Matrix equation with several X
$begingroup$
Solving matrix equation
$A^2X-B^T = 3X$ (to find X), I'm trying to do next thing:
$A^2X-B^T = 3X$
$A^2X-3X = B^T$
$(A^2-3)X = B^T$
Can we do it in that way and, if yes, what should we do with $(A^2 - 3)$?
Thank you in advance!
linear-algebra matrices matrix-equations
asked Jan 21 at 17:01
Sergey MalinovSergey Malinov
174
$endgroup$
add a comment |
$begingroup$
Solving matrix equation
$A^2X-B^T = 3X$ (to find X), I'm trying to do next thing:
$A^2X-B^T = 3X$
$A^2X-3X = B^T$
$(A^2-3)X = B^T$
Can we do it in that way and, if yes, what should we do with $(A^2 - 3)$?
Thank you in advance!
linear-algebra matrices matrix-equations
asked Jan 21 at 17:01
Sergey MalinovSergey Malinov
174
$endgroup$
$begingroup$
The equation $(A^2-3)X = B^T$ is equivalent to $X^Tleft((A^T)^2-3I right) = B$. Then you have to make a discussion on the content of the kernel of $B$. For example, if $v$ is an eigenvector associated to the eigenvalue $sqrt{3}$ of $A^T$, you have to have $v in ker A^T$. If not the equation has no solution... and so on.
$endgroup$
– mathcounterexamples.net
Jan 21 at 17:15
add a comment |
$begingroup$
Solving matrix equation
$A^2X-B^T = 3X$ (to find X), I'm trying to do next thing:
$A^2X-B^T = 3X$
$A^2X-3X = B^T$
$(A^2-3)X = B^T$
Can we do it in that way and, if yes, what should we do with $(A^2 - 3)$?
Thank you in advance!
linear-algebra matrices matrix-equations
asked Jan 21 at 17:01
Sergey MalinovSergey Malinov
174
$endgroup$
Solving matrix equation
$A^2X-B^T = 3X$ (to find X), I'm trying to do next thing:
$A^2X-B^T = 3X$
$A^2X-3X = B^T$
$(A^2-3)X = B^T$
Can we do it in that way and, if yes, what should we do with $(A^2 - 3)$?
Thank you in advance!
linear-algebra matrices matrix-equations
linear-algebra matrices matrix-equations
asked Jan 21 at 17:01
Sergey MalinovSergey Malinov
174
asked Jan 21 at 17:01
Sergey MalinovSergey Malinov
174
asked Jan 21 at 17:01
Sergey MalinovSergey Malinov
174
asked Jan 21 at 17:01
Sergey MalinovSergey Malinov
174
asked Jan 21 at 17:01
Sergey MalinovSergey Malinov
174
174
$begingroup$
The equation $(A^2-3)X = B^T$ is equivalent to $X^Tleft((A^T)^2-3I right) = B$. Then you have to make a discussion on the content of the kernel of $B$. For example, if $v$ is an eigenvector associated to the eigenvalue $sqrt{3}$ of $A^T$, you have to have $v in ker A^T$. If not the equation has no solution... and so on.
$endgroup$
– mathcounterexamples.net
Jan 21 at 17:15
add a comment |
$begingroup$
The equation $(A^2-3)X = B^T$ is equivalent to $X^Tleft((A^T)^2-3I right) = B$. Then you have to make a discussion on the content of the kernel of $B$. For example, if $v$ is an eigenvector associated to the eigenvalue $sqrt{3}$ of $A^T$, you have to have $v in ker A^T$. If not the equation has no solution... and so on.
$endgroup$
– mathcounterexamples.net
Jan 21 at 17:15
$begingroup$
The equation $(A^2-3)X = B^T$ is equivalent to $X^Tleft((A^T)^2-3I right) = B$. Then you have to make a discussion on the content of the kernel of $B$. For example, if $v$ is an eigenvector associated to the eigenvalue $sqrt{3}$ of $A^T$, you have to have $v in ker A^T$. If not the equation has no solution... and so on.
$endgroup$
– mathcounterexamples.net
Jan 21 at 17:15
$begingroup$
The equation $(A^2-3)X = B^T$ is equivalent to $X^Tleft((A^T)^2-3I right) = B$. Then you have to make a discussion on the content of the kernel of $B$. For example, if $v$ is an eigenvector associated to the eigenvalue $sqrt{3}$ of $A^T$, you have to have $v in ker A^T$. If not the equation has no solution... and so on.
$endgroup$
– mathcounterexamples.net
Jan 21 at 17:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$X = (A^2-3I)^{-1}B^T$$
If we assume that this matrix is invertible. Even if it is not, you may still be in the case that you have an underdetermined system, then you simply have infinitely many solutions. In the case where you have an overdetermined system you can use least squares.
answered Jan 21 at 17:06
lightxbulblightxbulb
945311
$endgroup$
$begingroup$
Thank you! So, if I understood it in a right way, instead of 3, we use identity matrix multiplied by 3?
$endgroup$
– Sergey Malinov
Jan 21 at 17:08
$begingroup$
Yes a matrix with 3 on the diagonal.
$endgroup$
– lightxbulb
Jan 21 at 17:08
$begingroup$
@Sergey Malinov You can accept it as an answer if you don't need anymore clarifications on it.
$endgroup$
– lightxbulb
Jan 21 at 17:13
$begingroup$
I know, thank you for reminding, but I can not do it yet. In couple of minutes the system will let me do it
$endgroup$
– Sergey Malinov
Jan 21 at 17:15
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$X = (A^2-3I)^{-1}B^T$$
If we assume that this matrix is invertible. Even if it is not, you may still be in the case that you have an underdetermined system, then you simply have infinitely many solutions. In the case where you have an overdetermined system you can use least squares.
answered Jan 21 at 17:06
lightxbulblightxbulb
945311
$endgroup$
$begingroup$
Thank you! So, if I understood it in a right way, instead of 3, we use identity matrix multiplied by 3?
$endgroup$
– Sergey Malinov
Jan 21 at 17:08
$begingroup$
Yes a matrix with 3 on the diagonal.
$endgroup$
– lightxbulb
Jan 21 at 17:08
$begingroup$
@Sergey Malinov You can accept it as an answer if you don't need anymore clarifications on it.
$endgroup$
– lightxbulb
Jan 21 at 17:13
$begingroup$
I know, thank you for reminding, but I can not do it yet. In couple of minutes the system will let me do it
$endgroup$
– Sergey Malinov
Jan 21 at 17:15
add a comment |
$begingroup$
$$X = (A^2-3I)^{-1}B^T$$
If we assume that this matrix is invertible. Even if it is not, you may still be in the case that you have an underdetermined system, then you simply have infinitely many solutions. In the case where you have an overdetermined system you can use least squares.
answered Jan 21 at 17:06
lightxbulblightxbulb
945311
$endgroup$
$begingroup$
Thank you! So, if I understood it in a right way, instead of 3, we use identity matrix multiplied by 3?
$endgroup$
– Sergey Malinov
Jan 21 at 17:08
$begingroup$
Yes a matrix with 3 on the diagonal.
$endgroup$
– lightxbulb
Jan 21 at 17:08
$begingroup$
@Sergey Malinov You can accept it as an answer if you don't need anymore clarifications on it.
$endgroup$
– lightxbulb
Jan 21 at 17:13
$begingroup$
I know, thank you for reminding, but I can not do it yet. In couple of minutes the system will let me do it
$endgroup$
– Sergey Malinov
Jan 21 at 17:15
add a comment |
$begingroup$
$$X = (A^2-3I)^{-1}B^T$$
If we assume that this matrix is invertible. Even if it is not, you may still be in the case that you have an underdetermined system, then you simply have infinitely many solutions. In the case where you have an overdetermined system you can use least squares.
answered Jan 21 at 17:06
lightxbulblightxbulb
945311
$endgroup$
$$X = (A^2-3I)^{-1}B^T$$
If we assume that this matrix is invertible. Even if it is not, you may still be in the case that you have an underdetermined system, then you simply have infinitely many solutions. In the case where you have an overdetermined system you can use least squares.
answered Jan 21 at 17:06
lightxbulblightxbulb
945311
edited Jan 21 at 17:08
answered Jan 21 at 17:06
lightxbulblightxbulb
945311
answered Jan 21 at 17:06
lightxbulblightxbulb
945311
answered Jan 21 at 17:06
lightxbulblightxbulb
945311
945311
$begingroup$
Thank you! So, if I understood it in a right way, instead of 3, we use identity matrix multiplied by 3?
$endgroup$
– Sergey Malinov
Jan 21 at 17:08
$begingroup$
Yes a matrix with 3 on the diagonal.
$endgroup$
– lightxbulb
Jan 21 at 17:08
$begingroup$
@Sergey Malinov You can accept it as an answer if you don't need anymore clarifications on it.
$endgroup$
– lightxbulb
Jan 21 at 17:13
$begingroup$
I know, thank you for reminding, but I can not do it yet. In couple of minutes the system will let me do it
$endgroup$
– Sergey Malinov
Jan 21 at 17:15
add a comment |
$begingroup$
Thank you! So, if I understood it in a right way, instead of 3, we use identity matrix multiplied by 3?
$endgroup$
– Sergey Malinov
Jan 21 at 17:08
$begingroup$
Yes a matrix with 3 on the diagonal.
$endgroup$
– lightxbulb
Jan 21 at 17:08
$begingroup$
@Sergey Malinov You can accept it as an answer if you don't need anymore clarifications on it.
$endgroup$
– lightxbulb
Jan 21 at 17:13
$begingroup$
I know, thank you for reminding, but I can not do it yet. In couple of minutes the system will let me do it
$endgroup$
– Sergey Malinov
Jan 21 at 17:15
$begingroup$
Thank you! So, if I understood it in a right way, instead of 3, we use identity matrix multiplied by 3?
$endgroup$
– Sergey Malinov
Jan 21 at 17:08
$begingroup$
Thank you! So, if I understood it in a right way, instead of 3, we use identity matrix multiplied by 3?
$endgroup$
– Sergey Malinov
Jan 21 at 17:08
$begingroup$
Yes a matrix with 3 on the diagonal.
$endgroup$
– lightxbulb
Jan 21 at 17:08
$begingroup$
Yes a matrix with 3 on the diagonal.
$endgroup$
– lightxbulb
Jan 21 at 17:08
$begingroup$
@Sergey Malinov You can accept it as an answer if you don't need anymore clarifications on it.
$endgroup$
– lightxbulb
Jan 21 at 17:13
$begingroup$
@Sergey Malinov You can accept it as an answer if you don't need anymore clarifications on it.
$endgroup$
– lightxbulb
Jan 21 at 17:13
$begingroup$
I know, thank you for reminding, but I can not do it yet. In couple of minutes the system will let me do it
$endgroup$
– Sergey Malinov
Jan 21 at 17:15
$begingroup$
I know, thank you for reminding, but I can not do it yet. In couple of minutes the system will let me do it
$endgroup$
– Sergey Malinov
Jan 21 at 17:15
add a comment |
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$begingroup$
The equation $(A^2-3)X = B^T$ is equivalent to $X^Tleft((A^T)^2-3I right) = B$. Then you have to make a discussion on the content of the kernel of $B$. For example, if $v$ is an eigenvector associated to the eigenvalue $sqrt{3}$ of $A^T$, you have to have $v in ker A^T$. If not the equation has no solution... and so on.
$endgroup$
– mathcounterexamples.net
Jan 21 at 17:15