Mixing
Find $P_2$ and $Q_2$ where $P_2cdot A cdot Q_2=B$ knowing that $Pcdot A cdot Q= I_r $
$begingroup$
Lets assume we have three matrix $A$,$P$ and $Q$, where:
$$Pcdot A cdot Q=
left[ {begin{array}{cc}
I_r & 0\
0 & 0\
end{array} } right]
$$
Now find $P_2$ and $Q_2$ where:
$$P_2cdot A cdot Q_2=B$$
And $B$ is a known matrix, how can I do this?
Example: (This is the exercise the problem comes from)
$$A=
left[ {begin{array}{cc}
1 & 0 & 2\
-1 & 1 & -1\
-1 & 4 & 2\
end{array} } right]
$$
Find $P$ and $Q$ where $Pcdot A cdot Q=
left[ {begin{array}{cc}
I_r & 0\
0 & 0\
end{array} } right]
$
I start defining $A=M_{beta_{k}}^{beta_{k}}(f)$
Then I find $Q=M_{beta}^{beta_{k}}=
left[ {begin{array}{cc}
1 & 0 & -2\
0 & 1 & -1\
0 & 0 & 1\
end{array} } right]$ and $P=M_{beta_{k}}^{beta^{'}}=left[ {begin{array}{cc}
0 & -4/3 & 1/3\
0 & -1/3 & 1/3\
1 & 4/3 & -1/3\
end{array} } right]$
Where $beta = {(1,0,0),(0,1,0),(-2,-1,1)}$ and $beta^{'}={(1,-1,1),(0,1,4),(1,0,0)}$
Now find $P_2$ and $Q_2$ where:
$P_2cdot A cdot Q_2=B=left[ {begin{array}{cc}
1 & 0 & 1\
1 & 1 & 0\
0 & 1 & -1\
end{array} } right]$
Those 2 matrices exist as $rg(A)=rg(B)$, how to find them?
linear-algebra
asked Jan 18 at 11:51
user605734 MBSuser605734 MBS
3029
$endgroup$
add a comment |
$begingroup$
Lets assume we have three matrix $A$,$P$ and $Q$, where:
$$Pcdot A cdot Q=
left[ {begin{array}{cc}
I_r & 0\
0 & 0\
end{array} } right]
$$
Now find $P_2$ and $Q_2$ where:
$$P_2cdot A cdot Q_2=B$$
And $B$ is a known matrix, how can I do this?
Example: (This is the exercise the problem comes from)
$$A=
left[ {begin{array}{cc}
1 & 0 & 2\
-1 & 1 & -1\
-1 & 4 & 2\
end{array} } right]
$$
Find $P$ and $Q$ where $Pcdot A cdot Q=
left[ {begin{array}{cc}
I_r & 0\
0 & 0\
end{array} } right]
$
I start defining $A=M_{beta_{k}}^{beta_{k}}(f)$
Then I find $Q=M_{beta}^{beta_{k}}=
left[ {begin{array}{cc}
1 & 0 & -2\
0 & 1 & -1\
0 & 0 & 1\
end{array} } right]$ and $P=M_{beta_{k}}^{beta^{'}}=left[ {begin{array}{cc}
0 & -4/3 & 1/3\
0 & -1/3 & 1/3\
1 & 4/3 & -1/3\
end{array} } right]$
Where $beta = {(1,0,0),(0,1,0),(-2,-1,1)}$ and $beta^{'}={(1,-1,1),(0,1,4),(1,0,0)}$
Now find $P_2$ and $Q_2$ where:
$P_2cdot A cdot Q_2=B=left[ {begin{array}{cc}
1 & 0 & 1\
1 & 1 & 0\
0 & 1 & -1\
end{array} } right]$
Those 2 matrices exist as $rg(A)=rg(B)$, how to find them?
linear-algebra
asked Jan 18 at 11:51
user605734 MBSuser605734 MBS
3029
$endgroup$
2
$begingroup$
If you find $P_3BQ_3=begin{bmatrix}I_r & 0\0 & 0end{bmatrix}$ then it is straigthforward from $P_3BQ_3=PAQ$.
$endgroup$
– A.Γ.
Jan 18 at 11:56
$begingroup$
Can I define $B=M_{beta_{k}}^{beta_{k}}(f)$ after defining $A=M_{beta_{k}}^{beta_{k}}(f)$? Or is it another function and $B=M_{beta_{k}}^{beta_{k}}(g)$?
$endgroup$
– user605734 MBS
Jan 18 at 12:05
$begingroup$
For the approach that you’re taking, assume that $A$ and $B$ represent the same map, but with expressed w/r different pairs of bases.
$endgroup$
– amd
Jan 21 at 0:13
add a comment |
$begingroup$
Lets assume we have three matrix $A$,$P$ and $Q$, where:
$$Pcdot A cdot Q=
left[ {begin{array}{cc}
I_r & 0\
0 & 0\
end{array} } right]
$$
Now find $P_2$ and $Q_2$ where:
$$P_2cdot A cdot Q_2=B$$
And $B$ is a known matrix, how can I do this?
Example: (This is the exercise the problem comes from)
$$A=
left[ {begin{array}{cc}
1 & 0 & 2\
-1 & 1 & -1\
-1 & 4 & 2\
end{array} } right]
$$
Find $P$ and $Q$ where $Pcdot A cdot Q=
left[ {begin{array}{cc}
I_r & 0\
0 & 0\
end{array} } right]
$
I start defining $A=M_{beta_{k}}^{beta_{k}}(f)$
Then I find $Q=M_{beta}^{beta_{k}}=
left[ {begin{array}{cc}
1 & 0 & -2\
0 & 1 & -1\
0 & 0 & 1\
end{array} } right]$ and $P=M_{beta_{k}}^{beta^{'}}=left[ {begin{array}{cc}
0 & -4/3 & 1/3\
0 & -1/3 & 1/3\
1 & 4/3 & -1/3\
end{array} } right]$
Where $beta = {(1,0,0),(0,1,0),(-2,-1,1)}$ and $beta^{'}={(1,-1,1),(0,1,4),(1,0,0)}$
Now find $P_2$ and $Q_2$ where:
$P_2cdot A cdot Q_2=B=left[ {begin{array}{cc}
1 & 0 & 1\
1 & 1 & 0\
0 & 1 & -1\
end{array} } right]$
Those 2 matrices exist as $rg(A)=rg(B)$, how to find them?
linear-algebra
asked Jan 18 at 11:51
user605734 MBSuser605734 MBS
3029
$endgroup$
Lets assume we have three matrix $A$,$P$ and $Q$, where:
$$Pcdot A cdot Q=
left[ {begin{array}{cc}
I_r & 0\
0 & 0\
end{array} } right]
$$
Now find $P_2$ and $Q_2$ where:
$$P_2cdot A cdot Q_2=B$$
And $B$ is a known matrix, how can I do this?
Example: (This is the exercise the problem comes from)
$$A=
left[ {begin{array}{cc}
1 & 0 & 2\
-1 & 1 & -1\
-1 & 4 & 2\
end{array} } right]
$$
Find $P$ and $Q$ where $Pcdot A cdot Q=
left[ {begin{array}{cc}
I_r & 0\
0 & 0\
end{array} } right]
$
I start defining $A=M_{beta_{k}}^{beta_{k}}(f)$
Then I find $Q=M_{beta}^{beta_{k}}=
left[ {begin{array}{cc}
1 & 0 & -2\
0 & 1 & -1\
0 & 0 & 1\
end{array} } right]$ and $P=M_{beta_{k}}^{beta^{'}}=left[ {begin{array}{cc}
0 & -4/3 & 1/3\
0 & -1/3 & 1/3\
1 & 4/3 & -1/3\
end{array} } right]$
Where $beta = {(1,0,0),(0,1,0),(-2,-1,1)}$ and $beta^{'}={(1,-1,1),(0,1,4),(1,0,0)}$
Now find $P_2$ and $Q_2$ where:
$P_2cdot A cdot Q_2=B=left[ {begin{array}{cc}
1 & 0 & 1\
1 & 1 & 0\
0 & 1 & -1\
end{array} } right]$
Those 2 matrices exist as $rg(A)=rg(B)$, how to find them?
linear-algebra
linear-algebra
asked Jan 18 at 11:51
user605734 MBSuser605734 MBS
3029
asked Jan 18 at 11:51
user605734 MBSuser605734 MBS
3029
asked Jan 18 at 11:51
user605734 MBSuser605734 MBS
3029
asked Jan 18 at 11:51
user605734 MBSuser605734 MBS
3029
asked Jan 18 at 11:51
user605734 MBSuser605734 MBS
3029
3029
2
$begingroup$
If you find $P_3BQ_3=begin{bmatrix}I_r & 0\0 & 0end{bmatrix}$ then it is straigthforward from $P_3BQ_3=PAQ$.
$endgroup$
– A.Γ.
Jan 18 at 11:56
$begingroup$
Can I define $B=M_{beta_{k}}^{beta_{k}}(f)$ after defining $A=M_{beta_{k}}^{beta_{k}}(f)$? Or is it another function and $B=M_{beta_{k}}^{beta_{k}}(g)$?
$endgroup$
– user605734 MBS
Jan 18 at 12:05
$begingroup$
For the approach that you’re taking, assume that $A$ and $B$ represent the same map, but with expressed w/r different pairs of bases.
$endgroup$
– amd
Jan 21 at 0:13
add a comment |
2
$begingroup$
If you find $P_3BQ_3=begin{bmatrix}I_r & 0\0 & 0end{bmatrix}$ then it is straigthforward from $P_3BQ_3=PAQ$.
$endgroup$
– A.Γ.
Jan 18 at 11:56
$begingroup$
Can I define $B=M_{beta_{k}}^{beta_{k}}(f)$ after defining $A=M_{beta_{k}}^{beta_{k}}(f)$? Or is it another function and $B=M_{beta_{k}}^{beta_{k}}(g)$?
$endgroup$
– user605734 MBS
Jan 18 at 12:05
$begingroup$
For the approach that you’re taking, assume that $A$ and $B$ represent the same map, but with expressed w/r different pairs of bases.
$endgroup$
– amd
Jan 21 at 0:13
2
2
$begingroup$
If you find $P_3BQ_3=begin{bmatrix}I_r & 0\0 & 0end{bmatrix}$ then it is straigthforward from $P_3BQ_3=PAQ$.
$endgroup$
– A.Γ.
Jan 18 at 11:56
$begingroup$
If you find $P_3BQ_3=begin{bmatrix}I_r & 0\0 & 0end{bmatrix}$ then it is straigthforward from $P_3BQ_3=PAQ$.
$endgroup$
– A.Γ.
Jan 18 at 11:56
$begingroup$
Can I define $B=M_{beta_{k}}^{beta_{k}}(f)$ after defining $A=M_{beta_{k}}^{beta_{k}}(f)$? Or is it another function and $B=M_{beta_{k}}^{beta_{k}}(g)$?
$endgroup$
– user605734 MBS
Jan 18 at 12:05
$begingroup$
Can I define $B=M_{beta_{k}}^{beta_{k}}(f)$ after defining $A=M_{beta_{k}}^{beta_{k}}(f)$? Or is it another function and $B=M_{beta_{k}}^{beta_{k}}(g)$?
$endgroup$
– user605734 MBS
Jan 18 at 12:05
$begingroup$
For the approach that you’re taking, assume that $A$ and $B$ represent the same map, but with expressed w/r different pairs of bases.
$endgroup$
– amd
Jan 21 at 0:13
$begingroup$
For the approach that you’re taking, assume that $A$ and $B$ represent the same map, but with expressed w/r different pairs of bases.
$endgroup$
– amd
Jan 21 at 0:13
add a comment |
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2
$begingroup$
If you find $P_3BQ_3=begin{bmatrix}I_r & 0\0 & 0end{bmatrix}$ then it is straigthforward from $P_3BQ_3=PAQ$.
$endgroup$
– A.Γ.
Jan 18 at 11:56
$begingroup$
Can I define $B=M_{beta_{k}}^{beta_{k}}(f)$ after defining $A=M_{beta_{k}}^{beta_{k}}(f)$? Or is it another function and $B=M_{beta_{k}}^{beta_{k}}(g)$?
$endgroup$
– user605734 MBS
Jan 18 at 12:05
$begingroup$
For the approach that you’re taking, assume that $A$ and $B$ represent the same map, but with expressed w/r different pairs of bases.
$endgroup$
– amd
Jan 21 at 0:13