Having a given matrix and a linear transformation, how to find the bases with whom the given matrix is the...
$begingroup$
I am given the linear transformation $f(x,y,z)=(x-2y+2z,-x+3y-z,x+4z)$. I have to find the bases $beta_1$ and $beta_2$ with whom this matrix $
A=
left[ {begin{array}{cc}
1 & -1 & 0\
0 & 2 & 0 \
0 & 0 & 0 \
end{array} } right]
$.
Well, I have deduced that as $rg(A)=dimImf=2$ this bases must exist.
I have tried to solve the problem this way:
First find the bases with whom the associated matrix of $f$ is $ left[ {begin{array}{cc}
I_2 & 0\
0 & 0\
end{array} } right] =
left[ {begin{array}{cc}
1 & 0 & 0\
0 & 1 & 0 \
0 & 0 & 0 \
end{array} } right] $. Then try to fing $P$ and $Q$ that meet $PAQ=left[ {begin{array}{cc}
1 & 0 & 0\
0 & 1 & 0 \
0 & 0 & 0 \
end{array} } right]$. And from them get the bases. I would get $P$ and $Q$ building another transformation that has $A$ as the associated matrix with the standard basis. I have tried this but it is quite long and I couldn't get anything. Do you know a better way to get those bases? Thanks in advance.
linear-algebra linear-transformations
asked Dec 2 '18 at 15:42
AndarrkorAndarrkor
446
$endgroup$
add a comment |
$begingroup$
I am given the linear transformation $f(x,y,z)=(x-2y+2z,-x+3y-z,x+4z)$. I have to find the bases $beta_1$ and $beta_2$ with whom this matrix $
A=
left[ {begin{array}{cc}
1 & -1 & 0\
0 & 2 & 0 \
0 & 0 & 0 \
end{array} } right]
$.
Well, I have deduced that as $rg(A)=dimImf=2$ this bases must exist.
I have tried to solve the problem this way:
First find the bases with whom the associated matrix of $f$ is $ left[ {begin{array}{cc}
I_2 & 0\
0 & 0\
end{array} } right] =
left[ {begin{array}{cc}
1 & 0 & 0\
0 & 1 & 0 \
0 & 0 & 0 \
end{array} } right] $. Then try to fing $P$ and $Q$ that meet $PAQ=left[ {begin{array}{cc}
1 & 0 & 0\
0 & 1 & 0 \
0 & 0 & 0 \
end{array} } right]$. And from them get the bases. I would get $P$ and $Q$ building another transformation that has $A$ as the associated matrix with the standard basis. I have tried this but it is quite long and I couldn't get anything. Do you know a better way to get those bases? Thanks in advance.
linear-algebra linear-transformations
asked Dec 2 '18 at 15:42
AndarrkorAndarrkor
446
$endgroup$
add a comment |
$begingroup$
I am given the linear transformation $f(x,y,z)=(x-2y+2z,-x+3y-z,x+4z)$. I have to find the bases $beta_1$ and $beta_2$ with whom this matrix $
A=
left[ {begin{array}{cc}
1 & -1 & 0\
0 & 2 & 0 \
0 & 0 & 0 \
end{array} } right]
$.
Well, I have deduced that as $rg(A)=dimImf=2$ this bases must exist.
I have tried to solve the problem this way:
First find the bases with whom the associated matrix of $f$ is $ left[ {begin{array}{cc}
I_2 & 0\
0 & 0\
end{array} } right] =
left[ {begin{array}{cc}
1 & 0 & 0\
0 & 1 & 0 \
0 & 0 & 0 \
end{array} } right] $. Then try to fing $P$ and $Q$ that meet $PAQ=left[ {begin{array}{cc}
1 & 0 & 0\
0 & 1 & 0 \
0 & 0 & 0 \
end{array} } right]$. And from them get the bases. I would get $P$ and $Q$ building another transformation that has $A$ as the associated matrix with the standard basis. I have tried this but it is quite long and I couldn't get anything. Do you know a better way to get those bases? Thanks in advance.
linear-algebra linear-transformations
asked Dec 2 '18 at 15:42
AndarrkorAndarrkor
446
$endgroup$
I am given the linear transformation $f(x,y,z)=(x-2y+2z,-x+3y-z,x+4z)$. I have to find the bases $beta_1$ and $beta_2$ with whom this matrix $
A=
left[ {begin{array}{cc}
1 & -1 & 0\
0 & 2 & 0 \
0 & 0 & 0 \
end{array} } right]
$.
Well, I have deduced that as $rg(A)=dimImf=2$ this bases must exist.
I have tried to solve the problem this way:
First find the bases with whom the associated matrix of $f$ is $ left[ {begin{array}{cc}
I_2 & 0\
0 & 0\
end{array} } right] =
left[ {begin{array}{cc}
1 & 0 & 0\
0 & 1 & 0 \
0 & 0 & 0 \
end{array} } right] $. Then try to fing $P$ and $Q$ that meet $PAQ=left[ {begin{array}{cc}
1 & 0 & 0\
0 & 1 & 0 \
0 & 0 & 0 \
end{array} } right]$. And from them get the bases. I would get $P$ and $Q$ building another transformation that has $A$ as the associated matrix with the standard basis. I have tried this but it is quite long and I couldn't get anything. Do you know a better way to get those bases? Thanks in advance.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Dec 2 '18 at 15:42
AndarrkorAndarrkor
446
asked Dec 2 '18 at 15:42
AndarrkorAndarrkor
446
asked Dec 2 '18 at 15:42
AndarrkorAndarrkor
446
asked Dec 2 '18 at 15:42
AndarrkorAndarrkor
446
asked Dec 2 '18 at 15:42
AndarrkorAndarrkor
446
446
add a comment |
add a comment |
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