Mixing
Given the rank of a matrix, determine matrices B and C such that BAC=H
$begingroup$
Matrix $H=
begin{bmatrix}
I_{3} & (0) \
(0) & (0)
end{bmatrix}$
Matrix $A=
begin{bmatrix}
2 & 0 & 1 & -1 \
1 & -1 & 1 & -1 \
1 & -1 & 2 & 0 \
2 & 0 & 0 & -2
end{bmatrix}$
Use elementary transformations to transform $A$ to $H$. Given these matrices, determine matrices $B$ and $C$ so that $BAC=H$. The following identity may be important: $A=B^{-1}HC^{-1}$
When I try to transform $A$ into $H$, I get the following:
Matrix $A=
begin{bmatrix}
1 & 0 & 0 & -1 \
0 & 1 & 0 & 1 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 0
end{bmatrix}$
How do I get $A$ to $H$ and complete the problem from here?
Thank you.
matrices
asked Jan 25 at 4:12
MatthewMatthew
687
$endgroup$
add a comment |
$begingroup$
Matrix $H=
begin{bmatrix}
I_{3} & (0) \
(0) & (0)
end{bmatrix}$
Matrix $A=
begin{bmatrix}
2 & 0 & 1 & -1 \
1 & -1 & 1 & -1 \
1 & -1 & 2 & 0 \
2 & 0 & 0 & -2
end{bmatrix}$
Use elementary transformations to transform $A$ to $H$. Given these matrices, determine matrices $B$ and $C$ so that $BAC=H$. The following identity may be important: $A=B^{-1}HC^{-1}$
When I try to transform $A$ into $H$, I get the following:
Matrix $A=
begin{bmatrix}
1 & 0 & 0 & -1 \
0 & 1 & 0 & 1 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 0
end{bmatrix}$
How do I get $A$ to $H$ and complete the problem from here?
Thank you.
matrices
asked Jan 25 at 4:12
MatthewMatthew
687
$endgroup$
$begingroup$
Note that $ker A$ is the span of $(1,-1,-1,1)^T$.
$endgroup$
– copper.hat
Jan 25 at 4:34
$begingroup$
@copper.hat can you please expand on this a little? I cannot see how this relates yet.
$endgroup$
– Matthew
Jan 26 at 20:50
$begingroup$
@copper.hat we have not yet learned what a ker is
$endgroup$
– Matthew
Jan 28 at 16:55
$begingroup$
Are you familiar with elementary row/column operations?
$endgroup$
– copper.hat
Jan 28 at 16:58
$begingroup$
@copper.hat yes, Gaussian elimination only
$endgroup$
– Matthew
Jan 28 at 16:59
add a comment |
$begingroup$
Matrix $H=
begin{bmatrix}
I_{3} & (0) \
(0) & (0)
end{bmatrix}$
Matrix $A=
begin{bmatrix}
2 & 0 & 1 & -1 \
1 & -1 & 1 & -1 \
1 & -1 & 2 & 0 \
2 & 0 & 0 & -2
end{bmatrix}$
Use elementary transformations to transform $A$ to $H$. Given these matrices, determine matrices $B$ and $C$ so that $BAC=H$. The following identity may be important: $A=B^{-1}HC^{-1}$
When I try to transform $A$ into $H$, I get the following:
Matrix $A=
begin{bmatrix}
1 & 0 & 0 & -1 \
0 & 1 & 0 & 1 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 0
end{bmatrix}$
How do I get $A$ to $H$ and complete the problem from here?
Thank you.
matrices
asked Jan 25 at 4:12
MatthewMatthew
687
$endgroup$
Matrix $H=
begin{bmatrix}
I_{3} & (0) \
(0) & (0)
end{bmatrix}$
Matrix $A=
begin{bmatrix}
2 & 0 & 1 & -1 \
1 & -1 & 1 & -1 \
1 & -1 & 2 & 0 \
2 & 0 & 0 & -2
end{bmatrix}$
Use elementary transformations to transform $A$ to $H$. Given these matrices, determine matrices $B$ and $C$ so that $BAC=H$. The following identity may be important: $A=B^{-1}HC^{-1}$
When I try to transform $A$ into $H$, I get the following:
Matrix $A=
begin{bmatrix}
1 & 0 & 0 & -1 \
0 & 1 & 0 & 1 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 0
end{bmatrix}$
How do I get $A$ to $H$ and complete the problem from here?
Thank you.
matrices
matrices
asked Jan 25 at 4:12
MatthewMatthew
687
asked Jan 25 at 4:12
MatthewMatthew
687
edited Jan 28 at 16:59
Matthew
asked Jan 25 at 4:12
MatthewMatthew
687
asked Jan 25 at 4:12
MatthewMatthew
687
asked Jan 25 at 4:12
MatthewMatthew
687
687
$begingroup$
Note that $ker A$ is the span of $(1,-1,-1,1)^T$.
$endgroup$
– copper.hat
Jan 25 at 4:34
$begingroup$
@copper.hat can you please expand on this a little? I cannot see how this relates yet.
$endgroup$
– Matthew
Jan 26 at 20:50
$begingroup$
@copper.hat we have not yet learned what a ker is
$endgroup$
– Matthew
Jan 28 at 16:55
$begingroup$
Are you familiar with elementary row/column operations?
$endgroup$
– copper.hat
Jan 28 at 16:58
$begingroup$
@copper.hat yes, Gaussian elimination only
$endgroup$
– Matthew
Jan 28 at 16:59
add a comment |
$begingroup$
Note that $ker A$ is the span of $(1,-1,-1,1)^T$.
$endgroup$
– copper.hat
Jan 25 at 4:34
$begingroup$
@copper.hat can you please expand on this a little? I cannot see how this relates yet.
$endgroup$
– Matthew
Jan 26 at 20:50
$begingroup$
@copper.hat we have not yet learned what a ker is
$endgroup$
– Matthew
Jan 28 at 16:55
$begingroup$
Are you familiar with elementary row/column operations?
$endgroup$
– copper.hat
Jan 28 at 16:58
$begingroup$
@copper.hat yes, Gaussian elimination only
$endgroup$
– Matthew
Jan 28 at 16:59
$begingroup$
Note that $ker A$ is the span of $(1,-1,-1,1)^T$.
$endgroup$
– copper.hat
Jan 25 at 4:34
$begingroup$
Note that $ker A$ is the span of $(1,-1,-1,1)^T$.
$endgroup$
– copper.hat
Jan 25 at 4:34
$begingroup$
@copper.hat can you please expand on this a little? I cannot see how this relates yet.
$endgroup$
– Matthew
Jan 26 at 20:50
$begingroup$
@copper.hat can you please expand on this a little? I cannot see how this relates yet.
$endgroup$
– Matthew
Jan 26 at 20:50
$begingroup$
@copper.hat we have not yet learned what a ker is
$endgroup$
– Matthew
Jan 28 at 16:55
$begingroup$
@copper.hat we have not yet learned what a ker is
$endgroup$
– Matthew
Jan 28 at 16:55
$begingroup$
Are you familiar with elementary row/column operations?
$endgroup$
– copper.hat
Jan 28 at 16:58
$begingroup$
Are you familiar with elementary row/column operations?
$endgroup$
– copper.hat
Jan 28 at 16:58
$begingroup$
@copper.hat yes, Gaussian elimination only
$endgroup$
– Matthew
Jan 28 at 16:59
$begingroup$
@copper.hat yes, Gaussian elimination only
$endgroup$
– Matthew
Jan 28 at 16:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
With a sequence of elementary row operations $E_3 E_2 E_1 A$ is an upper triangular matrix. With two elementary column operations $E_3 E_2 E_1 A C_1 C_2$ is a diagonal matrix
$Delta=operatorname{diag}{ 2, -1, 1, 0 }$. With $D=operatorname{diag}{ {1 over 2}, -1, 1, 1}$, we see that
$(D E_3 E_2 E_1 ) A ( C_1 C_2) = operatorname{diag}{1,1,1,0}$.
The $E$s are elementary row operations and the $C$s are elementary column operations. For example, I chose
$E_1 = begin{bmatrix} 1 & 0 & 0 & 0 \
-{1 over 2} & 1 & 0 & 0 \
-{1 over 2} & 0 & 1 & 0 \
-1 & 0 & 0 & 1 end{bmatrix}$,
$C_1 = begin{bmatrix} 1 & 0 & -{1 over 2} & 0 \
0 & 1 & {1 over 2} & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 end{bmatrix}$.
answered Jan 28 at 18:47
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
I just want to ensure that I understand this correctly, since some notation is new to me. What are $E_i$'s and $C_i$'s? Additionally, how do I find which operations to execute?
$endgroup$
– Matthew
Jan 29 at 2:15
$begingroup$
It appears that your designation of $E_3E_2E_1A$ is what I have obtained in my posted question. Is this correct? If so, which column operators do I execute to obtain $Delta=$diag{2,-1,1,0}?
$endgroup$
– Matthew
Jan 29 at 2:16
$begingroup$
Just apply the transpose of the operators used to do row operations. After performing the elementary row operations, I ended up with [2 0 1 -1 ; 0 -1 1/2 -1/2 ; 0 0 1 1 ; 0 0 0 1].
$endgroup$
– copper.hat
Jan 29 at 5:43
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
$begingroup$
With a sequence of elementary row operations $E_3 E_2 E_1 A$ is an upper triangular matrix. With two elementary column operations $E_3 E_2 E_1 A C_1 C_2$ is a diagonal matrix
$Delta=operatorname{diag}{ 2, -1, 1, 0 }$. With $D=operatorname{diag}{ {1 over 2}, -1, 1, 1}$, we see that
$(D E_3 E_2 E_1 ) A ( C_1 C_2) = operatorname{diag}{1,1,1,0}$.
The $E$s are elementary row operations and the $C$s are elementary column operations. For example, I chose
$E_1 = begin{bmatrix} 1 & 0 & 0 & 0 \
-{1 over 2} & 1 & 0 & 0 \
-{1 over 2} & 0 & 1 & 0 \
-1 & 0 & 0 & 1 end{bmatrix}$,
$C_1 = begin{bmatrix} 1 & 0 & -{1 over 2} & 0 \
0 & 1 & {1 over 2} & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 end{bmatrix}$.
answered Jan 28 at 18:47
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
I just want to ensure that I understand this correctly, since some notation is new to me. What are $E_i$'s and $C_i$'s? Additionally, how do I find which operations to execute?
$endgroup$
– Matthew
Jan 29 at 2:15
$begingroup$
It appears that your designation of $E_3E_2E_1A$ is what I have obtained in my posted question. Is this correct? If so, which column operators do I execute to obtain $Delta=$diag{2,-1,1,0}?
$endgroup$
– Matthew
Jan 29 at 2:16
$begingroup$
Just apply the transpose of the operators used to do row operations. After performing the elementary row operations, I ended up with [2 0 1 -1 ; 0 -1 1/2 -1/2 ; 0 0 1 1 ; 0 0 0 1].
$endgroup$
– copper.hat
Jan 29 at 5:43
add a comment |
$begingroup$
With a sequence of elementary row operations $E_3 E_2 E_1 A$ is an upper triangular matrix. With two elementary column operations $E_3 E_2 E_1 A C_1 C_2$ is a diagonal matrix
$Delta=operatorname{diag}{ 2, -1, 1, 0 }$. With $D=operatorname{diag}{ {1 over 2}, -1, 1, 1}$, we see that
$(D E_3 E_2 E_1 ) A ( C_1 C_2) = operatorname{diag}{1,1,1,0}$.
The $E$s are elementary row operations and the $C$s are elementary column operations. For example, I chose
$E_1 = begin{bmatrix} 1 & 0 & 0 & 0 \
-{1 over 2} & 1 & 0 & 0 \
-{1 over 2} & 0 & 1 & 0 \
-1 & 0 & 0 & 1 end{bmatrix}$,
$C_1 = begin{bmatrix} 1 & 0 & -{1 over 2} & 0 \
0 & 1 & {1 over 2} & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 end{bmatrix}$.
answered Jan 28 at 18:47
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
I just want to ensure that I understand this correctly, since some notation is new to me. What are $E_i$'s and $C_i$'s? Additionally, how do I find which operations to execute?
$endgroup$
– Matthew
Jan 29 at 2:15
$begingroup$
It appears that your designation of $E_3E_2E_1A$ is what I have obtained in my posted question. Is this correct? If so, which column operators do I execute to obtain $Delta=$diag{2,-1,1,0}?
$endgroup$
– Matthew
Jan 29 at 2:16
$begingroup$
Just apply the transpose of the operators used to do row operations. After performing the elementary row operations, I ended up with [2 0 1 -1 ; 0 -1 1/2 -1/2 ; 0 0 1 1 ; 0 0 0 1].
$endgroup$
– copper.hat
Jan 29 at 5:43
add a comment |
$begingroup$
With a sequence of elementary row operations $E_3 E_2 E_1 A$ is an upper triangular matrix. With two elementary column operations $E_3 E_2 E_1 A C_1 C_2$ is a diagonal matrix
$Delta=operatorname{diag}{ 2, -1, 1, 0 }$. With $D=operatorname{diag}{ {1 over 2}, -1, 1, 1}$, we see that
$(D E_3 E_2 E_1 ) A ( C_1 C_2) = operatorname{diag}{1,1,1,0}$.
The $E$s are elementary row operations and the $C$s are elementary column operations. For example, I chose
$E_1 = begin{bmatrix} 1 & 0 & 0 & 0 \
-{1 over 2} & 1 & 0 & 0 \
-{1 over 2} & 0 & 1 & 0 \
-1 & 0 & 0 & 1 end{bmatrix}$,
$C_1 = begin{bmatrix} 1 & 0 & -{1 over 2} & 0 \
0 & 1 & {1 over 2} & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 end{bmatrix}$.
answered Jan 28 at 18:47
copper.hatcopper.hat
127k559160
$endgroup$
With a sequence of elementary row operations $E_3 E_2 E_1 A$ is an upper triangular matrix. With two elementary column operations $E_3 E_2 E_1 A C_1 C_2$ is a diagonal matrix
$Delta=operatorname{diag}{ 2, -1, 1, 0 }$. With $D=operatorname{diag}{ {1 over 2}, -1, 1, 1}$, we see that
$(D E_3 E_2 E_1 ) A ( C_1 C_2) = operatorname{diag}{1,1,1,0}$.
The $E$s are elementary row operations and the $C$s are elementary column operations. For example, I chose
$E_1 = begin{bmatrix} 1 & 0 & 0 & 0 \
-{1 over 2} & 1 & 0 & 0 \
-{1 over 2} & 0 & 1 & 0 \
-1 & 0 & 0 & 1 end{bmatrix}$,
$C_1 = begin{bmatrix} 1 & 0 & -{1 over 2} & 0 \
0 & 1 & {1 over 2} & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 end{bmatrix}$.
answered Jan 28 at 18:47
copper.hatcopper.hat
127k559160
edited Jan 29 at 14:22
answered Jan 28 at 18:47
copper.hatcopper.hat
127k559160
answered Jan 28 at 18:47
copper.hatcopper.hat
127k559160
answered Jan 28 at 18:47
copper.hatcopper.hat
127k559160
127k559160
$begingroup$
I just want to ensure that I understand this correctly, since some notation is new to me. What are $E_i$'s and $C_i$'s? Additionally, how do I find which operations to execute?
$endgroup$
– Matthew
Jan 29 at 2:15
$begingroup$
It appears that your designation of $E_3E_2E_1A$ is what I have obtained in my posted question. Is this correct? If so, which column operators do I execute to obtain $Delta=$diag{2,-1,1,0}?
$endgroup$
– Matthew
Jan 29 at 2:16
$begingroup$
Just apply the transpose of the operators used to do row operations. After performing the elementary row operations, I ended up with [2 0 1 -1 ; 0 -1 1/2 -1/2 ; 0 0 1 1 ; 0 0 0 1].
$endgroup$
– copper.hat
Jan 29 at 5:43
add a comment |
$begingroup$
I just want to ensure that I understand this correctly, since some notation is new to me. What are $E_i$'s and $C_i$'s? Additionally, how do I find which operations to execute?
$endgroup$
– Matthew
Jan 29 at 2:15
$begingroup$
It appears that your designation of $E_3E_2E_1A$ is what I have obtained in my posted question. Is this correct? If so, which column operators do I execute to obtain $Delta=$diag{2,-1,1,0}?
$endgroup$
– Matthew
Jan 29 at 2:16
$begingroup$
Just apply the transpose of the operators used to do row operations. After performing the elementary row operations, I ended up with [2 0 1 -1 ; 0 -1 1/2 -1/2 ; 0 0 1 1 ; 0 0 0 1].
$endgroup$
– copper.hat
Jan 29 at 5:43
$begingroup$
I just want to ensure that I understand this correctly, since some notation is new to me. What are $E_i$'s and $C_i$'s? Additionally, how do I find which operations to execute?
$endgroup$
– Matthew
Jan 29 at 2:15
$begingroup$
I just want to ensure that I understand this correctly, since some notation is new to me. What are $E_i$'s and $C_i$'s? Additionally, how do I find which operations to execute?
$endgroup$
– Matthew
Jan 29 at 2:15
$begingroup$
It appears that your designation of $E_3E_2E_1A$ is what I have obtained in my posted question. Is this correct? If so, which column operators do I execute to obtain $Delta=$diag{2,-1,1,0}?
$endgroup$
– Matthew
Jan 29 at 2:16
$begingroup$
It appears that your designation of $E_3E_2E_1A$ is what I have obtained in my posted question. Is this correct? If so, which column operators do I execute to obtain $Delta=$diag{2,-1,1,0}?
$endgroup$
– Matthew
Jan 29 at 2:16
$begingroup$
Just apply the transpose of the operators used to do row operations. After performing the elementary row operations, I ended up with [2 0 1 -1 ; 0 -1 1/2 -1/2 ; 0 0 1 1 ; 0 0 0 1].
$endgroup$
– copper.hat
Jan 29 at 5:43
$begingroup$
Just apply the transpose of the operators used to do row operations. After performing the elementary row operations, I ended up with [2 0 1 -1 ; 0 -1 1/2 -1/2 ; 0 0 1 1 ; 0 0 0 1].
$endgroup$
– copper.hat
Jan 29 at 5:43
add a comment |
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$begingroup$
Note that $ker A$ is the span of $(1,-1,-1,1)^T$.
$endgroup$
– copper.hat
Jan 25 at 4:34
$begingroup$
@copper.hat can you please expand on this a little? I cannot see how this relates yet.
$endgroup$
– Matthew
Jan 26 at 20:50
$begingroup$
@copper.hat we have not yet learned what a ker is
$endgroup$
– Matthew
Jan 28 at 16:55
$begingroup$
Are you familiar with elementary row/column operations?
$endgroup$
– copper.hat
Jan 28 at 16:58
$begingroup$
@copper.hat yes, Gaussian elimination only
$endgroup$
– Matthew
Jan 28 at 16:59