Finding bases for a Linear Transformation of a Matrix The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraHow to find basis using the transformation matrixChange Bases of Linear TransformationLinear maps, matrix transformationFinding new matrix representation of a linear transformation with respect to new basisFind matrix of linear transformation relative to new basesHow to find bases $alpha$ and $beta$ for the following $P_3(BbbR)$ and $P_2(BbbR)$?Consider the matrix of the transformationHow to find the matrix of a linear transformation with respect to two bases?Verifying the row-rank and column-rank of a matrix are equal by finding bases for eachHow to find basesDetermine if bases for R2 and R3 exist, given a linear transformation matrix with respect to said bases
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Finding bases for a Linear Transformation of a Matrix
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraHow to find basis using the transformation matrixChange Bases of Linear TransformationLinear maps, matrix transformationFinding new matrix representation of a linear transformation with respect to new basisFind matrix of linear transformation relative to new basesHow to find bases $alpha$ and $beta$ for the following $P_3(BbbR)$ and $P_2(BbbR)$?Consider the matrix of the transformationHow to find the matrix of a linear transformation with respect to two bases?Verifying the row-rank and column-rank of a matrix are equal by finding bases for eachHow to find basesDetermine if bases for R2 and R3 exist, given a linear transformation matrix with respect to said bases
$begingroup$
Suppose a linear transformation $T $: $P_3(Bbb R)$ to $ P_2(Bbb R))$ has the matrix $$A=beginpmatrix 1 & 2 & 0 & 0 \ 0 & 1 & 2 & 1 \ 1& 1 & 1 & 1 endpmatrix$$
relative to the standard bases of $P_3(Bbb R)$ and $ P_2(Bbb R))$.
Find bases $alpha$ of $P_3(Bbb R)$ and $beta$ of $ P_2(Bbb R))$ such that the matrix $T$ relative to $alpha$ and $beta$ is the reduced row echelon form of A.
I have been looking everywhere for a similar example. I am struggling with where to begin.
I have calculated that the reduced row echelon form of A is:
$$beginpmatrix 1 & 0 & 0 & frac23 \ 0 & 1 & 0 & -frac13 \ 0& 0 & 1 & frac23 endpmatrix$$
linear-algebra matrices linear-transformations
$endgroup$
add a comment |
$begingroup$
Suppose a linear transformation $T $: $P_3(Bbb R)$ to $ P_2(Bbb R))$ has the matrix $$A=beginpmatrix 1 & 2 & 0 & 0 \ 0 & 1 & 2 & 1 \ 1& 1 & 1 & 1 endpmatrix$$
relative to the standard bases of $P_3(Bbb R)$ and $ P_2(Bbb R))$.
Find bases $alpha$ of $P_3(Bbb R)$ and $beta$ of $ P_2(Bbb R))$ such that the matrix $T$ relative to $alpha$ and $beta$ is the reduced row echelon form of A.
I have been looking everywhere for a similar example. I am struggling with where to begin.
I have calculated that the reduced row echelon form of A is:
$$beginpmatrix 1 & 0 & 0 & frac23 \ 0 & 1 & 0 & -frac13 \ 0& 0 & 1 & frac23 endpmatrix$$
linear-algebra matrices linear-transformations
$endgroup$
1
$begingroup$
Hint: Can you find an invertible matrix $B$ such that $BA$ is the rref of $A$?
$endgroup$
– amd
Mar 4 '16 at 19:47
$begingroup$
Thank you! I have figured out B. Is that all that has to be done?
$endgroup$
– user319635
Mar 4 '16 at 20:17
$begingroup$
Oops, I still have to find my bases alpha and beta, how do i go about this?
$endgroup$
– user319635
Mar 4 '16 at 20:27
$begingroup$
Interpret $B$ as a change-of-basis matrix. Note that the solution isn’t unique. You can start with pretty much any basis for $P_3(mathbb R)$ and find a corresponding basis for $P_2(mathbb R)$ so that the matrix of $A$ has the desired form.
$endgroup$
– amd
Mar 4 '16 at 20:37
add a comment |
$begingroup$
Suppose a linear transformation $T $: $P_3(Bbb R)$ to $ P_2(Bbb R))$ has the matrix $$A=beginpmatrix 1 & 2 & 0 & 0 \ 0 & 1 & 2 & 1 \ 1& 1 & 1 & 1 endpmatrix$$
relative to the standard bases of $P_3(Bbb R)$ and $ P_2(Bbb R))$.
Find bases $alpha$ of $P_3(Bbb R)$ and $beta$ of $ P_2(Bbb R))$ such that the matrix $T$ relative to $alpha$ and $beta$ is the reduced row echelon form of A.
I have been looking everywhere for a similar example. I am struggling with where to begin.
I have calculated that the reduced row echelon form of A is:
$$beginpmatrix 1 & 0 & 0 & frac23 \ 0 & 1 & 0 & -frac13 \ 0& 0 & 1 & frac23 endpmatrix$$
linear-algebra matrices linear-transformations
$endgroup$
Suppose a linear transformation $T $: $P_3(Bbb R)$ to $ P_2(Bbb R))$ has the matrix $$A=beginpmatrix 1 & 2 & 0 & 0 \ 0 & 1 & 2 & 1 \ 1& 1 & 1 & 1 endpmatrix$$
relative to the standard bases of $P_3(Bbb R)$ and $ P_2(Bbb R))$.
Find bases $alpha$ of $P_3(Bbb R)$ and $beta$ of $ P_2(Bbb R))$ such that the matrix $T$ relative to $alpha$ and $beta$ is the reduced row echelon form of A.
I have been looking everywhere for a similar example. I am struggling with where to begin.
I have calculated that the reduced row echelon form of A is:
$$beginpmatrix 1 & 0 & 0 & frac23 \ 0 & 1 & 0 & -frac13 \ 0& 0 & 1 & frac23 endpmatrix$$
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
edited Mar 4 '16 at 20:15
asked Mar 4 '16 at 16:06
user319635
1
$begingroup$
Hint: Can you find an invertible matrix $B$ such that $BA$ is the rref of $A$?
$endgroup$
– amd
Mar 4 '16 at 19:47
$begingroup$
Thank you! I have figured out B. Is that all that has to be done?
$endgroup$
– user319635
Mar 4 '16 at 20:17
$begingroup$
Oops, I still have to find my bases alpha and beta, how do i go about this?
$endgroup$
– user319635
Mar 4 '16 at 20:27
$begingroup$
Interpret $B$ as a change-of-basis matrix. Note that the solution isn’t unique. You can start with pretty much any basis for $P_3(mathbb R)$ and find a corresponding basis for $P_2(mathbb R)$ so that the matrix of $A$ has the desired form.
$endgroup$
– amd
Mar 4 '16 at 20:37
add a comment |
1
$begingroup$
Hint: Can you find an invertible matrix $B$ such that $BA$ is the rref of $A$?
$endgroup$
– amd
Mar 4 '16 at 19:47
$begingroup$
Thank you! I have figured out B. Is that all that has to be done?
$endgroup$
– user319635
Mar 4 '16 at 20:17
$begingroup$
Oops, I still have to find my bases alpha and beta, how do i go about this?
$endgroup$
– user319635
Mar 4 '16 at 20:27
$begingroup$
Interpret $B$ as a change-of-basis matrix. Note that the solution isn’t unique. You can start with pretty much any basis for $P_3(mathbb R)$ and find a corresponding basis for $P_2(mathbb R)$ so that the matrix of $A$ has the desired form.
$endgroup$
– amd
Mar 4 '16 at 20:37
1
1
$begingroup$
Hint: Can you find an invertible matrix $B$ such that $BA$ is the rref of $A$?
$endgroup$
– amd
Mar 4 '16 at 19:47
$begingroup$
Hint: Can you find an invertible matrix $B$ such that $BA$ is the rref of $A$?
$endgroup$
– amd
Mar 4 '16 at 19:47
$begingroup$
Thank you! I have figured out B. Is that all that has to be done?
$endgroup$
– user319635
Mar 4 '16 at 20:17
$begingroup$
Thank you! I have figured out B. Is that all that has to be done?
$endgroup$
– user319635
Mar 4 '16 at 20:17
$begingroup$
Oops, I still have to find my bases alpha and beta, how do i go about this?
$endgroup$
– user319635
Mar 4 '16 at 20:27
$begingroup$
Oops, I still have to find my bases alpha and beta, how do i go about this?
$endgroup$
– user319635
Mar 4 '16 at 20:27
$begingroup$
Interpret $B$ as a change-of-basis matrix. Note that the solution isn’t unique. You can start with pretty much any basis for $P_3(mathbb R)$ and find a corresponding basis for $P_2(mathbb R)$ so that the matrix of $A$ has the desired form.
$endgroup$
– amd
Mar 4 '16 at 20:37
$begingroup$
Interpret $B$ as a change-of-basis matrix. Note that the solution isn’t unique. You can start with pretty much any basis for $P_3(mathbb R)$ and find a corresponding basis for $P_2(mathbb R)$ so that the matrix of $A$ has the desired form.
$endgroup$
– amd
Mar 4 '16 at 20:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have found a product of elementary matrices $B$ such that $operatornamerref(A) = BA$. If we denote the matrix of a transformation $T$ in the bases $B,C$ (for the domain, codomain respectively) by $[T]_B^C$, we have in our situation $$[mathitid]_mathitst^beta [T]_mathitst^mathitst = [T]_mathitst^beta = operatornamerref(A) = BA $$
for some unknown basis $beta$. (And where $mathitst$ represents either of the standard bases, and $mathitid$ is the identity transformation.) This suggests that we take $alpha = mathitst$, and $beta$ to be the unique basis such that $$[mathitid]_mathitst^beta = B,$$ or $$[mathitid]_beta^mathitst = B^-1.$$
This means that the coordinates of the vectors in $beta$ in the standard basis are simply the columns of $B^-1$.
answered Mar 4 '16 at 20:38
Alex ProvostAlex Provost
15.6k32351
$endgroup$
$begingroup$
So then does this mean $beta = [1, 0, 1], [2, 1, 1], [0, 2, 1]$ ? I am a bit confused what $alpha$ is
$endgroup$
– user319635
Mar 4 '16 at 23:49
$begingroup$
Wait but this isn't a polynomial of degree 2?
$endgroup$
– user319635
Mar 5 '16 at 0:03
$begingroup$
@user319635 $alpha$ is the standard basis $(1,x,x^2,x^3)$. And not quite; these are the coordinates of the $beta$-vectors in the standard basis. Thus $beta = (1 + x^2,2 + x + x^2, 2x + x^2)$.
$endgroup$
– Alex Provost
Mar 5 '16 at 0:12
$begingroup$
Right! Thank you so much!
$endgroup$
– user319635
Mar 5 '16 at 0:29
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
You have found a product of elementary matrices $B$ such that $operatornamerref(A) = BA$. If we denote the matrix of a transformation $T$ in the bases $B,C$ (for the domain, codomain respectively) by $[T]_B^C$, we have in our situation $$[mathitid]_mathitst^beta [T]_mathitst^mathitst = [T]_mathitst^beta = operatornamerref(A) = BA $$
for some unknown basis $beta$. (And where $mathitst$ represents either of the standard bases, and $mathitid$ is the identity transformation.) This suggests that we take $alpha = mathitst$, and $beta$ to be the unique basis such that $$[mathitid]_mathitst^beta = B,$$ or $$[mathitid]_beta^mathitst = B^-1.$$
This means that the coordinates of the vectors in $beta$ in the standard basis are simply the columns of $B^-1$.
answered Mar 4 '16 at 20:38
Alex ProvostAlex Provost
15.6k32351
$endgroup$
$begingroup$
So then does this mean $beta = [1, 0, 1], [2, 1, 1], [0, 2, 1]$ ? I am a bit confused what $alpha$ is
$endgroup$
– user319635
Mar 4 '16 at 23:49
$begingroup$
Wait but this isn't a polynomial of degree 2?
$endgroup$
– user319635
Mar 5 '16 at 0:03
$begingroup$
@user319635 $alpha$ is the standard basis $(1,x,x^2,x^3)$. And not quite; these are the coordinates of the $beta$-vectors in the standard basis. Thus $beta = (1 + x^2,2 + x + x^2, 2x + x^2)$.
$endgroup$
– Alex Provost
Mar 5 '16 at 0:12
$begingroup$
Right! Thank you so much!
$endgroup$
– user319635
Mar 5 '16 at 0:29
add a comment |
$begingroup$
You have found a product of elementary matrices $B$ such that $operatornamerref(A) = BA$. If we denote the matrix of a transformation $T$ in the bases $B,C$ (for the domain, codomain respectively) by $[T]_B^C$, we have in our situation $$[mathitid]_mathitst^beta [T]_mathitst^mathitst = [T]_mathitst^beta = operatornamerref(A) = BA $$
for some unknown basis $beta$. (And where $mathitst$ represents either of the standard bases, and $mathitid$ is the identity transformation.) This suggests that we take $alpha = mathitst$, and $beta$ to be the unique basis such that $$[mathitid]_mathitst^beta = B,$$ or $$[mathitid]_beta^mathitst = B^-1.$$
This means that the coordinates of the vectors in $beta$ in the standard basis are simply the columns of $B^-1$.
answered Mar 4 '16 at 20:38
Alex ProvostAlex Provost
15.6k32351
$endgroup$
$begingroup$
So then does this mean $beta = [1, 0, 1], [2, 1, 1], [0, 2, 1]$ ? I am a bit confused what $alpha$ is
$endgroup$
– user319635
Mar 4 '16 at 23:49
$begingroup$
Wait but this isn't a polynomial of degree 2?
$endgroup$
– user319635
Mar 5 '16 at 0:03
$begingroup$
@user319635 $alpha$ is the standard basis $(1,x,x^2,x^3)$. And not quite; these are the coordinates of the $beta$-vectors in the standard basis. Thus $beta = (1 + x^2,2 + x + x^2, 2x + x^2)$.
$endgroup$
– Alex Provost
Mar 5 '16 at 0:12
$begingroup$
Right! Thank you so much!
$endgroup$
– user319635
Mar 5 '16 at 0:29
add a comment |
$begingroup$
You have found a product of elementary matrices $B$ such that $operatornamerref(A) = BA$. If we denote the matrix of a transformation $T$ in the bases $B,C$ (for the domain, codomain respectively) by $[T]_B^C$, we have in our situation $$[mathitid]_mathitst^beta [T]_mathitst^mathitst = [T]_mathitst^beta = operatornamerref(A) = BA $$
for some unknown basis $beta$. (And where $mathitst$ represents either of the standard bases, and $mathitid$ is the identity transformation.) This suggests that we take $alpha = mathitst$, and $beta$ to be the unique basis such that $$[mathitid]_mathitst^beta = B,$$ or $$[mathitid]_beta^mathitst = B^-1.$$
This means that the coordinates of the vectors in $beta$ in the standard basis are simply the columns of $B^-1$.
answered Mar 4 '16 at 20:38
Alex ProvostAlex Provost
15.6k32351
$endgroup$
You have found a product of elementary matrices $B$ such that $operatornamerref(A) = BA$. If we denote the matrix of a transformation $T$ in the bases $B,C$ (for the domain, codomain respectively) by $[T]_B^C$, we have in our situation $$[mathitid]_mathitst^beta [T]_mathitst^mathitst = [T]_mathitst^beta = operatornamerref(A) = BA $$
for some unknown basis $beta$. (And where $mathitst$ represents either of the standard bases, and $mathitid$ is the identity transformation.) This suggests that we take $alpha = mathitst$, and $beta$ to be the unique basis such that $$[mathitid]_mathitst^beta = B,$$ or $$[mathitid]_beta^mathitst = B^-1.$$
This means that the coordinates of the vectors in $beta$ in the standard basis are simply the columns of $B^-1$.
answered Mar 4 '16 at 20:38
Alex ProvostAlex Provost
15.6k32351
answered Mar 4 '16 at 20:38
Alex ProvostAlex Provost
15.6k32351
answered Mar 4 '16 at 20:38
Alex ProvostAlex Provost
15.6k32351
answered Mar 4 '16 at 20:38
Alex ProvostAlex Provost
15.6k32351
15.6k32351
$begingroup$
So then does this mean $beta = [1, 0, 1], [2, 1, 1], [0, 2, 1]$ ? I am a bit confused what $alpha$ is
$endgroup$
– user319635
Mar 4 '16 at 23:49
$begingroup$
Wait but this isn't a polynomial of degree 2?
$endgroup$
– user319635
Mar 5 '16 at 0:03
$begingroup$
@user319635 $alpha$ is the standard basis $(1,x,x^2,x^3)$. And not quite; these are the coordinates of the $beta$-vectors in the standard basis. Thus $beta = (1 + x^2,2 + x + x^2, 2x + x^2)$.
$endgroup$
– Alex Provost
Mar 5 '16 at 0:12
$begingroup$
Right! Thank you so much!
$endgroup$
– user319635
Mar 5 '16 at 0:29
add a comment |
$begingroup$
So then does this mean $beta = [1, 0, 1], [2, 1, 1], [0, 2, 1]$ ? I am a bit confused what $alpha$ is
$endgroup$
– user319635
Mar 4 '16 at 23:49
$begingroup$
Wait but this isn't a polynomial of degree 2?
$endgroup$
– user319635
Mar 5 '16 at 0:03
$begingroup$
@user319635 $alpha$ is the standard basis $(1,x,x^2,x^3)$. And not quite; these are the coordinates of the $beta$-vectors in the standard basis. Thus $beta = (1 + x^2,2 + x + x^2, 2x + x^2)$.
$endgroup$
– Alex Provost
Mar 5 '16 at 0:12
$begingroup$
Right! Thank you so much!
$endgroup$
– user319635
Mar 5 '16 at 0:29
$begingroup$
So then does this mean $beta = [1, 0, 1], [2, 1, 1], [0, 2, 1]$ ? I am a bit confused what $alpha$ is
$endgroup$
– user319635
Mar 4 '16 at 23:49
$begingroup$
So then does this mean $beta = [1, 0, 1], [2, 1, 1], [0, 2, 1]$ ? I am a bit confused what $alpha$ is
$endgroup$
– user319635
Mar 4 '16 at 23:49
$begingroup$
Wait but this isn't a polynomial of degree 2?
$endgroup$
– user319635
Mar 5 '16 at 0:03
$begingroup$
Wait but this isn't a polynomial of degree 2?
$endgroup$
– user319635
Mar 5 '16 at 0:03
$begingroup$
@user319635 $alpha$ is the standard basis $(1,x,x^2,x^3)$. And not quite; these are the coordinates of the $beta$-vectors in the standard basis. Thus $beta = (1 + x^2,2 + x + x^2, 2x + x^2)$.
$endgroup$
– Alex Provost
Mar 5 '16 at 0:12
$begingroup$
@user319635 $alpha$ is the standard basis $(1,x,x^2,x^3)$. And not quite; these are the coordinates of the $beta$-vectors in the standard basis. Thus $beta = (1 + x^2,2 + x + x^2, 2x + x^2)$.
$endgroup$
– Alex Provost
Mar 5 '16 at 0:12
$begingroup$
Right! Thank you so much!
$endgroup$
– user319635
Mar 5 '16 at 0:29
$begingroup$
Right! Thank you so much!
$endgroup$
– user319635
Mar 5 '16 at 0:29
add a comment |
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1
$begingroup$
Hint: Can you find an invertible matrix $B$ such that $BA$ is the rref of $A$?
$endgroup$
– amd
Mar 4 '16 at 19:47
$begingroup$
Thank you! I have figured out B. Is that all that has to be done?
$endgroup$
– user319635
Mar 4 '16 at 20:17
$begingroup$
Oops, I still have to find my bases alpha and beta, how do i go about this?
$endgroup$
– user319635
Mar 4 '16 at 20:27
$begingroup$
Interpret $B$ as a change-of-basis matrix. Note that the solution isn’t unique. You can start with pretty much any basis for $P_3(mathbb R)$ and find a corresponding basis for $P_2(mathbb R)$ so that the matrix of $A$ has the desired form.
$endgroup$
– amd
Mar 4 '16 at 20:37