Mixing
For a non homogeneous system of linear equation Ax=b, can we conclude any relation between rank of A and...
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For homogenous system of linear equation Ax=0 we can say that the nulity of the matrix A and the dimension of the solution space is same.But I cannot find any relation for non homogeneous system of linear equation.So please help me.Thank you in advance
linear-algebra
asked Jan 28 at 17:17
Debprasad KunduDebprasad Kundu
112
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add a comment |
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For homogenous system of linear equation Ax=0 we can say that the nulity of the matrix A and the dimension of the solution space is same.But I cannot find any relation for non homogeneous system of linear equation.So please help me.Thank you in advance
linear-algebra
asked Jan 28 at 17:17
Debprasad KunduDebprasad Kundu
112
$endgroup$
$begingroup$
When $b neq 0$, the solution space to $Ax = b$ will not be a linear subspace. It will, however, be an affine subspace.
$endgroup$
– Omnomnomnom
Jan 29 at 0:20
add a comment |
$begingroup$
For homogenous system of linear equation Ax=0 we can say that the nulity of the matrix A and the dimension of the solution space is same.But I cannot find any relation for non homogeneous system of linear equation.So please help me.Thank you in advance
linear-algebra
asked Jan 28 at 17:17
Debprasad KunduDebprasad Kundu
112
$endgroup$
For homogenous system of linear equation Ax=0 we can say that the nulity of the matrix A and the dimension of the solution space is same.But I cannot find any relation for non homogeneous system of linear equation.So please help me.Thank you in advance
linear-algebra
linear-algebra
asked Jan 28 at 17:17
Debprasad KunduDebprasad Kundu
112
asked Jan 28 at 17:17
Debprasad KunduDebprasad Kundu
112
asked Jan 28 at 17:17
Debprasad KunduDebprasad Kundu
112
asked Jan 28 at 17:17
Debprasad KunduDebprasad Kundu
112
asked Jan 28 at 17:17
Debprasad KunduDebprasad Kundu
112
112
$begingroup$
When $b neq 0$, the solution space to $Ax = b$ will not be a linear subspace. It will, however, be an affine subspace.
$endgroup$
– Omnomnomnom
Jan 29 at 0:20
add a comment |
$begingroup$
When $b neq 0$, the solution space to $Ax = b$ will not be a linear subspace. It will, however, be an affine subspace.
$endgroup$
– Omnomnomnom
Jan 29 at 0:20
$begingroup$
When $b neq 0$, the solution space to $Ax = b$ will not be a linear subspace. It will, however, be an affine subspace.
$endgroup$
– Omnomnomnom
Jan 29 at 0:20
$begingroup$
When $b neq 0$, the solution space to $Ax = b$ will not be a linear subspace. It will, however, be an affine subspace.
$endgroup$
– Omnomnomnom
Jan 29 at 0:20
add a comment |
1 Answer
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Hint:
If $x_0$ is any solution of $Ax_0 = b$, then $Ax = b$ if and only if $A(x - x_0) = 0$.
answered Jan 29 at 0:14
Paul SinclairPaul Sinclair
20.7k21543
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint:
If $x_0$ is any solution of $Ax_0 = b$, then $Ax = b$ if and only if $A(x - x_0) = 0$.
answered Jan 29 at 0:14
Paul SinclairPaul Sinclair
20.7k21543
$endgroup$
add a comment |
$begingroup$
Hint:
If $x_0$ is any solution of $Ax_0 = b$, then $Ax = b$ if and only if $A(x - x_0) = 0$.
answered Jan 29 at 0:14
Paul SinclairPaul Sinclair
20.7k21543
$endgroup$
add a comment |
$begingroup$
Hint:
If $x_0$ is any solution of $Ax_0 = b$, then $Ax = b$ if and only if $A(x - x_0) = 0$.
answered Jan 29 at 0:14
Paul SinclairPaul Sinclair
20.7k21543
$endgroup$
Hint:
If $x_0$ is any solution of $Ax_0 = b$, then $Ax = b$ if and only if $A(x - x_0) = 0$.
answered Jan 29 at 0:14
Paul SinclairPaul Sinclair
20.7k21543
answered Jan 29 at 0:14
Paul SinclairPaul Sinclair
20.7k21543
answered Jan 29 at 0:14
Paul SinclairPaul Sinclair
20.7k21543
answered Jan 29 at 0:14
Paul SinclairPaul Sinclair
20.7k21543
20.7k21543
add a comment |
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$begingroup$
When $b neq 0$, the solution space to $Ax = b$ will not be a linear subspace. It will, however, be an affine subspace.
$endgroup$
– Omnomnomnom
Jan 29 at 0:20