Mixing
Finding the matrix $A$ of $T$ with respect to the basis
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Let $T:mathbbR^3rightarrowmathbbR^3$ be given by
$$Tleft(beginbmatrix x \ y \ z endbmatrixright)=beginbmatrix x-y \ y-z \ z-x endbmatrix$$
Find the matrix $A$ of $T$ with respect to the basis
$$beginbmatrix -1 \ 0 \ 1 endbmatrixbeginbmatrix 1 \ 1 \ 1 endbmatrixbeginbmatrix 0 \ 1 \ 0 endbmatrix$$
My try:
I found $Tleft(beginbmatrix x \ y \ z endbmatrixright)=xbeginbmatrix 1 \ 0 \ -1 endbmatrix+ybeginbmatrix -1 \ 1 \ 0 endbmatrix+zbeginbmatrix 0 \ -1 \ 1 endbmatrix$
Now,
$$beginbmatrix -1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 1 & 0 endbmatrixbeginbmatrix -1 \ 0 \ 1 endbmatrix=beginbmatrix 1 \ 1 \ -1 endbmatrix$$
In the same way I found the other matrices as
$$beginbmatrix 0 \ 2 \ 2 endbmatrix,beginbmatrix 1 \ 1 \ 1 endbmatrix$$
Now,
$$beginbmatrix 1 \ 1 \ -1 endbmatrix=abeginbmatrix -1 \ 0 \ 1 endbmatrix+bbeginbmatrix 1 \ 1 \ 1 endbmatrix+cbeginbmatrix 0 \ 1 \ 0 endbmatrix$$
So, if I find $a,b,c$ then that will become the first row of my required matrix right? Can anyone tell whether this method is correct or not?
linear-algebra
asked Jul 18 at 4:09
philip
1228
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up vote
1
down vote
favorite
Let $T:mathbbR^3rightarrowmathbbR^3$ be given by
$$Tleft(beginbmatrix x \ y \ z endbmatrixright)=beginbmatrix x-y \ y-z \ z-x endbmatrix$$
Find the matrix $A$ of $T$ with respect to the basis
$$beginbmatrix -1 \ 0 \ 1 endbmatrixbeginbmatrix 1 \ 1 \ 1 endbmatrixbeginbmatrix 0 \ 1 \ 0 endbmatrix$$
My try:
I found $Tleft(beginbmatrix x \ y \ z endbmatrixright)=xbeginbmatrix 1 \ 0 \ -1 endbmatrix+ybeginbmatrix -1 \ 1 \ 0 endbmatrix+zbeginbmatrix 0 \ -1 \ 1 endbmatrix$
Now,
$$beginbmatrix -1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 1 & 0 endbmatrixbeginbmatrix -1 \ 0 \ 1 endbmatrix=beginbmatrix 1 \ 1 \ -1 endbmatrix$$
In the same way I found the other matrices as
$$beginbmatrix 0 \ 2 \ 2 endbmatrix,beginbmatrix 1 \ 1 \ 1 endbmatrix$$
Now,
$$beginbmatrix 1 \ 1 \ -1 endbmatrix=abeginbmatrix -1 \ 0 \ 1 endbmatrix+bbeginbmatrix 1 \ 1 \ 1 endbmatrix+cbeginbmatrix 0 \ 1 \ 0 endbmatrix$$
So, if I find $a,b,c$ then that will become the first row of my required matrix right? Can anyone tell whether this method is correct or not?
linear-algebra
asked Jul 18 at 4:09
philip
1228
Recall that the i-th column of the matrix should be the i-th basis vector passed through the transformation and than shown in the given basis
– GuySa
Jul 18 at 16:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $T:mathbbR^3rightarrowmathbbR^3$ be given by
$$Tleft(beginbmatrix x \ y \ z endbmatrixright)=beginbmatrix x-y \ y-z \ z-x endbmatrix$$
Find the matrix $A$ of $T$ with respect to the basis
$$beginbmatrix -1 \ 0 \ 1 endbmatrixbeginbmatrix 1 \ 1 \ 1 endbmatrixbeginbmatrix 0 \ 1 \ 0 endbmatrix$$
My try:
I found $Tleft(beginbmatrix x \ y \ z endbmatrixright)=xbeginbmatrix 1 \ 0 \ -1 endbmatrix+ybeginbmatrix -1 \ 1 \ 0 endbmatrix+zbeginbmatrix 0 \ -1 \ 1 endbmatrix$
Now,
$$beginbmatrix -1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 1 & 0 endbmatrixbeginbmatrix -1 \ 0 \ 1 endbmatrix=beginbmatrix 1 \ 1 \ -1 endbmatrix$$
In the same way I found the other matrices as
$$beginbmatrix 0 \ 2 \ 2 endbmatrix,beginbmatrix 1 \ 1 \ 1 endbmatrix$$
Now,
$$beginbmatrix 1 \ 1 \ -1 endbmatrix=abeginbmatrix -1 \ 0 \ 1 endbmatrix+bbeginbmatrix 1 \ 1 \ 1 endbmatrix+cbeginbmatrix 0 \ 1 \ 0 endbmatrix$$
So, if I find $a,b,c$ then that will become the first row of my required matrix right? Can anyone tell whether this method is correct or not?
linear-algebra
asked Jul 18 at 4:09
philip
1228
Let $T:mathbbR^3rightarrowmathbbR^3$ be given by
$$Tleft(beginbmatrix x \ y \ z endbmatrixright)=beginbmatrix x-y \ y-z \ z-x endbmatrix$$
Find the matrix $A$ of $T$ with respect to the basis
$$beginbmatrix -1 \ 0 \ 1 endbmatrixbeginbmatrix 1 \ 1 \ 1 endbmatrixbeginbmatrix 0 \ 1 \ 0 endbmatrix$$
My try:
I found $Tleft(beginbmatrix x \ y \ z endbmatrixright)=xbeginbmatrix 1 \ 0 \ -1 endbmatrix+ybeginbmatrix -1 \ 1 \ 0 endbmatrix+zbeginbmatrix 0 \ -1 \ 1 endbmatrix$
Now,
$$beginbmatrix -1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 1 & 0 endbmatrixbeginbmatrix -1 \ 0 \ 1 endbmatrix=beginbmatrix 1 \ 1 \ -1 endbmatrix$$
In the same way I found the other matrices as
$$beginbmatrix 0 \ 2 \ 2 endbmatrix,beginbmatrix 1 \ 1 \ 1 endbmatrix$$
Now,
$$beginbmatrix 1 \ 1 \ -1 endbmatrix=abeginbmatrix -1 \ 0 \ 1 endbmatrix+bbeginbmatrix 1 \ 1 \ 1 endbmatrix+cbeginbmatrix 0 \ 1 \ 0 endbmatrix$$
So, if I find $a,b,c$ then that will become the first row of my required matrix right? Can anyone tell whether this method is correct or not?
linear-algebra
asked Jul 18 at 4:09
philip
1228
asked Jul 18 at 4:09
philip
1228
asked Jul 18 at 4:09
philip
1228
asked Jul 18 at 4:09
philip
1228
1228
Recall that the i-th column of the matrix should be the i-th basis vector passed through the transformation and than shown in the given basis
– GuySa
Jul 18 at 16:37
add a comment |Â
Recall that the i-th column of the matrix should be the i-th basis vector passed through the transformation and than shown in the given basis
– GuySa
Jul 18 at 16:37
Recall that the i-th column of the matrix should be the i-th basis vector passed through the transformation and than shown in the given basis
– GuySa
Jul 18 at 16:37
Recall that the i-th column of the matrix should be the i-th basis vector passed through the transformation and than shown in the given basis
– GuySa
Jul 18 at 16:37
add a comment |Â
1 Answer
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up vote
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You haven't actually applied $T$ to any of the basis vectors (the matrix you have there is the matrix with the basis vectors as columns, and as such it has little to do with $T$).
If you apply $T$ to the first basis vector, you get $[-1,-1,2]^T$. Now that is the vector you should express as a linear combination of the basis vectors, and the coefficients make up the first column of $A$.
answered Jul 18 at 5:44
Arthur
98.8k793175
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You haven't actually applied $T$ to any of the basis vectors (the matrix you have there is the matrix with the basis vectors as columns, and as such it has little to do with $T$).
If you apply $T$ to the first basis vector, you get $[-1,-1,2]^T$. Now that is the vector you should express as a linear combination of the basis vectors, and the coefficients make up the first column of $A$.
answered Jul 18 at 5:44
Arthur
98.8k793175
add a comment |Â
up vote
2
down vote
accepted
You haven't actually applied $T$ to any of the basis vectors (the matrix you have there is the matrix with the basis vectors as columns, and as such it has little to do with $T$).
If you apply $T$ to the first basis vector, you get $[-1,-1,2]^T$. Now that is the vector you should express as a linear combination of the basis vectors, and the coefficients make up the first column of $A$.
answered Jul 18 at 5:44
Arthur
98.8k793175
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You haven't actually applied $T$ to any of the basis vectors (the matrix you have there is the matrix with the basis vectors as columns, and as such it has little to do with $T$).
If you apply $T$ to the first basis vector, you get $[-1,-1,2]^T$. Now that is the vector you should express as a linear combination of the basis vectors, and the coefficients make up the first column of $A$.
answered Jul 18 at 5:44
Arthur
98.8k793175
You haven't actually applied $T$ to any of the basis vectors (the matrix you have there is the matrix with the basis vectors as columns, and as such it has little to do with $T$).
If you apply $T$ to the first basis vector, you get $[-1,-1,2]^T$. Now that is the vector you should express as a linear combination of the basis vectors, and the coefficients make up the first column of $A$.
answered Jul 18 at 5:44
Arthur
98.8k793175
answered Jul 18 at 5:44
Arthur
98.8k793175
answered Jul 18 at 5:44
Arthur
98.8k793175
answered Jul 18 at 5:44
Arthur
98.8k793175
98.8k793175
add a comment |Â
add a comment |Â
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Recall that the i-th column of the matrix should be the i-th basis vector passed through the transformation and than shown in the given basis
– GuySa
Jul 18 at 16:37