Mixing
Find a basis for the orthogonal complement of the column space of the following matrix
$begingroup$
I was assigned this problem for homework but don't know if I'm tackling it properly..
Find a basis for the orthogonal complement of the column
space of the following matrix
$
M=
begin{bmatrix}
1 & 1 \
1 & -1 \
1 & 1 \
1 & -1 \
end{bmatrix}
$
I'm guessing I need to find $vec{x}$ such that $M^Tvec{x}=vec{0}$.
I have $M^T$ in rref which turns out to be
$M^T= begin{bmatrix}
1 & 0 & 1 & 0 \
0 & 1 & 0 & 1 \
end{bmatrix}$
I am confused as to where to go from here. Are the 3rd and 4th columns free variables?
linear-algebra
asked Dec 1 '15 at 20:35
hairspraygirliehairspraygirlie
307
$endgroup$
add a comment |
$begingroup$
I was assigned this problem for homework but don't know if I'm tackling it properly..
Find a basis for the orthogonal complement of the column
space of the following matrix
$
M=
begin{bmatrix}
1 & 1 \
1 & -1 \
1 & 1 \
1 & -1 \
end{bmatrix}
$
I'm guessing I need to find $vec{x}$ such that $M^Tvec{x}=vec{0}$.
I have $M^T$ in rref which turns out to be
$M^T= begin{bmatrix}
1 & 0 & 1 & 0 \
0 & 1 & 0 & 1 \
end{bmatrix}$
I am confused as to where to go from here. Are the 3rd and 4th columns free variables?
linear-algebra
asked Dec 1 '15 at 20:35
hairspraygirliehairspraygirlie
307
$endgroup$
$begingroup$
Yes. You can let $x_3$ and $x_4$ be free variables.
$endgroup$
– user137731
Dec 1 '15 at 20:41
add a comment |
$begingroup$
I was assigned this problem for homework but don't know if I'm tackling it properly..
Find a basis for the orthogonal complement of the column
space of the following matrix
$
M=
begin{bmatrix}
1 & 1 \
1 & -1 \
1 & 1 \
1 & -1 \
end{bmatrix}
$
I'm guessing I need to find $vec{x}$ such that $M^Tvec{x}=vec{0}$.
I have $M^T$ in rref which turns out to be
$M^T= begin{bmatrix}
1 & 0 & 1 & 0 \
0 & 1 & 0 & 1 \
end{bmatrix}$
I am confused as to where to go from here. Are the 3rd and 4th columns free variables?
linear-algebra
asked Dec 1 '15 at 20:35
hairspraygirliehairspraygirlie
307
$endgroup$
I was assigned this problem for homework but don't know if I'm tackling it properly..
Find a basis for the orthogonal complement of the column
space of the following matrix
$
M=
begin{bmatrix}
1 & 1 \
1 & -1 \
1 & 1 \
1 & -1 \
end{bmatrix}
$
I'm guessing I need to find $vec{x}$ such that $M^Tvec{x}=vec{0}$.
I have $M^T$ in rref which turns out to be
$M^T= begin{bmatrix}
1 & 0 & 1 & 0 \
0 & 1 & 0 & 1 \
end{bmatrix}$
I am confused as to where to go from here. Are the 3rd and 4th columns free variables?
linear-algebra
linear-algebra
asked Dec 1 '15 at 20:35
hairspraygirliehairspraygirlie
307
asked Dec 1 '15 at 20:35
hairspraygirliehairspraygirlie
307
asked Dec 1 '15 at 20:35
hairspraygirliehairspraygirlie
307
asked Dec 1 '15 at 20:35
hairspraygirliehairspraygirlie
307
asked Dec 1 '15 at 20:35
hairspraygirliehairspraygirlie
307
307
$begingroup$
Yes. You can let $x_3$ and $x_4$ be free variables.
$endgroup$
– user137731
Dec 1 '15 at 20:41
add a comment |
$begingroup$
Yes. You can let $x_3$ and $x_4$ be free variables.
$endgroup$
– user137731
Dec 1 '15 at 20:41
$begingroup$
Yes. You can let $x_3$ and $x_4$ be free variables.
$endgroup$
– user137731
Dec 1 '15 at 20:41
$begingroup$
Yes. You can let $x_3$ and $x_4$ be free variables.
$endgroup$
– user137731
Dec 1 '15 at 20:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$M^Tx=0 \ iff begin{bmatrix}
1 & 0 & 1 & 0 \
0 & 1 & 0 & 1 \
end{bmatrix}begin{bmatrix}x_1 \ x_2 \ x_3 \ x_4end{bmatrix} = begin{bmatrix}0 \ 0end{bmatrix} \ iff begin{cases}x_1 + x_3 = 0 \ x_2 + x_4 = 0end{cases}$$
Let $x_3=s$ and $x_4=t$ where $s,tinBbb R$, then $$begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4end{bmatrix} = begin{bmatrix}-s \ -t \ s \ tend{bmatrix} = sbegin{bmatrix}-1 \ 0 \ 1 \ 0end{bmatrix} + tbegin{bmatrix}0 \ -1 \ 0 \ 1end{bmatrix}$$
Thus $left{begin{bmatrix}-1 \ 0 \ 1 \ 0end{bmatrix}, begin{bmatrix}0 \ -1 \ 0 \ 1end{bmatrix}right}$ is a basis for the orthogonal complement of the column space of $M$.
$endgroup$
$begingroup$
Well done. After writing up the problem I got the same thing. Thanks for the reassurance!!
$endgroup$
– hairspraygirlie
Dec 1 '15 at 20:48
$begingroup$
No problem. :-)
$endgroup$
– user137731
Dec 1 '15 at 20:50
add a comment |
$begingroup$
In general, you can use the fact that the kernel of $M^*$ annihilates the image of $M$. If $M$ is relative to the standard basis, then $M^*=M^T$, so all you need to do is find the kernel (nullspace) of $M^T$, which is equivalent to solving the system of linear equations in Bye_World’s answer.
answered Dec 1 '15 at 22:27
amdamd
31.3k21052
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add a comment |
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2 Answers
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active
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2 Answers
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active
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votes
$begingroup$
$$M^Tx=0 \ iff begin{bmatrix}
1 & 0 & 1 & 0 \
0 & 1 & 0 & 1 \
end{bmatrix}begin{bmatrix}x_1 \ x_2 \ x_3 \ x_4end{bmatrix} = begin{bmatrix}0 \ 0end{bmatrix} \ iff begin{cases}x_1 + x_3 = 0 \ x_2 + x_4 = 0end{cases}$$
Let $x_3=s$ and $x_4=t$ where $s,tinBbb R$, then $$begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4end{bmatrix} = begin{bmatrix}-s \ -t \ s \ tend{bmatrix} = sbegin{bmatrix}-1 \ 0 \ 1 \ 0end{bmatrix} + tbegin{bmatrix}0 \ -1 \ 0 \ 1end{bmatrix}$$
Thus $left{begin{bmatrix}-1 \ 0 \ 1 \ 0end{bmatrix}, begin{bmatrix}0 \ -1 \ 0 \ 1end{bmatrix}right}$ is a basis for the orthogonal complement of the column space of $M$.
$endgroup$
$begingroup$
Well done. After writing up the problem I got the same thing. Thanks for the reassurance!!
$endgroup$
– hairspraygirlie
Dec 1 '15 at 20:48
$begingroup$
No problem. :-)
$endgroup$
– user137731
Dec 1 '15 at 20:50
add a comment |
$begingroup$
$$M^Tx=0 \ iff begin{bmatrix}
1 & 0 & 1 & 0 \
0 & 1 & 0 & 1 \
end{bmatrix}begin{bmatrix}x_1 \ x_2 \ x_3 \ x_4end{bmatrix} = begin{bmatrix}0 \ 0end{bmatrix} \ iff begin{cases}x_1 + x_3 = 0 \ x_2 + x_4 = 0end{cases}$$
Let $x_3=s$ and $x_4=t$ where $s,tinBbb R$, then $$begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4end{bmatrix} = begin{bmatrix}-s \ -t \ s \ tend{bmatrix} = sbegin{bmatrix}-1 \ 0 \ 1 \ 0end{bmatrix} + tbegin{bmatrix}0 \ -1 \ 0 \ 1end{bmatrix}$$
Thus $left{begin{bmatrix}-1 \ 0 \ 1 \ 0end{bmatrix}, begin{bmatrix}0 \ -1 \ 0 \ 1end{bmatrix}right}$ is a basis for the orthogonal complement of the column space of $M$.
$endgroup$
$begingroup$
Well done. After writing up the problem I got the same thing. Thanks for the reassurance!!
$endgroup$
– hairspraygirlie
Dec 1 '15 at 20:48
$begingroup$
No problem. :-)
$endgroup$
– user137731
Dec 1 '15 at 20:50
add a comment |
$begingroup$
$$M^Tx=0 \ iff begin{bmatrix}
1 & 0 & 1 & 0 \
0 & 1 & 0 & 1 \
end{bmatrix}begin{bmatrix}x_1 \ x_2 \ x_3 \ x_4end{bmatrix} = begin{bmatrix}0 \ 0end{bmatrix} \ iff begin{cases}x_1 + x_3 = 0 \ x_2 + x_4 = 0end{cases}$$
Let $x_3=s$ and $x_4=t$ where $s,tinBbb R$, then $$begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4end{bmatrix} = begin{bmatrix}-s \ -t \ s \ tend{bmatrix} = sbegin{bmatrix}-1 \ 0 \ 1 \ 0end{bmatrix} + tbegin{bmatrix}0 \ -1 \ 0 \ 1end{bmatrix}$$
Thus $left{begin{bmatrix}-1 \ 0 \ 1 \ 0end{bmatrix}, begin{bmatrix}0 \ -1 \ 0 \ 1end{bmatrix}right}$ is a basis for the orthogonal complement of the column space of $M$.
$endgroup$
$$M^Tx=0 \ iff begin{bmatrix}
1 & 0 & 1 & 0 \
0 & 1 & 0 & 1 \
end{bmatrix}begin{bmatrix}x_1 \ x_2 \ x_3 \ x_4end{bmatrix} = begin{bmatrix}0 \ 0end{bmatrix} \ iff begin{cases}x_1 + x_3 = 0 \ x_2 + x_4 = 0end{cases}$$
Let $x_3=s$ and $x_4=t$ where $s,tinBbb R$, then $$begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4end{bmatrix} = begin{bmatrix}-s \ -t \ s \ tend{bmatrix} = sbegin{bmatrix}-1 \ 0 \ 1 \ 0end{bmatrix} + tbegin{bmatrix}0 \ -1 \ 0 \ 1end{bmatrix}$$
Thus $left{begin{bmatrix}-1 \ 0 \ 1 \ 0end{bmatrix}, begin{bmatrix}0 \ -1 \ 0 \ 1end{bmatrix}right}$ is a basis for the orthogonal complement of the column space of $M$.
answered Dec 1 '15 at 20:46
user137731
$begingroup$
Well done. After writing up the problem I got the same thing. Thanks for the reassurance!!
$endgroup$
– hairspraygirlie
Dec 1 '15 at 20:48
$begingroup$
No problem. :-)
$endgroup$
– user137731
Dec 1 '15 at 20:50
add a comment |
$begingroup$
Well done. After writing up the problem I got the same thing. Thanks for the reassurance!!
$endgroup$
– hairspraygirlie
Dec 1 '15 at 20:48
$begingroup$
No problem. :-)
$endgroup$
– user137731
Dec 1 '15 at 20:50
$begingroup$
Well done. After writing up the problem I got the same thing. Thanks for the reassurance!!
$endgroup$
– hairspraygirlie
Dec 1 '15 at 20:48
$begingroup$
Well done. After writing up the problem I got the same thing. Thanks for the reassurance!!
$endgroup$
– hairspraygirlie
Dec 1 '15 at 20:48
$begingroup$
No problem. :-)
$endgroup$
– user137731
Dec 1 '15 at 20:50
$begingroup$
No problem. :-)
$endgroup$
– user137731
Dec 1 '15 at 20:50
add a comment |
$begingroup$
In general, you can use the fact that the kernel of $M^*$ annihilates the image of $M$. If $M$ is relative to the standard basis, then $M^*=M^T$, so all you need to do is find the kernel (nullspace) of $M^T$, which is equivalent to solving the system of linear equations in Bye_World’s answer.
answered Dec 1 '15 at 22:27
amdamd
31.3k21052
$endgroup$
add a comment |
$begingroup$
In general, you can use the fact that the kernel of $M^*$ annihilates the image of $M$. If $M$ is relative to the standard basis, then $M^*=M^T$, so all you need to do is find the kernel (nullspace) of $M^T$, which is equivalent to solving the system of linear equations in Bye_World’s answer.
answered Dec 1 '15 at 22:27
amdamd
31.3k21052
$endgroup$
add a comment |
$begingroup$
In general, you can use the fact that the kernel of $M^*$ annihilates the image of $M$. If $M$ is relative to the standard basis, then $M^*=M^T$, so all you need to do is find the kernel (nullspace) of $M^T$, which is equivalent to solving the system of linear equations in Bye_World’s answer.
answered Dec 1 '15 at 22:27
amdamd
31.3k21052
$endgroup$
In general, you can use the fact that the kernel of $M^*$ annihilates the image of $M$. If $M$ is relative to the standard basis, then $M^*=M^T$, so all you need to do is find the kernel (nullspace) of $M^T$, which is equivalent to solving the system of linear equations in Bye_World’s answer.
answered Dec 1 '15 at 22:27
amdamd
31.3k21052
answered Dec 1 '15 at 22:27
amdamd
31.3k21052
answered Dec 1 '15 at 22:27
amdamd
31.3k21052
answered Dec 1 '15 at 22:27
amdamd
31.3k21052
31.3k21052
add a comment |
add a comment |
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$begingroup$
Yes. You can let $x_3$ and $x_4$ be free variables.
$endgroup$
– user137731
Dec 1 '15 at 20:41