Find a change-of-coordinates matrix $P$ and use it to show that $A$ is similar to $begin{bmatrix} a & -b...
Let $A = begin{bmatrix}
1 & 5 \
-1 & 3
end{bmatrix}$ act on $mathbb{C}^2$. An eigenvector of $A$ is $begin{bmatrix}
1-2i \
1
end{bmatrix}$, corresponding to the eigenvalue $2+2i$. Find a change-of-coordinates matrix $P$ and use it to show that $A$ is similar to a matrix of the form $begin{bmatrix}
a & -b \
b & a
end{bmatrix}$.
I know that Eigenvectors for $begin{bmatrix}
a & -b \
b & a
end{bmatrix}$. are $begin{bmatrix}1\-iend{bmatrix}$ and $begin{bmatrix}1\iend{bmatrix}$. I believe I need to make a augmented matrix and row reduce yet still unsure how to approach it.
linear-algebra change-of-basis
edited Nov 29 at 9:38
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 at 8:29
HighSchool15
1,065517
add a comment |
Let $A = begin{bmatrix}
1 & 5 \
-1 & 3
end{bmatrix}$ act on $mathbb{C}^2$. An eigenvector of $A$ is $begin{bmatrix}
1-2i \
1
end{bmatrix}$, corresponding to the eigenvalue $2+2i$. Find a change-of-coordinates matrix $P$ and use it to show that $A$ is similar to a matrix of the form $begin{bmatrix}
a & -b \
b & a
end{bmatrix}$.
I know that Eigenvectors for $begin{bmatrix}
a & -b \
b & a
end{bmatrix}$. are $begin{bmatrix}1\-iend{bmatrix}$ and $begin{bmatrix}1\iend{bmatrix}$. I believe I need to make a augmented matrix and row reduce yet still unsure how to approach it.
linear-algebra change-of-basis
edited Nov 29 at 9:38
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 at 8:29
HighSchool15
1,065517
A real polynomial has conjugate roots.
– GNUSupporter 8964民主女神 地下教會
Nov 29 at 9:40
1
They will be similar if they have common eigenvalues. From this you obtain $a,b.$ Then one can go through the diagonalization on each of matrices. There can exist a more elegant method...this one will works as well.
– user376343
Nov 29 at 10:01
Play around with the real and imaginary parts of the eigenvectors.
– amd
Nov 29 at 20:00
add a comment |
Let $A = begin{bmatrix}
1 & 5 \
-1 & 3
end{bmatrix}$ act on $mathbb{C}^2$. An eigenvector of $A$ is $begin{bmatrix}
1-2i \
1
end{bmatrix}$, corresponding to the eigenvalue $2+2i$. Find a change-of-coordinates matrix $P$ and use it to show that $A$ is similar to a matrix of the form $begin{bmatrix}
a & -b \
b & a
end{bmatrix}$.
I know that Eigenvectors for $begin{bmatrix}
a & -b \
b & a
end{bmatrix}$. are $begin{bmatrix}1\-iend{bmatrix}$ and $begin{bmatrix}1\iend{bmatrix}$. I believe I need to make a augmented matrix and row reduce yet still unsure how to approach it.
linear-algebra change-of-basis
edited Nov 29 at 9:38
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 at 8:29
HighSchool15
1,065517
Let $A = begin{bmatrix}
1 & 5 \
-1 & 3
end{bmatrix}$ act on $mathbb{C}^2$. An eigenvector of $A$ is $begin{bmatrix}
1-2i \
1
end{bmatrix}$, corresponding to the eigenvalue $2+2i$. Find a change-of-coordinates matrix $P$ and use it to show that $A$ is similar to a matrix of the form $begin{bmatrix}
a & -b \
b & a
end{bmatrix}$.
I know that Eigenvectors for $begin{bmatrix}
a & -b \
b & a
end{bmatrix}$. are $begin{bmatrix}1\-iend{bmatrix}$ and $begin{bmatrix}1\iend{bmatrix}$. I believe I need to make a augmented matrix and row reduce yet still unsure how to approach it.
linear-algebra change-of-basis
linear-algebra change-of-basis
edited Nov 29 at 9:38
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 at 8:29
HighSchool15
1,065517
edited Nov 29 at 9:38
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 at 8:29
HighSchool15
1,065517
edited Nov 29 at 9:38
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 29 at 9:38
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 29 at 9:38
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Nov 29 at 8:29
HighSchool15
1,065517
asked Nov 29 at 8:29
HighSchool15
1,065517
asked Nov 29 at 8:29
HighSchool15
1,065517
1,065517
A real polynomial has conjugate roots.
– GNUSupporter 8964民主女神 地下教會
Nov 29 at 9:40
1
They will be similar if they have common eigenvalues. From this you obtain $a,b.$ Then one can go through the diagonalization on each of matrices. There can exist a more elegant method...this one will works as well.
– user376343
Nov 29 at 10:01
Play around with the real and imaginary parts of the eigenvectors.
– amd
Nov 29 at 20:00
add a comment |
A real polynomial has conjugate roots.
– GNUSupporter 8964民主女神 地下教會
Nov 29 at 9:40
1
They will be similar if they have common eigenvalues. From this you obtain $a,b.$ Then one can go through the diagonalization on each of matrices. There can exist a more elegant method...this one will works as well.
– user376343
Nov 29 at 10:01
Play around with the real and imaginary parts of the eigenvectors.
– amd
Nov 29 at 20:00
A real polynomial has conjugate roots.
– GNUSupporter 8964民主女神 地下教會
Nov 29 at 9:40
A real polynomial has conjugate roots.
– GNUSupporter 8964民主女神 地下教會
Nov 29 at 9:40
1
1
They will be similar if they have common eigenvalues. From this you obtain $a,b.$ Then one can go through the diagonalization on each of matrices. There can exist a more elegant method...this one will works as well.
– user376343
Nov 29 at 10:01
They will be similar if they have common eigenvalues. From this you obtain $a,b.$ Then one can go through the diagonalization on each of matrices. There can exist a more elegant method...this one will works as well.
– user376343
Nov 29 at 10:01
Play around with the real and imaginary parts of the eigenvectors.
– amd
Nov 29 at 20:00
Play around with the real and imaginary parts of the eigenvectors.
– amd
Nov 29 at 20:00
add a comment |
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A real polynomial has conjugate roots.
– GNUSupporter 8964民主女神 地下教會
Nov 29 at 9:40
1
They will be similar if they have common eigenvalues. From this you obtain $a,b.$ Then one can go through the diagonalization on each of matrices. There can exist a more elegant method...this one will works as well.
– user376343
Nov 29 at 10:01
Play around with the real and imaginary parts of the eigenvectors.
– amd
Nov 29 at 20:00