 11:32
dmtri
1,7612521
asked Dec 21 '18 at 10:43
Jay PatelJay Patel
656
$endgroup$
add a comment |
$begingroup$
We found the orthonormal basis for the eigen spaces.
We got $C$ to be the matrix
[ 1/squareroot(2) 1/squareroot(6) 1/squareroot(3)
-1/squareroot(2) 1/squareroot(6) 1/squareroot(3)
0 -2/squareroot(6) 1/squareroot(3) ]
And the original matrix $A $ is
[4 2 2
2 4 2
2 2 4]
After finding $C$, my notes jump to:
therefore $C^-1 A C = $
[2 0 0
0 2 0
0 0 8]
They do not show any steps on how to calculate the inverse of $C$. Is there an easy way of calculating it? How would I start off reducing it to RREF? How would I get rid of the square roots? (normally, I'm used to just dealing with regular integers).
Thanks in advance!
matrices diagonalization
edited Dec 21 '18 at 11:32
dmtri
1,7612521
asked Dec 21 '18 at 10:43
Jay PatelJay Patel
656
$endgroup$
We found the orthonormal basis for the eigen spaces.
We got $C$ to be the matrix
[ 1/squareroot(2) 1/squareroot(6) 1/squareroot(3)
-1/squareroot(2) 1/squareroot(6) 1/squareroot(3)
0 -2/squareroot(6) 1/squareroot(3) ]
And the original matrix $A $ is
[4 2 2
2 4 2
2 2 4]
After finding $C$, my notes jump to:
therefore $C^-1 A C = $
[2 0 0
0 2 0
0 0 8]
They do not show any steps on how to calculate the inverse of $C$. Is there an easy way of calculating it? How would I start off reducing it to RREF? How would I get rid of the square roots? (normally, I'm used to just dealing with regular integers).
Thanks in advance!
matrices diagonalization
matrices diagonalization
edited Dec 21 '18 at 11:32
dmtri
1,7612521
asked Dec 21 '18 at 10:43
Jay PatelJay Patel
656
edited Dec 21 '18 at 11:32
dmtri
1,7612521
asked Dec 21 '18 at 10:43
Jay PatelJay Patel
656
edited Dec 21 '18 at 11:32
dmtri
1,7612521
edited Dec 21 '18 at 11:32
dmtri
1,7612521
edited Dec 21 '18 at 11:32
dmtri
1,7612521
1,7612521
asked Dec 21 '18 at 10:43
Jay PatelJay Patel
656
asked Dec 21 '18 at 10:43
Jay PatelJay Patel
656
asked Dec 21 '18 at 10:43
Jay PatelJay Patel
656
656
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The matrix $C$ is orthogonal. Hence $C^{-1}=C^T$.
answered Dec 21 '18 at 10:56
nicomezinicomezi
4,26411020
$endgroup$
add a comment |
$begingroup$
Notice that $ A = begin{bmatrix}4 & 2 & 2 \ 2 & 4 & 2 \ 2 & 2 & 4 end{bmatrix}$
has eigenvalues $2$ and $8$. Since there exists a basis of eigenvectors $v_1,v_2,v_3$ define the change of basis matrix $C = begin{bmatrix} vec{v_1} & vec{v}_2 &vec{v}_3\end{bmatrix}$. Then $$ C^{-1} A C$$ is the expression of $A$ in the basis $ {v_1, v_2,v_3}$. Since this is a basis of eigenvectors $C^{-1} AC$ is a diagonal matrix with the eigenvalues on the diagonal
$$ begin{bmatrix} lambda_1 & 0& 0 \ 0 & lambda_2 & 0 \ 0 & 0 & lambda_3 end{bmatrix}$$
Where $lambda_i$ is the eigenvalue corresponding to the eigenvector $v_i$. It is not necessary to explicitly calculate the inverse of $C$ and multiply the three matrices together. If you still wish to do it check out nicomezi's answer.
answered Dec 21 '18 at 11:09
DigitalisDigitalis
554216
$endgroup$
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
$begingroup$
The matrix $C$ is orthogonal. Hence $C^{-1}=C^T$.
answered Dec 21 '18 at 10:56
nicomezinicomezi
4,26411020
$endgroup$
add a comment |
$begingroup$
The matrix $C$ is orthogonal. Hence $C^{-1}=C^T$.
answered Dec 21 '18 at 10:56
nicomezinicomezi
4,26411020
$endgroup$
add a comment |
$begingroup$
The matrix $C$ is orthogonal. Hence $C^{-1}=C^T$.
answered Dec 21 '18 at 10:56
nicomezinicomezi
4,26411020
$endgroup$
The matrix $C$ is orthogonal. Hence $C^{-1}=C^T$.
answered Dec 21 '18 at 10:56
nicomezinicomezi
4,26411020
answered Dec 21 '18 at 10:56
nicomezinicomezi
4,26411020
answered Dec 21 '18 at 10:56
nicomezinicomezi
4,26411020
answered Dec 21 '18 at 10:56
nicomezinicomezi
4,26411020
4,26411020
add a comment |
add a comment |
$begingroup$
Notice that $ A = begin{bmatrix}4 & 2 & 2 \ 2 & 4 & 2 \ 2 & 2 & 4 end{bmatrix}$
has eigenvalues $2$ and $8$. Since there exists a basis of eigenvectors $v_1,v_2,v_3$ define the change of basis matrix $C = begin{bmatrix} vec{v_1} & vec{v}_2 &vec{v}_3\end{bmatrix}$. Then $$ C^{-1} A C$$ is the expression of $A$ in the basis $ {v_1, v_2,v_3}$. Since this is a basis of eigenvectors $C^{-1} AC$ is a diagonal matrix with the eigenvalues on the diagonal
$$ begin{bmatrix} lambda_1 & 0& 0 \ 0 & lambda_2 & 0 \ 0 & 0 & lambda_3 end{bmatrix}$$
Where $lambda_i$ is the eigenvalue corresponding to the eigenvector $v_i$. It is not necessary to explicitly calculate the inverse of $C$ and multiply the three matrices together. If you still wish to do it check out nicomezi's answer.
answered Dec 21 '18 at 11:09
DigitalisDigitalis
554216
$endgroup$
add a comment |
$begingroup$
Notice that $ A = begin{bmatrix}4 & 2 & 2 \ 2 & 4 & 2 \ 2 & 2 & 4 end{bmatrix}$
has eigenvalues $2$ and $8$. Since there exists a basis of eigenvectors $v_1,v_2,v_3$ define the change of basis matrix $C = begin{bmatrix} vec{v_1} & vec{v}_2 &vec{v}_3\end{bmatrix}$. Then $$ C^{-1} A C$$ is the expression of $A$ in the basis $ {v_1, v_2,v_3}$. Since this is a basis of eigenvectors $C^{-1} AC$ is a diagonal matrix with the eigenvalues on the diagonal
$$ begin{bmatrix} lambda_1 & 0& 0 \ 0 & lambda_2 & 0 \ 0 & 0 & lambda_3 end{bmatrix}$$
Where $lambda_i$ is the eigenvalue corresponding to the eigenvector $v_i$. It is not necessary to explicitly calculate the inverse of $C$ and multiply the three matrices together. If you still wish to do it check out nicomezi's answer.
answered Dec 21 '18 at 11:09
DigitalisDigitalis
554216
$endgroup$
add a comment |
$begingroup$
Notice that $ A = begin{bmatrix}4 & 2 & 2 \ 2 & 4 & 2 \ 2 & 2 & 4 end{bmatrix}$
has eigenvalues $2$ and $8$. Since there exists a basis of eigenvectors $v_1,v_2,v_3$ define the change of basis matrix $C = begin{bmatrix} vec{v_1} & vec{v}_2 &vec{v}_3\end{bmatrix}$. Then $$ C^{-1} A C$$ is the expression of $A$ in the basis $ {v_1, v_2,v_3}$. Since this is a basis of eigenvectors $C^{-1} AC$ is a diagonal matrix with the eigenvalues on the diagonal
$$ begin{bmatrix} lambda_1 & 0& 0 \ 0 & lambda_2 & 0 \ 0 & 0 & lambda_3 end{bmatrix}$$
Where $lambda_i$ is the eigenvalue corresponding to the eigenvector $v_i$. It is not necessary to explicitly calculate the inverse of $C$ and multiply the three matrices together. If you still wish to do it check out nicomezi's answer.
answered Dec 21 '18 at 11:09
DigitalisDigitalis
554216
$endgroup$
Notice that $ A = begin{bmatrix}4 & 2 & 2 \ 2 & 4 & 2 \ 2 & 2 & 4 end{bmatrix}$
has eigenvalues $2$ and $8$. Since there exists a basis of eigenvectors $v_1,v_2,v_3$ define the change of basis matrix $C = begin{bmatrix} vec{v_1} & vec{v}_2 &vec{v}_3\end{bmatrix}$. Then $$ C^{-1} A C$$ is the expression of $A$ in the basis $ {v_1, v_2,v_3}$. Since this is a basis of eigenvectors $C^{-1} AC$ is a diagonal matrix with the eigenvalues on the diagonal
$$ begin{bmatrix} lambda_1 & 0& 0 \ 0 & lambda_2 & 0 \ 0 & 0 & lambda_3 end{bmatrix}$$
Where $lambda_i$ is the eigenvalue corresponding to the eigenvector $v_i$. It is not necessary to explicitly calculate the inverse of $C$ and multiply the three matrices together. If you still wish to do it check out nicomezi's answer.
answered Dec 21 '18 at 11:09
DigitalisDigitalis
554216
answered Dec 21 '18 at 11:09
DigitalisDigitalis
554216
answered Dec 21 '18 at 11:09
DigitalisDigitalis
554216
answered Dec 21 '18 at 11:09
DigitalisDigitalis
554216
554216
add a comment |
add a comment |
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