Question on matrix algebra
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1
down vote
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Consider the matrix equation
$$
ABx = Ay
$$
where $A,B in mathbbR^n times n$ are square matrices, and $x,y in mathbbR^n$ are column vectors. If $A$ is invertible, then I can multiply the equation by $A^-1$ and find
$$
Bx = y
$$
However, now if I modify $B$ to be $n times m$ with $n > m$ and $x in mathbbR^m$, it seems to be no longer true that I can multiply by $A^-1$ and obtain $Bx = y$ (I've verified this with some matlab experiments).
So my question is under what circumstances can I multiply a matrix equation by $A^-1$ and still have the equation be true? This results seems a bit counter-intuitive.
linear-algebra matrix-equations
asked Aug 30 at 1:27
vibe
1438
add a comment |Â
up vote
1
down vote
favorite
Consider the matrix equation
$$
ABx = Ay
$$
where $A,B in mathbbR^n times n$ are square matrices, and $x,y in mathbbR^n$ are column vectors. If $A$ is invertible, then I can multiply the equation by $A^-1$ and find
$$
Bx = y
$$
However, now if I modify $B$ to be $n times m$ with $n > m$ and $x in mathbbR^m$, it seems to be no longer true that I can multiply by $A^-1$ and obtain $Bx = y$ (I've verified this with some matlab experiments).
So my question is under what circumstances can I multiply a matrix equation by $A^-1$ and still have the equation be true? This results seems a bit counter-intuitive.
linear-algebra matrix-equations
asked Aug 30 at 1:27
vibe
1438
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the matrix equation
$$
ABx = Ay
$$
where $A,B in mathbbR^n times n$ are square matrices, and $x,y in mathbbR^n$ are column vectors. If $A$ is invertible, then I can multiply the equation by $A^-1$ and find
$$
Bx = y
$$
However, now if I modify $B$ to be $n times m$ with $n > m$ and $x in mathbbR^m$, it seems to be no longer true that I can multiply by $A^-1$ and obtain $Bx = y$ (I've verified this with some matlab experiments).
So my question is under what circumstances can I multiply a matrix equation by $A^-1$ and still have the equation be true? This results seems a bit counter-intuitive.
linear-algebra matrix-equations
asked Aug 30 at 1:27
vibe
1438
Consider the matrix equation
$$
ABx = Ay
$$
where $A,B in mathbbR^n times n$ are square matrices, and $x,y in mathbbR^n$ are column vectors. If $A$ is invertible, then I can multiply the equation by $A^-1$ and find
$$
Bx = y
$$
However, now if I modify $B$ to be $n times m$ with $n > m$ and $x in mathbbR^m$, it seems to be no longer true that I can multiply by $A^-1$ and obtain $Bx = y$ (I've verified this with some matlab experiments).
So my question is under what circumstances can I multiply a matrix equation by $A^-1$ and still have the equation be true? This results seems a bit counter-intuitive.
linear-algebra matrix-equations
linear-algebra matrix-equations
asked Aug 30 at 1:27
vibe
1438
asked Aug 30 at 1:27
vibe
1438
asked Aug 30 at 1:27
vibe
1438
asked Aug 30 at 1:27
vibe
1438
asked Aug 30 at 1:27
vibe
1438
1438
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
$$ABx=Ay$$
$$A^-1ABx=A^-1Ay$$
$$I_nBx=I_ny$$
$$Bx=y$$
You can always do that. Perhaps you might want to check the code.
answered Aug 30 at 1:37
Siong Thye Goh
81.2k1453103
add a comment |Â
up vote
0
down vote
As you mentioned, it is anti-intuitive and something is wrong because if A is invertible , you should get the same results by multiplying both sides by $A^-1$
answered Aug 30 at 1:46
Mohammad Riazi-Kermani
31k41853
No, $y$ is $n times 1$
– vibe
Aug 30 at 2:13
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$$ABx=Ay$$
$$A^-1ABx=A^-1Ay$$
$$I_nBx=I_ny$$
$$Bx=y$$
You can always do that. Perhaps you might want to check the code.
answered Aug 30 at 1:37
Siong Thye Goh
81.2k1453103
add a comment |Â
up vote
1
down vote
$$ABx=Ay$$
$$A^-1ABx=A^-1Ay$$
$$I_nBx=I_ny$$
$$Bx=y$$
You can always do that. Perhaps you might want to check the code.
answered Aug 30 at 1:37
Siong Thye Goh
81.2k1453103
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$ABx=Ay$$
$$A^-1ABx=A^-1Ay$$
$$I_nBx=I_ny$$
$$Bx=y$$
You can always do that. Perhaps you might want to check the code.
answered Aug 30 at 1:37
Siong Thye Goh
81.2k1453103
$$ABx=Ay$$
$$A^-1ABx=A^-1Ay$$
$$I_nBx=I_ny$$
$$Bx=y$$
You can always do that. Perhaps you might want to check the code.
answered Aug 30 at 1:37
Siong Thye Goh
81.2k1453103
answered Aug 30 at 1:37
Siong Thye Goh
81.2k1453103
answered Aug 30 at 1:37
Siong Thye Goh
81.2k1453103
answered Aug 30 at 1:37
Siong Thye Goh
81.2k1453103
81.2k1453103
add a comment |Â
add a comment |Â
up vote
0
down vote
As you mentioned, it is anti-intuitive and something is wrong because if A is invertible , you should get the same results by multiplying both sides by $A^-1$
answered Aug 30 at 1:46
Mohammad Riazi-Kermani
31k41853
No, $y$ is $n times 1$
– vibe
Aug 30 at 2:13
add a comment |Â
up vote
0
down vote
As you mentioned, it is anti-intuitive and something is wrong because if A is invertible , you should get the same results by multiplying both sides by $A^-1$
answered Aug 30 at 1:46
Mohammad Riazi-Kermani
31k41853
No, $y$ is $n times 1$
– vibe
Aug 30 at 2:13
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As you mentioned, it is anti-intuitive and something is wrong because if A is invertible , you should get the same results by multiplying both sides by $A^-1$
answered Aug 30 at 1:46
Mohammad Riazi-Kermani
31k41853
As you mentioned, it is anti-intuitive and something is wrong because if A is invertible , you should get the same results by multiplying both sides by $A^-1$
answered Aug 30 at 1:46
Mohammad Riazi-Kermani
31k41853
edited Aug 30 at 2:22
answered Aug 30 at 1:46
Mohammad Riazi-Kermani
31k41853
answered Aug 30 at 1:46
Mohammad Riazi-Kermani
31k41853
answered Aug 30 at 1:46
Mohammad Riazi-Kermani
31k41853
31k41853
No, $y$ is $n times 1$
– vibe
Aug 30 at 2:13
add a comment |Â
No, $y$ is $n times 1$
– vibe
Aug 30 at 2:13
No, $y$ is $n times 1$
– vibe
Aug 30 at 2:13
No, $y$ is $n times 1$
– vibe
Aug 30 at 2:13
add a comment |Â
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