If the system Ax=b is consistent for every n x 1 matrix in b, then A is invertible. [closed]
$begingroup$
I have trouble understanding the proof here. I spent an hour trying to understand it but I give up. Can anyone help me with it?
linear-algebra systems-of-equations
edited Nov 30 '18 at 21:45
Martin Sleziak
44.7k9117272
asked Oct 7 '15 at 2:14
Zhi J TeohZhi J Teoh
5619
$endgroup$
closed as off-topic by user1551, Saad, Adrian Keister, user10354138, José Carlos Santos Nov 30 '18 at 23:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user1551, Saad, Adrian Keister, user10354138, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have trouble understanding the proof here. I spent an hour trying to understand it but I give up. Can anyone help me with it?
linear-algebra systems-of-equations
edited Nov 30 '18 at 21:45
Martin Sleziak
44.7k9117272
asked Oct 7 '15 at 2:14
Zhi J TeohZhi J Teoh
5619
$endgroup$
closed as off-topic by user1551, Saad, Adrian Keister, user10354138, José Carlos Santos Nov 30 '18 at 23:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user1551, Saad, Adrian Keister, user10354138, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
1) Yes, that's exactly what it is. The point is that if you can solve $A{mathbf x}={mathbf b}$ for any ${mathbf b}$, you can solve it for ${mathbf b}$ equal to any of the various columns of the identity matrix. 2) They mean to build a matrix $C$ out of the columns that form the solutions in part 1). Since matrix multiplication is done one column at a time, that gives you a matrix that gives you all the right columns to get the identity matrix, so they produced an inverse for $A$.
$endgroup$
– John Brevik
Oct 7 '15 at 2:23
$begingroup$
I think I kind of get it. In matrix C, it isn't just 1 x n matrix right? It's n x n matrix, where each column has n rows. Basically n unknowns. And we take matrix A multiply by this n column to give us a column on the identity matrix. And then we repeat again to get column of the identity matrix. And then we combine them to form an augmented matrix, which really is an identity matrix. And from there we prove the rest. Am I right?
$endgroup$
– Zhi J Teoh
Oct 7 '15 at 3:29
$begingroup$
Yeah, I think that's pretty much the point. I wouldn't really call it an augmented matrix, but otherwise you're right on.
$endgroup$
– John Brevik
Oct 7 '15 at 11:54
add a comment |
$begingroup$
I have trouble understanding the proof here. I spent an hour trying to understand it but I give up. Can anyone help me with it?
linear-algebra systems-of-equations
edited Nov 30 '18 at 21:45
Martin Sleziak
44.7k9117272
asked Oct 7 '15 at 2:14
Zhi J TeohZhi J Teoh
5619
$endgroup$
I have trouble understanding the proof here. I spent an hour trying to understand it but I give up. Can anyone help me with it?
linear-algebra systems-of-equations
linear-algebra systems-of-equations
edited Nov 30 '18 at 21:45
Martin Sleziak
44.7k9117272
asked Oct 7 '15 at 2:14
Zhi J TeohZhi J Teoh
5619
edited Nov 30 '18 at 21:45
Martin Sleziak
44.7k9117272
asked Oct 7 '15 at 2:14
Zhi J TeohZhi J Teoh
5619
edited Nov 30 '18 at 21:45
Martin Sleziak
44.7k9117272
edited Nov 30 '18 at 21:45
Martin Sleziak
44.7k9117272
edited Nov 30 '18 at 21:45
Martin Sleziak
44.7k9117272
44.7k9117272
asked Oct 7 '15 at 2:14
Zhi J TeohZhi J Teoh
5619
asked Oct 7 '15 at 2:14
Zhi J TeohZhi J Teoh
5619
asked Oct 7 '15 at 2:14
Zhi J TeohZhi J Teoh
5619
5619
closed as off-topic by user1551, Saad, Adrian Keister, user10354138, José Carlos Santos Nov 30 '18 at 23:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user1551, Saad, Adrian Keister, user10354138, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user1551, Saad, Adrian Keister, user10354138, José Carlos Santos Nov 30 '18 at 23:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user1551, Saad, Adrian Keister, user10354138, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
1) Yes, that's exactly what it is. The point is that if you can solve $A{mathbf x}={mathbf b}$ for any ${mathbf b}$, you can solve it for ${mathbf b}$ equal to any of the various columns of the identity matrix. 2) They mean to build a matrix $C$ out of the columns that form the solutions in part 1). Since matrix multiplication is done one column at a time, that gives you a matrix that gives you all the right columns to get the identity matrix, so they produced an inverse for $A$.
$endgroup$
– John Brevik
Oct 7 '15 at 2:23
$begingroup$
I think I kind of get it. In matrix C, it isn't just 1 x n matrix right? It's n x n matrix, where each column has n rows. Basically n unknowns. And we take matrix A multiply by this n column to give us a column on the identity matrix. And then we repeat again to get column of the identity matrix. And then we combine them to form an augmented matrix, which really is an identity matrix. And from there we prove the rest. Am I right?
$endgroup$
– Zhi J Teoh
Oct 7 '15 at 3:29
$begingroup$
Yeah, I think that's pretty much the point. I wouldn't really call it an augmented matrix, but otherwise you're right on.
$endgroup$
– John Brevik
Oct 7 '15 at 11:54
add a comment |
$begingroup$
1) Yes, that's exactly what it is. The point is that if you can solve $A{mathbf x}={mathbf b}$ for any ${mathbf b}$, you can solve it for ${mathbf b}$ equal to any of the various columns of the identity matrix. 2) They mean to build a matrix $C$ out of the columns that form the solutions in part 1). Since matrix multiplication is done one column at a time, that gives you a matrix that gives you all the right columns to get the identity matrix, so they produced an inverse for $A$.
$endgroup$
– John Brevik
Oct 7 '15 at 2:23
$begingroup$
I think I kind of get it. In matrix C, it isn't just 1 x n matrix right? It's n x n matrix, where each column has n rows. Basically n unknowns. And we take matrix A multiply by this n column to give us a column on the identity matrix. And then we repeat again to get column of the identity matrix. And then we combine them to form an augmented matrix, which really is an identity matrix. And from there we prove the rest. Am I right?
$endgroup$
– Zhi J Teoh
Oct 7 '15 at 3:29
$begingroup$
Yeah, I think that's pretty much the point. I wouldn't really call it an augmented matrix, but otherwise you're right on.
$endgroup$
– John Brevik
Oct 7 '15 at 11:54
$begingroup$
1) Yes, that's exactly what it is. The point is that if you can solve $A{mathbf x}={mathbf b}$ for any ${mathbf b}$, you can solve it for ${mathbf b}$ equal to any of the various columns of the identity matrix. 2) They mean to build a matrix $C$ out of the columns that form the solutions in part 1). Since matrix multiplication is done one column at a time, that gives you a matrix that gives you all the right columns to get the identity matrix, so they produced an inverse for $A$.
$endgroup$
– John Brevik
Oct 7 '15 at 2:23
$begingroup$
1) Yes, that's exactly what it is. The point is that if you can solve $A{mathbf x}={mathbf b}$ for any ${mathbf b}$, you can solve it for ${mathbf b}$ equal to any of the various columns of the identity matrix. 2) They mean to build a matrix $C$ out of the columns that form the solutions in part 1). Since matrix multiplication is done one column at a time, that gives you a matrix that gives you all the right columns to get the identity matrix, so they produced an inverse for $A$.
$endgroup$
– John Brevik
Oct 7 '15 at 2:23
$begingroup$
I think I kind of get it. In matrix C, it isn't just 1 x n matrix right? It's n x n matrix, where each column has n rows. Basically n unknowns. And we take matrix A multiply by this n column to give us a column on the identity matrix. And then we repeat again to get column of the identity matrix. And then we combine them to form an augmented matrix, which really is an identity matrix. And from there we prove the rest. Am I right?
$endgroup$
– Zhi J Teoh
Oct 7 '15 at 3:29
$begingroup$
I think I kind of get it. In matrix C, it isn't just 1 x n matrix right? It's n x n matrix, where each column has n rows. Basically n unknowns. And we take matrix A multiply by this n column to give us a column on the identity matrix. And then we repeat again to get column of the identity matrix. And then we combine them to form an augmented matrix, which really is an identity matrix. And from there we prove the rest. Am I right?
$endgroup$
– Zhi J Teoh
Oct 7 '15 at 3:29
$begingroup$
Yeah, I think that's pretty much the point. I wouldn't really call it an augmented matrix, but otherwise you're right on.
$endgroup$
– John Brevik
Oct 7 '15 at 11:54
$begingroup$
Yeah, I think that's pretty much the point. I wouldn't really call it an augmented matrix, but otherwise you're right on.
$endgroup$
– John Brevik
Oct 7 '15 at 11:54
add a comment |
1 Answer
1
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oldest
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$begingroup$
At the beginning, they're saying that since $Ax = b$ is consistent no matter which vector $b$ is, then in particular the system
begin{equation}
Ax = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 end{bmatrix}
end{equation}
is consistent. In other words, there exists a vector $x_1$ such that
begin{equation}
Ax_1 = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 end{bmatrix}.
end{equation}
And, since $Ax = b$ is consistent no matter what $b$ is, it must be true that the system
begin{equation}
Ax = begin{bmatrix} 0 \ 1 \ 0 \ vdots \ 0 end{bmatrix}
end{equation}
is consistent. In other words, there exists a vector $x_2$ such that
begin{equation}
Ax_2 = begin{bmatrix} 0 \ 1 \ 0 \ vdots \ 0 end{bmatrix}.
end{equation}
And so on. Now take these vectors $x_1, x_2,ldots$ and make them the columns of a matrix $C$:
begin{equation}
C = begin{bmatrix} x_1 & x_2 & cdots & x_n end{bmatrix}.
end{equation}
You can check that $AC = I$, the identity matrix. So $A$ is invertible.
answered Oct 7 '15 at 2:25
littleOlittleO
29.5k645109
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
At the beginning, they're saying that since $Ax = b$ is consistent no matter which vector $b$ is, then in particular the system
begin{equation}
Ax = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 end{bmatrix}
end{equation}
is consistent. In other words, there exists a vector $x_1$ such that
begin{equation}
Ax_1 = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 end{bmatrix}.
end{equation}
And, since $Ax = b$ is consistent no matter what $b$ is, it must be true that the system
begin{equation}
Ax = begin{bmatrix} 0 \ 1 \ 0 \ vdots \ 0 end{bmatrix}
end{equation}
is consistent. In other words, there exists a vector $x_2$ such that
begin{equation}
Ax_2 = begin{bmatrix} 0 \ 1 \ 0 \ vdots \ 0 end{bmatrix}.
end{equation}
And so on. Now take these vectors $x_1, x_2,ldots$ and make them the columns of a matrix $C$:
begin{equation}
C = begin{bmatrix} x_1 & x_2 & cdots & x_n end{bmatrix}.
end{equation}
You can check that $AC = I$, the identity matrix. So $A$ is invertible.
answered Oct 7 '15 at 2:25
littleOlittleO
29.5k645109
$endgroup$
add a comment |
$begingroup$
At the beginning, they're saying that since $Ax = b$ is consistent no matter which vector $b$ is, then in particular the system
begin{equation}
Ax = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 end{bmatrix}
end{equation}
is consistent. In other words, there exists a vector $x_1$ such that
begin{equation}
Ax_1 = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 end{bmatrix}.
end{equation}
And, since $Ax = b$ is consistent no matter what $b$ is, it must be true that the system
begin{equation}
Ax = begin{bmatrix} 0 \ 1 \ 0 \ vdots \ 0 end{bmatrix}
end{equation}
is consistent. In other words, there exists a vector $x_2$ such that
begin{equation}
Ax_2 = begin{bmatrix} 0 \ 1 \ 0 \ vdots \ 0 end{bmatrix}.
end{equation}
And so on. Now take these vectors $x_1, x_2,ldots$ and make them the columns of a matrix $C$:
begin{equation}
C = begin{bmatrix} x_1 & x_2 & cdots & x_n end{bmatrix}.
end{equation}
You can check that $AC = I$, the identity matrix. So $A$ is invertible.
answered Oct 7 '15 at 2:25
littleOlittleO
29.5k645109
$endgroup$
add a comment |
$begingroup$
At the beginning, they're saying that since $Ax = b$ is consistent no matter which vector $b$ is, then in particular the system
begin{equation}
Ax = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 end{bmatrix}
end{equation}
is consistent. In other words, there exists a vector $x_1$ such that
begin{equation}
Ax_1 = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 end{bmatrix}.
end{equation}
And, since $Ax = b$ is consistent no matter what $b$ is, it must be true that the system
begin{equation}
Ax = begin{bmatrix} 0 \ 1 \ 0 \ vdots \ 0 end{bmatrix}
end{equation}
is consistent. In other words, there exists a vector $x_2$ such that
begin{equation}
Ax_2 = begin{bmatrix} 0 \ 1 \ 0 \ vdots \ 0 end{bmatrix}.
end{equation}
And so on. Now take these vectors $x_1, x_2,ldots$ and make them the columns of a matrix $C$:
begin{equation}
C = begin{bmatrix} x_1 & x_2 & cdots & x_n end{bmatrix}.
end{equation}
You can check that $AC = I$, the identity matrix. So $A$ is invertible.
answered Oct 7 '15 at 2:25
littleOlittleO
29.5k645109
$endgroup$
At the beginning, they're saying that since $Ax = b$ is consistent no matter which vector $b$ is, then in particular the system
begin{equation}
Ax = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 end{bmatrix}
end{equation}
is consistent. In other words, there exists a vector $x_1$ such that
begin{equation}
Ax_1 = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 end{bmatrix}.
end{equation}
And, since $Ax = b$ is consistent no matter what $b$ is, it must be true that the system
begin{equation}
Ax = begin{bmatrix} 0 \ 1 \ 0 \ vdots \ 0 end{bmatrix}
end{equation}
is consistent. In other words, there exists a vector $x_2$ such that
begin{equation}
Ax_2 = begin{bmatrix} 0 \ 1 \ 0 \ vdots \ 0 end{bmatrix}.
end{equation}
And so on. Now take these vectors $x_1, x_2,ldots$ and make them the columns of a matrix $C$:
begin{equation}
C = begin{bmatrix} x_1 & x_2 & cdots & x_n end{bmatrix}.
end{equation}
You can check that $AC = I$, the identity matrix. So $A$ is invertible.
answered Oct 7 '15 at 2:25
littleOlittleO
29.5k645109
answered Oct 7 '15 at 2:25
littleOlittleO
29.5k645109
answered Oct 7 '15 at 2:25
littleOlittleO
29.5k645109
answered Oct 7 '15 at 2:25
littleOlittleO
29.5k645109
29.5k645109
add a comment |
add a comment |
$begingroup$
1) Yes, that's exactly what it is. The point is that if you can solve $A{mathbf x}={mathbf b}$ for any ${mathbf b}$, you can solve it for ${mathbf b}$ equal to any of the various columns of the identity matrix. 2) They mean to build a matrix $C$ out of the columns that form the solutions in part 1). Since matrix multiplication is done one column at a time, that gives you a matrix that gives you all the right columns to get the identity matrix, so they produced an inverse for $A$.
$endgroup$
– John Brevik
Oct 7 '15 at 2:23
$begingroup$
I think I kind of get it. In matrix C, it isn't just 1 x n matrix right? It's n x n matrix, where each column has n rows. Basically n unknowns. And we take matrix A multiply by this n column to give us a column on the identity matrix. And then we repeat again to get column of the identity matrix. And then we combine them to form an augmented matrix, which really is an identity matrix. And from there we prove the rest. Am I right?
$endgroup$
– Zhi J Teoh
Oct 7 '15 at 3:29
$begingroup$
Yeah, I think that's pretty much the point. I wouldn't really call it an augmented matrix, but otherwise you're right on.
$endgroup$
– John Brevik
Oct 7 '15 at 11:54