Linear mapping and linear spanProduct of linear mappingWhat are the basis that span $mathbbR^2$ and $mathbbR^3$Linear algebra (Coordinates)how to show that a linear transformation exists between two vectors?Finding the formula of a linear transformationWhat am I doing wrong? - Change of basis matrixLinear mapping matrix with paramters.How do you find the matrix M(T)?How to find $[T]_beta ^beta$ for $beta$ based on the following condition?Linear Transformation of a vector from $R^3$ to $R^2$
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Linear mapping and linear span
Product of linear mappingWhat are the basis that span $mathbbR^2$ and $mathbbR^3$Linear algebra (Coordinates)how to show that a linear transformation exists between two vectors?Finding the formula of a linear transformationWhat am I doing wrong? - Change of basis matrixLinear mapping matrix with paramters.How do you find the matrix M(T)?How to find $[T]_beta ^beta$ for $beta$ based on the following condition?Linear Transformation of a vector from $R^3$ to $R^2$
$begingroup$
This is about a linear algebra problem.
We have $fin L(mathbbR^3,mathbbR^2)$ such that :
- $f(1,0,0)=(0,1)$
- $f(1,1,0)=(1,0)$
- $f(1,1,1)=(1,1)$
We need to find $f(x,y,z)$.
My professor started with $(x,y,z)=alpha(1,0,0)+beta(1,1,0)+gamma(1,1,1)$ but isn't this supposing that $f$ is mapping from $<F>$ not $mathbbR^3$ (with $F=(1,0,0), (1,1,0), (1,1,1)$)?
linear-algebra linear-transformations
asked 2 days ago
LuywLuyw
315
$endgroup$
add a comment |
$begingroup$
This is about a linear algebra problem.
We have $fin L(mathbbR^3,mathbbR^2)$ such that :
- $f(1,0,0)=(0,1)$
- $f(1,1,0)=(1,0)$
- $f(1,1,1)=(1,1)$
We need to find $f(x,y,z)$.
My professor started with $(x,y,z)=alpha(1,0,0)+beta(1,1,0)+gamma(1,1,1)$ but isn't this supposing that $f$ is mapping from $<F>$ not $mathbbR^3$ (with $F=(1,0,0), (1,1,0), (1,1,1)$)?
linear-algebra linear-transformations
asked 2 days ago
LuywLuyw
315
$endgroup$
2
$begingroup$
You can show that those three vectors span $mathbbR^3$. So $langle Frangle = mathbbR^3$.
$endgroup$
– Minus One-Twelfth
2 days ago
add a comment |
$begingroup$
This is about a linear algebra problem.
We have $fin L(mathbbR^3,mathbbR^2)$ such that :
- $f(1,0,0)=(0,1)$
- $f(1,1,0)=(1,0)$
- $f(1,1,1)=(1,1)$
We need to find $f(x,y,z)$.
My professor started with $(x,y,z)=alpha(1,0,0)+beta(1,1,0)+gamma(1,1,1)$ but isn't this supposing that $f$ is mapping from $<F>$ not $mathbbR^3$ (with $F=(1,0,0), (1,1,0), (1,1,1)$)?
linear-algebra linear-transformations
asked 2 days ago
LuywLuyw
315
$endgroup$
This is about a linear algebra problem.
We have $fin L(mathbbR^3,mathbbR^2)$ such that :
- $f(1,0,0)=(0,1)$
- $f(1,1,0)=(1,0)$
- $f(1,1,1)=(1,1)$
We need to find $f(x,y,z)$.
My professor started with $(x,y,z)=alpha(1,0,0)+beta(1,1,0)+gamma(1,1,1)$ but isn't this supposing that $f$ is mapping from $<F>$ not $mathbbR^3$ (with $F=(1,0,0), (1,1,0), (1,1,1)$)?
linear-algebra linear-transformations
linear-algebra linear-transformations
asked 2 days ago
LuywLuyw
315
asked 2 days ago
LuywLuyw
315
edited 2 days ago
Luyw
asked 2 days ago
LuywLuyw
315
asked 2 days ago
LuywLuyw
315
asked 2 days ago
LuywLuyw
315
315
2
$begingroup$
You can show that those three vectors span $mathbbR^3$. So $langle Frangle = mathbbR^3$.
$endgroup$
– Minus One-Twelfth
2 days ago
add a comment |
2
$begingroup$
You can show that those three vectors span $mathbbR^3$. So $langle Frangle = mathbbR^3$.
$endgroup$
– Minus One-Twelfth
2 days ago
2
2
$begingroup$
You can show that those three vectors span $mathbbR^3$. So $langle Frangle = mathbbR^3$.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
You can show that those three vectors span $mathbbR^3$. So $langle Frangle = mathbbR^3$.
$endgroup$
– Minus One-Twelfth
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Find $f(0,1,0)$ and $f(0,0,1)$ by linearity of $f$.
answered 2 days ago
BerciBerci
61.3k23674
$endgroup$
$begingroup$
Can you please explain? I have found them but nothing is jumping out except $f(0,0,1)=f(1,0,0)$.
$endgroup$
– Luyw
2 days ago
1
$begingroup$
Still use linearity for the $f$-image of $(x, y, z) =x(1,0,0)+y(0,1,0)+z(0,0,1)$.
$endgroup$
– Berci
2 days ago
$begingroup$
Much thanks Berci!
$endgroup$
– Luyw
2 days ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Find $f(0,1,0)$ and $f(0,0,1)$ by linearity of $f$.
answered 2 days ago
BerciBerci
61.3k23674
$endgroup$
$begingroup$
Can you please explain? I have found them but nothing is jumping out except $f(0,0,1)=f(1,0,0)$.
$endgroup$
– Luyw
2 days ago
1
$begingroup$
Still use linearity for the $f$-image of $(x, y, z) =x(1,0,0)+y(0,1,0)+z(0,0,1)$.
$endgroup$
– Berci
2 days ago
$begingroup$
Much thanks Berci!
$endgroup$
– Luyw
2 days ago
add a comment |
$begingroup$
Hint: Find $f(0,1,0)$ and $f(0,0,1)$ by linearity of $f$.
answered 2 days ago
BerciBerci
61.3k23674
$endgroup$
$begingroup$
Can you please explain? I have found them but nothing is jumping out except $f(0,0,1)=f(1,0,0)$.
$endgroup$
– Luyw
2 days ago
1
$begingroup$
Still use linearity for the $f$-image of $(x, y, z) =x(1,0,0)+y(0,1,0)+z(0,0,1)$.
$endgroup$
– Berci
2 days ago
$begingroup$
Much thanks Berci!
$endgroup$
– Luyw
2 days ago
add a comment |
$begingroup$
Hint: Find $f(0,1,0)$ and $f(0,0,1)$ by linearity of $f$.
answered 2 days ago
BerciBerci
61.3k23674
$endgroup$
Hint: Find $f(0,1,0)$ and $f(0,0,1)$ by linearity of $f$.
answered 2 days ago
BerciBerci
61.3k23674
answered 2 days ago
BerciBerci
61.3k23674
answered 2 days ago
BerciBerci
61.3k23674
answered 2 days ago
BerciBerci
61.3k23674
61.3k23674
$begingroup$
Can you please explain? I have found them but nothing is jumping out except $f(0,0,1)=f(1,0,0)$.
$endgroup$
– Luyw
2 days ago
1
$begingroup$
Still use linearity for the $f$-image of $(x, y, z) =x(1,0,0)+y(0,1,0)+z(0,0,1)$.
$endgroup$
– Berci
2 days ago
$begingroup$
Much thanks Berci!
$endgroup$
– Luyw
2 days ago
add a comment |
$begingroup$
Can you please explain? I have found them but nothing is jumping out except $f(0,0,1)=f(1,0,0)$.
$endgroup$
– Luyw
2 days ago
1
$begingroup$
Still use linearity for the $f$-image of $(x, y, z) =x(1,0,0)+y(0,1,0)+z(0,0,1)$.
$endgroup$
– Berci
2 days ago
$begingroup$
Much thanks Berci!
$endgroup$
– Luyw
2 days ago
$begingroup$
Can you please explain? I have found them but nothing is jumping out except $f(0,0,1)=f(1,0,0)$.
$endgroup$
– Luyw
2 days ago
$begingroup$
Can you please explain? I have found them but nothing is jumping out except $f(0,0,1)=f(1,0,0)$.
$endgroup$
– Luyw
2 days ago
1
1
$begingroup$
Still use linearity for the $f$-image of $(x, y, z) =x(1,0,0)+y(0,1,0)+z(0,0,1)$.
$endgroup$
– Berci
2 days ago
$begingroup$
Still use linearity for the $f$-image of $(x, y, z) =x(1,0,0)+y(0,1,0)+z(0,0,1)$.
$endgroup$
– Berci
2 days ago
$begingroup$
Much thanks Berci!
$endgroup$
– Luyw
2 days ago
$begingroup$
Much thanks Berci!
$endgroup$
– Luyw
2 days ago
add a comment |
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2
$begingroup$
You can show that those three vectors span $mathbbR^3$. So $langle Frangle = mathbbR^3$.
$endgroup$
– Minus One-Twelfth
2 days ago