Mixing
Proving the particular case of change of basis theorem over two matrix without using canonical basis.
Let $A in M_{m times n}(mathbb{F})$, and let $beta$ and $gamma$ ordinated basis for $mathbb{F}^n$ and $mathbb{F}^{m}$ respectively. Let $B=[L_{A}]_{beta} ^{gamma}$ . Show $B=P^{-1}AQ$ where $P in M_{m times m}$ is a matrix such their $j$-th column is the $j$-th vector of $gamma$ and $Q in M_{n times n}$ is a matrix such their $j$-th column is the $j$-th vector of $beta$.
I know $L_{A}:mathbb{F}^n to mathbb{F}^m$ where $L_{A}(v)=Av$ for every $v in mathbb{F}^n$. Then having $beta= lbrace beta_{1}, beta_{2},...,beta_{n} rbrace$ where $beta_{i} in mathbb{F}^{n}$ for each $1 leq i leq n$. We can think each vector $beta_{i}$ as $beta_{i}=(beta_{i_{1}},beta_{i_{2}},...,beta_{i_{n}})$ for every $1 leq i leq n$.
Then as $Q=begin{pmatrix} beta_{1}, beta_{2},....,beta_{n} end{pmatrix}$ thinking the vectors $beta_{i}$ vertically. Calculating $AQ$ we have $AQ=(aq)_{ij}$ where $$aq_{ij}=a_{i_{1}}beta_{j_{1}}+a_{i_{2}}beta_{j_{2}}+...+a_{i_{n}}beta_{j_{n}}$$ where $1 leq i leq m$ and $1 leq j leq n$ Now Im struggled to end up the proof. I wold aprecciate any help proving the statement.
linear-algebra abstract-algebra matrices vector-spaces representation-theory
asked Nov 21 '18 at 22:52
CosCos
1126
add a comment |
Let $A in M_{m times n}(mathbb{F})$, and let $beta$ and $gamma$ ordinated basis for $mathbb{F}^n$ and $mathbb{F}^{m}$ respectively. Let $B=[L_{A}]_{beta} ^{gamma}$ . Show $B=P^{-1}AQ$ where $P in M_{m times m}$ is a matrix such their $j$-th column is the $j$-th vector of $gamma$ and $Q in M_{n times n}$ is a matrix such their $j$-th column is the $j$-th vector of $beta$.
I know $L_{A}:mathbb{F}^n to mathbb{F}^m$ where $L_{A}(v)=Av$ for every $v in mathbb{F}^n$. Then having $beta= lbrace beta_{1}, beta_{2},...,beta_{n} rbrace$ where $beta_{i} in mathbb{F}^{n}$ for each $1 leq i leq n$. We can think each vector $beta_{i}$ as $beta_{i}=(beta_{i_{1}},beta_{i_{2}},...,beta_{i_{n}})$ for every $1 leq i leq n$.
Then as $Q=begin{pmatrix} beta_{1}, beta_{2},....,beta_{n} end{pmatrix}$ thinking the vectors $beta_{i}$ vertically. Calculating $AQ$ we have $AQ=(aq)_{ij}$ where $$aq_{ij}=a_{i_{1}}beta_{j_{1}}+a_{i_{2}}beta_{j_{2}}+...+a_{i_{n}}beta_{j_{n}}$$ where $1 leq i leq m$ and $1 leq j leq n$ Now Im struggled to end up the proof. I wold aprecciate any help proving the statement.
linear-algebra abstract-algebra matrices vector-spaces representation-theory
asked Nov 21 '18 at 22:52
CosCos
1126
I might start by expanding $v$ in terms of $beta$ and $L_A(v)$ in terms of $gamma$ and then regrouping.
– amd
Nov 22 '18 at 0:07
I´ve already tried but I only go in circles trying to prove the equality! @amd
– Cos
Nov 22 '18 at 1:01
add a comment |
Let $A in M_{m times n}(mathbb{F})$, and let $beta$ and $gamma$ ordinated basis for $mathbb{F}^n$ and $mathbb{F}^{m}$ respectively. Let $B=[L_{A}]_{beta} ^{gamma}$ . Show $B=P^{-1}AQ$ where $P in M_{m times m}$ is a matrix such their $j$-th column is the $j$-th vector of $gamma$ and $Q in M_{n times n}$ is a matrix such their $j$-th column is the $j$-th vector of $beta$.
I know $L_{A}:mathbb{F}^n to mathbb{F}^m$ where $L_{A}(v)=Av$ for every $v in mathbb{F}^n$. Then having $beta= lbrace beta_{1}, beta_{2},...,beta_{n} rbrace$ where $beta_{i} in mathbb{F}^{n}$ for each $1 leq i leq n$. We can think each vector $beta_{i}$ as $beta_{i}=(beta_{i_{1}},beta_{i_{2}},...,beta_{i_{n}})$ for every $1 leq i leq n$.
Then as $Q=begin{pmatrix} beta_{1}, beta_{2},....,beta_{n} end{pmatrix}$ thinking the vectors $beta_{i}$ vertically. Calculating $AQ$ we have $AQ=(aq)_{ij}$ where $$aq_{ij}=a_{i_{1}}beta_{j_{1}}+a_{i_{2}}beta_{j_{2}}+...+a_{i_{n}}beta_{j_{n}}$$ where $1 leq i leq m$ and $1 leq j leq n$ Now Im struggled to end up the proof. I wold aprecciate any help proving the statement.
linear-algebra abstract-algebra matrices vector-spaces representation-theory
asked Nov 21 '18 at 22:52
CosCos
1126
Let $A in M_{m times n}(mathbb{F})$, and let $beta$ and $gamma$ ordinated basis for $mathbb{F}^n$ and $mathbb{F}^{m}$ respectively. Let $B=[L_{A}]_{beta} ^{gamma}$ . Show $B=P^{-1}AQ$ where $P in M_{m times m}$ is a matrix such their $j$-th column is the $j$-th vector of $gamma$ and $Q in M_{n times n}$ is a matrix such their $j$-th column is the $j$-th vector of $beta$.
I know $L_{A}:mathbb{F}^n to mathbb{F}^m$ where $L_{A}(v)=Av$ for every $v in mathbb{F}^n$. Then having $beta= lbrace beta_{1}, beta_{2},...,beta_{n} rbrace$ where $beta_{i} in mathbb{F}^{n}$ for each $1 leq i leq n$. We can think each vector $beta_{i}$ as $beta_{i}=(beta_{i_{1}},beta_{i_{2}},...,beta_{i_{n}})$ for every $1 leq i leq n$.
Then as $Q=begin{pmatrix} beta_{1}, beta_{2},....,beta_{n} end{pmatrix}$ thinking the vectors $beta_{i}$ vertically. Calculating $AQ$ we have $AQ=(aq)_{ij}$ where $$aq_{ij}=a_{i_{1}}beta_{j_{1}}+a_{i_{2}}beta_{j_{2}}+...+a_{i_{n}}beta_{j_{n}}$$ where $1 leq i leq m$ and $1 leq j leq n$ Now Im struggled to end up the proof. I wold aprecciate any help proving the statement.
linear-algebra abstract-algebra matrices vector-spaces representation-theory
linear-algebra abstract-algebra matrices vector-spaces representation-theory
asked Nov 21 '18 at 22:52
CosCos
1126
asked Nov 21 '18 at 22:52
CosCos
1126
asked Nov 21 '18 at 22:52
CosCos
1126
asked Nov 21 '18 at 22:52
CosCos
1126
asked Nov 21 '18 at 22:52
CosCos
1126
1126
I might start by expanding $v$ in terms of $beta$ and $L_A(v)$ in terms of $gamma$ and then regrouping.
– amd
Nov 22 '18 at 0:07
I´ve already tried but I only go in circles trying to prove the equality! @amd
– Cos
Nov 22 '18 at 1:01
add a comment |
I might start by expanding $v$ in terms of $beta$ and $L_A(v)$ in terms of $gamma$ and then regrouping.
– amd
Nov 22 '18 at 0:07
I´ve already tried but I only go in circles trying to prove the equality! @amd
– Cos
Nov 22 '18 at 1:01
I might start by expanding $v$ in terms of $beta$ and $L_A(v)$ in terms of $gamma$ and then regrouping.
– amd
Nov 22 '18 at 0:07
I might start by expanding $v$ in terms of $beta$ and $L_A(v)$ in terms of $gamma$ and then regrouping.
– amd
Nov 22 '18 at 0:07
I´ve already tried but I only go in circles trying to prove the equality! @amd
– Cos
Nov 22 '18 at 1:01
I´ve already tried but I only go in circles trying to prove the equality! @amd
– Cos
Nov 22 '18 at 1:01
add a comment |
1 Answer
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Expanding in terms of $beta$, we have $$v = sum_{k=1}^n v_kbeta_k = Qbegin{bmatrix}v_1\vdots\v_nend{bmatrix} = Q[v]_beta$$ so $$L_A(v) = Av = AQ[v]_beta.$$ On the other hand, we can expand $L_A(v)$ in terms of $gamma$: $$L_A(v) = sum_{k=1}^m w_kgamma_k = Pbegin{bmatrix}w_1\vdots\w_mend{bmatrix} = P[L_A(v)]_gamma.$$ Equating and multiplying both sides by $P^{-1}$ we get $$[L_A(v)]_gamma = P^{-1}AQ[v]_beta,$$ therefore $[L_A]_beta^gamma = P^{-1}AQ$.
answered Nov 22 '18 at 1:17
amdamd
29.3k21050
BTW, I prefer to put the “input” basis to a matrix as the superscript so that it “cancels” with the subscript to the right. E.g., the change-of-basis matrix from $beta$ to $gamma$ is $[operatorname{id}]_gamma^beta$ so that $[v]_gamma = [operatorname{id}]_gamma^beta [v]_beta$.
– amd
Nov 22 '18 at 1:19
Thank you so much! Very well explained.
– Cos
Nov 22 '18 at 4:36
add a comment |
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Expanding in terms of $beta$, we have $$v = sum_{k=1}^n v_kbeta_k = Qbegin{bmatrix}v_1\vdots\v_nend{bmatrix} = Q[v]_beta$$ so $$L_A(v) = Av = AQ[v]_beta.$$ On the other hand, we can expand $L_A(v)$ in terms of $gamma$: $$L_A(v) = sum_{k=1}^m w_kgamma_k = Pbegin{bmatrix}w_1\vdots\w_mend{bmatrix} = P[L_A(v)]_gamma.$$ Equating and multiplying both sides by $P^{-1}$ we get $$[L_A(v)]_gamma = P^{-1}AQ[v]_beta,$$ therefore $[L_A]_beta^gamma = P^{-1}AQ$.
answered Nov 22 '18 at 1:17
amdamd
29.3k21050
BTW, I prefer to put the “input” basis to a matrix as the superscript so that it “cancels” with the subscript to the right. E.g., the change-of-basis matrix from $beta$ to $gamma$ is $[operatorname{id}]_gamma^beta$ so that $[v]_gamma = [operatorname{id}]_gamma^beta [v]_beta$.
– amd
Nov 22 '18 at 1:19
Thank you so much! Very well explained.
– Cos
Nov 22 '18 at 4:36
add a comment |
Expanding in terms of $beta$, we have $$v = sum_{k=1}^n v_kbeta_k = Qbegin{bmatrix}v_1\vdots\v_nend{bmatrix} = Q[v]_beta$$ so $$L_A(v) = Av = AQ[v]_beta.$$ On the other hand, we can expand $L_A(v)$ in terms of $gamma$: $$L_A(v) = sum_{k=1}^m w_kgamma_k = Pbegin{bmatrix}w_1\vdots\w_mend{bmatrix} = P[L_A(v)]_gamma.$$ Equating and multiplying both sides by $P^{-1}$ we get $$[L_A(v)]_gamma = P^{-1}AQ[v]_beta,$$ therefore $[L_A]_beta^gamma = P^{-1}AQ$.
answered Nov 22 '18 at 1:17
amdamd
29.3k21050
BTW, I prefer to put the “input” basis to a matrix as the superscript so that it “cancels” with the subscript to the right. E.g., the change-of-basis matrix from $beta$ to $gamma$ is $[operatorname{id}]_gamma^beta$ so that $[v]_gamma = [operatorname{id}]_gamma^beta [v]_beta$.
– amd
Nov 22 '18 at 1:19
Thank you so much! Very well explained.
– Cos
Nov 22 '18 at 4:36
add a comment |
Expanding in terms of $beta$, we have $$v = sum_{k=1}^n v_kbeta_k = Qbegin{bmatrix}v_1\vdots\v_nend{bmatrix} = Q[v]_beta$$ so $$L_A(v) = Av = AQ[v]_beta.$$ On the other hand, we can expand $L_A(v)$ in terms of $gamma$: $$L_A(v) = sum_{k=1}^m w_kgamma_k = Pbegin{bmatrix}w_1\vdots\w_mend{bmatrix} = P[L_A(v)]_gamma.$$ Equating and multiplying both sides by $P^{-1}$ we get $$[L_A(v)]_gamma = P^{-1}AQ[v]_beta,$$ therefore $[L_A]_beta^gamma = P^{-1}AQ$.
answered Nov 22 '18 at 1:17
amdamd
29.3k21050
Expanding in terms of $beta$, we have $$v = sum_{k=1}^n v_kbeta_k = Qbegin{bmatrix}v_1\vdots\v_nend{bmatrix} = Q[v]_beta$$ so $$L_A(v) = Av = AQ[v]_beta.$$ On the other hand, we can expand $L_A(v)$ in terms of $gamma$: $$L_A(v) = sum_{k=1}^m w_kgamma_k = Pbegin{bmatrix}w_1\vdots\w_mend{bmatrix} = P[L_A(v)]_gamma.$$ Equating and multiplying both sides by $P^{-1}$ we get $$[L_A(v)]_gamma = P^{-1}AQ[v]_beta,$$ therefore $[L_A]_beta^gamma = P^{-1}AQ$.
answered Nov 22 '18 at 1:17
amdamd
29.3k21050
answered Nov 22 '18 at 1:17
amdamd
29.3k21050
answered Nov 22 '18 at 1:17
amdamd
29.3k21050
answered Nov 22 '18 at 1:17
amdamd
29.3k21050
29.3k21050
BTW, I prefer to put the “input” basis to a matrix as the superscript so that it “cancels” with the subscript to the right. E.g., the change-of-basis matrix from $beta$ to $gamma$ is $[operatorname{id}]_gamma^beta$ so that $[v]_gamma = [operatorname{id}]_gamma^beta [v]_beta$.
– amd
Nov 22 '18 at 1:19
Thank you so much! Very well explained.
– Cos
Nov 22 '18 at 4:36
add a comment |
BTW, I prefer to put the “input” basis to a matrix as the superscript so that it “cancels” with the subscript to the right. E.g., the change-of-basis matrix from $beta$ to $gamma$ is $[operatorname{id}]_gamma^beta$ so that $[v]_gamma = [operatorname{id}]_gamma^beta [v]_beta$.
– amd
Nov 22 '18 at 1:19
Thank you so much! Very well explained.
– Cos
Nov 22 '18 at 4:36
BTW, I prefer to put the “input” basis to a matrix as the superscript so that it “cancels” with the subscript to the right. E.g., the change-of-basis matrix from $beta$ to $gamma$ is $[operatorname{id}]_gamma^beta$ so that $[v]_gamma = [operatorname{id}]_gamma^beta [v]_beta$.
– amd
Nov 22 '18 at 1:19
BTW, I prefer to put the “input” basis to a matrix as the superscript so that it “cancels” with the subscript to the right. E.g., the change-of-basis matrix from $beta$ to $gamma$ is $[operatorname{id}]_gamma^beta$ so that $[v]_gamma = [operatorname{id}]_gamma^beta [v]_beta$.
– amd
Nov 22 '18 at 1:19
Thank you so much! Very well explained.
– Cos
Nov 22 '18 at 4:36
Thank you so much! Very well explained.
– Cos
Nov 22 '18 at 4:36
add a comment |
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I might start by expanding $v$ in terms of $beta$ and $L_A(v)$ in terms of $gamma$ and then regrouping.
– amd
Nov 22 '18 at 0:07
I´ve already tried but I only go in circles trying to prove the equality! @amd
– Cos
Nov 22 '18 at 1:01