Mixing
Determining the dimension of the kernel and the dimension of the image of the linear map $T:mathbbR^3rightarrowmathbbR^4$
Clash Royale CLAN TAG#URR8PPP
up vote
-1
down vote
favorite
Determine the dimension of the kernel and the dimension of the image of the linear map $T:mathbbR^3rightarrowmathbbR^4$ given by:
$$T(x,y,z)^T=(x-y+z,x+2y-z,2x+y,3y-2z)^T$$
My try:
$$xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix\beginbmatrix 1 & -1 & 1 \ 1 & 2 & -1 \ 2 & 1 & 0 \ 0 & 3 & -2 endbmatrix$$ I don't know how to proceed further. Can anyone please help me to solve this?
linear-algebra
asked Aug 2 at 2:57
philip
938
add a comment |Â
up vote
-1
down vote
favorite
Determine the dimension of the kernel and the dimension of the image of the linear map $T:mathbbR^3rightarrowmathbbR^4$ given by:
$$T(x,y,z)^T=(x-y+z,x+2y-z,2x+y,3y-2z)^T$$
My try:
$$xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix\beginbmatrix 1 & -1 & 1 \ 1 & 2 & -1 \ 2 & 1 & 0 \ 0 & 3 & -2 endbmatrix$$ I don't know how to proceed further. Can anyone please help me to solve this?
linear-algebra
asked Aug 2 at 2:57
philip
938
Do you know how to reduce the matrix?
– Dionel Jaime
Aug 2 at 3:07
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Determine the dimension of the kernel and the dimension of the image of the linear map $T:mathbbR^3rightarrowmathbbR^4$ given by:
$$T(x,y,z)^T=(x-y+z,x+2y-z,2x+y,3y-2z)^T$$
My try:
$$xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix\beginbmatrix 1 & -1 & 1 \ 1 & 2 & -1 \ 2 & 1 & 0 \ 0 & 3 & -2 endbmatrix$$ I don't know how to proceed further. Can anyone please help me to solve this?
linear-algebra
asked Aug 2 at 2:57
philip
938
Determine the dimension of the kernel and the dimension of the image of the linear map $T:mathbbR^3rightarrowmathbbR^4$ given by:
$$T(x,y,z)^T=(x-y+z,x+2y-z,2x+y,3y-2z)^T$$
My try:
$$xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix\beginbmatrix 1 & -1 & 1 \ 1 & 2 & -1 \ 2 & 1 & 0 \ 0 & 3 & -2 endbmatrix$$ I don't know how to proceed further. Can anyone please help me to solve this?
linear-algebra
asked Aug 2 at 2:57
philip
938
asked Aug 2 at 2:57
philip
938
asked Aug 2 at 2:57
philip
938
asked Aug 2 at 2:57
philip
938
938
Do you know how to reduce the matrix?
– Dionel Jaime
Aug 2 at 3:07
add a comment |Â
Do you know how to reduce the matrix?
– Dionel Jaime
Aug 2 at 3:07
Do you know how to reduce the matrix?
– Dionel Jaime
Aug 2 at 3:07
Do you know how to reduce the matrix?
– Dionel Jaime
Aug 2 at 3:07
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Remember what you're trying to solve here. The expression
$$xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix$$
is a vector in $mathbbR^4$. You have to ask yourself, what does it have to do with the question? What are you trying to find out here? What does the kernel even mean?
The kernel is the set of vectors in $mathbbR^3$ that map to the zero vector in $mathbbR^4$ under $T$. Our map $T$ does the following:
$$Tbeginbmatrixx \ y \ z endbmatrix=beginbmatrixx−y+z \x+2y−z \2x+y \3y−2zendbmatrix = xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix$$
When does it send the vector to $0$? When $x, y, z$ satisfy
$$beginbmatrixx−y+z \x+2y−z \2x+y \3y−2zendbmatrix = beginbmatrix 0 \ 0 \ 0 \ 0 endbmatrix.$$
Equating entries yields the system of equations,
beginalign*
x - y + z &= 0 \ x + 2y - z &= 0 \ 2x + y &= 0 \ 3y - 2z &= 0.
endalign*
Now it's a matter of solving this system, which we can do with an augmented matrix:
$$beginbmatrix 1 & -1 & 1 & 0 \ 1 & 2 & -1 & 0 \ 2 & 1 & 0 & 0 \ 0 & 3 & -2 & 0 endbmatrix$$
Or, if you're comfortable, you can omit the $0$ column. If you're comfortable, you can skip some of these steps and go straight to the matrix.
But, as is my point, only if you're comfortable! If you don't know/understand this stuff, then you can get lost along the shortcuts!
From here, Gauss-Jordan elimination gives:
$$beginbmatrix1 & 0 & frac13 & 0 \ 0 & 1 & -frac23 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix,$$
which translates to
beginalign*
x + fracz3 &= 0 \
y - frac2z3 &= 0.
endalign*
If we let $z = 3t$, for $t in mathbbR$, then $x = -t$ and $y = 2t$. This gives us a solution of
$$beginbmatrix x \ y \ z endbmatrix = beginbmatrix -t \ 2t \ 3t endbmatrix = tbeginbmatrix -1 \ 2 \ 3 endbmatrix$$
as $t$ ranges over $mathbbR$. This means our kernel is
$$operatornameKer(T) = operatornamespan beginbmatrix -1 \ 2 \ 3 endbmatrix.$$
It is the span of a single non-zero (hence, linearly independent) vector, hence the kernel is of dimension $1$ (i.e. $operatornamenull T = 1$). Using the rank-nullity theorem,
$$3 = operatornamedim mathbbR^3 = operatornamerank T + operatornamenull T = operatornamerank T + 1$$
which implies $operatornamerank T = 2$, which is the dimension of the image.
answered Aug 2 at 3:24
Theo Bendit
11.7k1841
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Remember what you're trying to solve here. The expression
$$xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix$$
is a vector in $mathbbR^4$. You have to ask yourself, what does it have to do with the question? What are you trying to find out here? What does the kernel even mean?
The kernel is the set of vectors in $mathbbR^3$ that map to the zero vector in $mathbbR^4$ under $T$. Our map $T$ does the following:
$$Tbeginbmatrixx \ y \ z endbmatrix=beginbmatrixx−y+z \x+2y−z \2x+y \3y−2zendbmatrix = xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix$$
When does it send the vector to $0$? When $x, y, z$ satisfy
$$beginbmatrixx−y+z \x+2y−z \2x+y \3y−2zendbmatrix = beginbmatrix 0 \ 0 \ 0 \ 0 endbmatrix.$$
Equating entries yields the system of equations,
beginalign*
x - y + z &= 0 \ x + 2y - z &= 0 \ 2x + y &= 0 \ 3y - 2z &= 0.
endalign*
Now it's a matter of solving this system, which we can do with an augmented matrix:
$$beginbmatrix 1 & -1 & 1 & 0 \ 1 & 2 & -1 & 0 \ 2 & 1 & 0 & 0 \ 0 & 3 & -2 & 0 endbmatrix$$
Or, if you're comfortable, you can omit the $0$ column. If you're comfortable, you can skip some of these steps and go straight to the matrix.
But, as is my point, only if you're comfortable! If you don't know/understand this stuff, then you can get lost along the shortcuts!
From here, Gauss-Jordan elimination gives:
$$beginbmatrix1 & 0 & frac13 & 0 \ 0 & 1 & -frac23 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix,$$
which translates to
beginalign*
x + fracz3 &= 0 \
y - frac2z3 &= 0.
endalign*
If we let $z = 3t$, for $t in mathbbR$, then $x = -t$ and $y = 2t$. This gives us a solution of
$$beginbmatrix x \ y \ z endbmatrix = beginbmatrix -t \ 2t \ 3t endbmatrix = tbeginbmatrix -1 \ 2 \ 3 endbmatrix$$
as $t$ ranges over $mathbbR$. This means our kernel is
$$operatornameKer(T) = operatornamespan beginbmatrix -1 \ 2 \ 3 endbmatrix.$$
It is the span of a single non-zero (hence, linearly independent) vector, hence the kernel is of dimension $1$ (i.e. $operatornamenull T = 1$). Using the rank-nullity theorem,
$$3 = operatornamedim mathbbR^3 = operatornamerank T + operatornamenull T = operatornamerank T + 1$$
which implies $operatornamerank T = 2$, which is the dimension of the image.
answered Aug 2 at 3:24
Theo Bendit
11.7k1841
add a comment |Â
up vote
4
down vote
accepted
Remember what you're trying to solve here. The expression
$$xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix$$
is a vector in $mathbbR^4$. You have to ask yourself, what does it have to do with the question? What are you trying to find out here? What does the kernel even mean?
The kernel is the set of vectors in $mathbbR^3$ that map to the zero vector in $mathbbR^4$ under $T$. Our map $T$ does the following:
$$Tbeginbmatrixx \ y \ z endbmatrix=beginbmatrixx−y+z \x+2y−z \2x+y \3y−2zendbmatrix = xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix$$
When does it send the vector to $0$? When $x, y, z$ satisfy
$$beginbmatrixx−y+z \x+2y−z \2x+y \3y−2zendbmatrix = beginbmatrix 0 \ 0 \ 0 \ 0 endbmatrix.$$
Equating entries yields the system of equations,
beginalign*
x - y + z &= 0 \ x + 2y - z &= 0 \ 2x + y &= 0 \ 3y - 2z &= 0.
endalign*
Now it's a matter of solving this system, which we can do with an augmented matrix:
$$beginbmatrix 1 & -1 & 1 & 0 \ 1 & 2 & -1 & 0 \ 2 & 1 & 0 & 0 \ 0 & 3 & -2 & 0 endbmatrix$$
Or, if you're comfortable, you can omit the $0$ column. If you're comfortable, you can skip some of these steps and go straight to the matrix.
But, as is my point, only if you're comfortable! If you don't know/understand this stuff, then you can get lost along the shortcuts!
From here, Gauss-Jordan elimination gives:
$$beginbmatrix1 & 0 & frac13 & 0 \ 0 & 1 & -frac23 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix,$$
which translates to
beginalign*
x + fracz3 &= 0 \
y - frac2z3 &= 0.
endalign*
If we let $z = 3t$, for $t in mathbbR$, then $x = -t$ and $y = 2t$. This gives us a solution of
$$beginbmatrix x \ y \ z endbmatrix = beginbmatrix -t \ 2t \ 3t endbmatrix = tbeginbmatrix -1 \ 2 \ 3 endbmatrix$$
as $t$ ranges over $mathbbR$. This means our kernel is
$$operatornameKer(T) = operatornamespan beginbmatrix -1 \ 2 \ 3 endbmatrix.$$
It is the span of a single non-zero (hence, linearly independent) vector, hence the kernel is of dimension $1$ (i.e. $operatornamenull T = 1$). Using the rank-nullity theorem,
$$3 = operatornamedim mathbbR^3 = operatornamerank T + operatornamenull T = operatornamerank T + 1$$
which implies $operatornamerank T = 2$, which is the dimension of the image.
answered Aug 2 at 3:24
Theo Bendit
11.7k1841
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Remember what you're trying to solve here. The expression
$$xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix$$
is a vector in $mathbbR^4$. You have to ask yourself, what does it have to do with the question? What are you trying to find out here? What does the kernel even mean?
The kernel is the set of vectors in $mathbbR^3$ that map to the zero vector in $mathbbR^4$ under $T$. Our map $T$ does the following:
$$Tbeginbmatrixx \ y \ z endbmatrix=beginbmatrixx−y+z \x+2y−z \2x+y \3y−2zendbmatrix = xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix$$
When does it send the vector to $0$? When $x, y, z$ satisfy
$$beginbmatrixx−y+z \x+2y−z \2x+y \3y−2zendbmatrix = beginbmatrix 0 \ 0 \ 0 \ 0 endbmatrix.$$
Equating entries yields the system of equations,
beginalign*
x - y + z &= 0 \ x + 2y - z &= 0 \ 2x + y &= 0 \ 3y - 2z &= 0.
endalign*
Now it's a matter of solving this system, which we can do with an augmented matrix:
$$beginbmatrix 1 & -1 & 1 & 0 \ 1 & 2 & -1 & 0 \ 2 & 1 & 0 & 0 \ 0 & 3 & -2 & 0 endbmatrix$$
Or, if you're comfortable, you can omit the $0$ column. If you're comfortable, you can skip some of these steps and go straight to the matrix.
But, as is my point, only if you're comfortable! If you don't know/understand this stuff, then you can get lost along the shortcuts!
From here, Gauss-Jordan elimination gives:
$$beginbmatrix1 & 0 & frac13 & 0 \ 0 & 1 & -frac23 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix,$$
which translates to
beginalign*
x + fracz3 &= 0 \
y - frac2z3 &= 0.
endalign*
If we let $z = 3t$, for $t in mathbbR$, then $x = -t$ and $y = 2t$. This gives us a solution of
$$beginbmatrix x \ y \ z endbmatrix = beginbmatrix -t \ 2t \ 3t endbmatrix = tbeginbmatrix -1 \ 2 \ 3 endbmatrix$$
as $t$ ranges over $mathbbR$. This means our kernel is
$$operatornameKer(T) = operatornamespan beginbmatrix -1 \ 2 \ 3 endbmatrix.$$
It is the span of a single non-zero (hence, linearly independent) vector, hence the kernel is of dimension $1$ (i.e. $operatornamenull T = 1$). Using the rank-nullity theorem,
$$3 = operatornamedim mathbbR^3 = operatornamerank T + operatornamenull T = operatornamerank T + 1$$
which implies $operatornamerank T = 2$, which is the dimension of the image.
answered Aug 2 at 3:24
Theo Bendit
11.7k1841
Remember what you're trying to solve here. The expression
$$xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix$$
is a vector in $mathbbR^4$. You have to ask yourself, what does it have to do with the question? What are you trying to find out here? What does the kernel even mean?
The kernel is the set of vectors in $mathbbR^3$ that map to the zero vector in $mathbbR^4$ under $T$. Our map $T$ does the following:
$$Tbeginbmatrixx \ y \ z endbmatrix=beginbmatrixx−y+z \x+2y−z \2x+y \3y−2zendbmatrix = xbeginbmatrix 1 \ 1 \ 2 \ 0 endbmatrix+ybeginbmatrix -1 \ 2 \ 1 \ 3 endbmatrix+zbeginbmatrix 1 \ -1 \ 0 \ -2 endbmatrix$$
When does it send the vector to $0$? When $x, y, z$ satisfy
$$beginbmatrixx−y+z \x+2y−z \2x+y \3y−2zendbmatrix = beginbmatrix 0 \ 0 \ 0 \ 0 endbmatrix.$$
Equating entries yields the system of equations,
beginalign*
x - y + z &= 0 \ x + 2y - z &= 0 \ 2x + y &= 0 \ 3y - 2z &= 0.
endalign*
Now it's a matter of solving this system, which we can do with an augmented matrix:
$$beginbmatrix 1 & -1 & 1 & 0 \ 1 & 2 & -1 & 0 \ 2 & 1 & 0 & 0 \ 0 & 3 & -2 & 0 endbmatrix$$
Or, if you're comfortable, you can omit the $0$ column. If you're comfortable, you can skip some of these steps and go straight to the matrix.
But, as is my point, only if you're comfortable! If you don't know/understand this stuff, then you can get lost along the shortcuts!
From here, Gauss-Jordan elimination gives:
$$beginbmatrix1 & 0 & frac13 & 0 \ 0 & 1 & -frac23 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix,$$
which translates to
beginalign*
x + fracz3 &= 0 \
y - frac2z3 &= 0.
endalign*
If we let $z = 3t$, for $t in mathbbR$, then $x = -t$ and $y = 2t$. This gives us a solution of
$$beginbmatrix x \ y \ z endbmatrix = beginbmatrix -t \ 2t \ 3t endbmatrix = tbeginbmatrix -1 \ 2 \ 3 endbmatrix$$
as $t$ ranges over $mathbbR$. This means our kernel is
$$operatornameKer(T) = operatornamespan beginbmatrix -1 \ 2 \ 3 endbmatrix.$$
It is the span of a single non-zero (hence, linearly independent) vector, hence the kernel is of dimension $1$ (i.e. $operatornamenull T = 1$). Using the rank-nullity theorem,
$$3 = operatornamedim mathbbR^3 = operatornamerank T + operatornamenull T = operatornamerank T + 1$$
which implies $operatornamerank T = 2$, which is the dimension of the image.
answered Aug 2 at 3:24
Theo Bendit
11.7k1841
answered Aug 2 at 3:24
Theo Bendit
11.7k1841
answered Aug 2 at 3:24
Theo Bendit
11.7k1841
answered Aug 2 at 3:24
Theo Bendit
11.7k1841
11.7k1841
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2869683%2fdetermining-the-dimension-of-the-kernel-and-the-dimension-of-the-image-of-the-li%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Do you know how to reduce the matrix?
– Dionel Jaime
Aug 2 at 3:07