Find the matrix representation of $psivarphi^{-2}+2varphi+I$
$begingroup$
Let $varphi$ and $psi$ be a linear transformation in vector space $V$, and given the inverse linear transformation of $varphi$ exists, and the matrix representation of $varphi$ and $psi$ on the first basis of $V$ to be matrices $A$ and $B$, respectively, the transition matrix of $V$ from the first basis to second basis is $P$. Find the matrix representation $psivarphi^{-2}+2varphi+I$ (where $I$ is the identity transformation on $V$) on the second basis of $V$.
I should find the matrix representation of $psivarphi^{-2}+2varphi+I$ separately, i.e. find the martix representation of $psivarphi^{-2}$, $2varphi$ and $I$, respectively, by linearity property. What makes me feel difficulty is to find the matrix representation of $psivarphi^{-2}$, any idea to deal with it?
[Transition matrix in here refers to the matrix associated with a change of basis for a vector space.(Source: Wikipedia)]
The answer given for this question is $P^{-1}BA^{-2}P+2P^{-1}AP+I_{n}$.
Thanks in advance!
linear-transformations
asked Dec 3 '18 at 14:33
weilam06weilam06
14511
$endgroup$
add a comment |
$begingroup$
Let $varphi$ and $psi$ be a linear transformation in vector space $V$, and given the inverse linear transformation of $varphi$ exists, and the matrix representation of $varphi$ and $psi$ on the first basis of $V$ to be matrices $A$ and $B$, respectively, the transition matrix of $V$ from the first basis to second basis is $P$. Find the matrix representation $psivarphi^{-2}+2varphi+I$ (where $I$ is the identity transformation on $V$) on the second basis of $V$.
I should find the matrix representation of $psivarphi^{-2}+2varphi+I$ separately, i.e. find the martix representation of $psivarphi^{-2}$, $2varphi$ and $I$, respectively, by linearity property. What makes me feel difficulty is to find the matrix representation of $psivarphi^{-2}$, any idea to deal with it?
[Transition matrix in here refers to the matrix associated with a change of basis for a vector space.(Source: Wikipedia)]
The answer given for this question is $P^{-1}BA^{-2}P+2P^{-1}AP+I_{n}$.
Thanks in advance!
linear-transformations
asked Dec 3 '18 at 14:33
weilam06weilam06
14511
$endgroup$
1
$begingroup$
Note that for any matrices $A$ and $B$ you have $(P^{-1} A P) (P^{-1} B P) = P^{-1} A (P P^{-1}) B P = P^{-1} (AB) P$; i.e., it doesn't matter if you change the basis of matrices before or after multiplying them. Does this help?
$endgroup$
– Connor Harris
Dec 3 '18 at 14:39
$begingroup$
Nope, I am struggling in the inverse linear transformation of $varphi$ and its matrix representation on second basis of $V$.
$endgroup$
– weilam06
Dec 3 '18 at 14:56
add a comment |
$begingroup$
Let $varphi$ and $psi$ be a linear transformation in vector space $V$, and given the inverse linear transformation of $varphi$ exists, and the matrix representation of $varphi$ and $psi$ on the first basis of $V$ to be matrices $A$ and $B$, respectively, the transition matrix of $V$ from the first basis to second basis is $P$. Find the matrix representation $psivarphi^{-2}+2varphi+I$ (where $I$ is the identity transformation on $V$) on the second basis of $V$.
I should find the matrix representation of $psivarphi^{-2}+2varphi+I$ separately, i.e. find the martix representation of $psivarphi^{-2}$, $2varphi$ and $I$, respectively, by linearity property. What makes me feel difficulty is to find the matrix representation of $psivarphi^{-2}$, any idea to deal with it?
[Transition matrix in here refers to the matrix associated with a change of basis for a vector space.(Source: Wikipedia)]
The answer given for this question is $P^{-1}BA^{-2}P+2P^{-1}AP+I_{n}$.
Thanks in advance!
linear-transformations
asked Dec 3 '18 at 14:33
weilam06weilam06
14511
$endgroup$
Let $varphi$ and $psi$ be a linear transformation in vector space $V$, and given the inverse linear transformation of $varphi$ exists, and the matrix representation of $varphi$ and $psi$ on the first basis of $V$ to be matrices $A$ and $B$, respectively, the transition matrix of $V$ from the first basis to second basis is $P$. Find the matrix representation $psivarphi^{-2}+2varphi+I$ (where $I$ is the identity transformation on $V$) on the second basis of $V$.
I should find the matrix representation of $psivarphi^{-2}+2varphi+I$ separately, i.e. find the martix representation of $psivarphi^{-2}$, $2varphi$ and $I$, respectively, by linearity property. What makes me feel difficulty is to find the matrix representation of $psivarphi^{-2}$, any idea to deal with it?
[Transition matrix in here refers to the matrix associated with a change of basis for a vector space.(Source: Wikipedia)]
The answer given for this question is $P^{-1}BA^{-2}P+2P^{-1}AP+I_{n}$.
Thanks in advance!
linear-transformations
linear-transformations
asked Dec 3 '18 at 14:33
weilam06weilam06
14511
asked Dec 3 '18 at 14:33
weilam06weilam06
14511
asked Dec 3 '18 at 14:33
weilam06weilam06
14511
asked Dec 3 '18 at 14:33
weilam06weilam06
14511
asked Dec 3 '18 at 14:33
weilam06weilam06
14511
14511
1
$begingroup$
Note that for any matrices $A$ and $B$ you have $(P^{-1} A P) (P^{-1} B P) = P^{-1} A (P P^{-1}) B P = P^{-1} (AB) P$; i.e., it doesn't matter if you change the basis of matrices before or after multiplying them. Does this help?
$endgroup$
– Connor Harris
Dec 3 '18 at 14:39
$begingroup$
Nope, I am struggling in the inverse linear transformation of $varphi$ and its matrix representation on second basis of $V$.
$endgroup$
– weilam06
Dec 3 '18 at 14:56
add a comment |
1
$begingroup$
Note that for any matrices $A$ and $B$ you have $(P^{-1} A P) (P^{-1} B P) = P^{-1} A (P P^{-1}) B P = P^{-1} (AB) P$; i.e., it doesn't matter if you change the basis of matrices before or after multiplying them. Does this help?
$endgroup$
– Connor Harris
Dec 3 '18 at 14:39
$begingroup$
Nope, I am struggling in the inverse linear transformation of $varphi$ and its matrix representation on second basis of $V$.
$endgroup$
– weilam06
Dec 3 '18 at 14:56
1
1
$begingroup$
Note that for any matrices $A$ and $B$ you have $(P^{-1} A P) (P^{-1} B P) = P^{-1} A (P P^{-1}) B P = P^{-1} (AB) P$; i.e., it doesn't matter if you change the basis of matrices before or after multiplying them. Does this help?
$endgroup$
– Connor Harris
Dec 3 '18 at 14:39
$begingroup$
Note that for any matrices $A$ and $B$ you have $(P^{-1} A P) (P^{-1} B P) = P^{-1} A (P P^{-1}) B P = P^{-1} (AB) P$; i.e., it doesn't matter if you change the basis of matrices before or after multiplying them. Does this help?
$endgroup$
– Connor Harris
Dec 3 '18 at 14:39
$begingroup$
Nope, I am struggling in the inverse linear transformation of $varphi$ and its matrix representation on second basis of $V$.
$endgroup$
– weilam06
Dec 3 '18 at 14:56
$begingroup$
Nope, I am struggling in the inverse linear transformation of $varphi$ and its matrix representation on second basis of $V$.
$endgroup$
– weilam06
Dec 3 '18 at 14:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
remember that, if the inverse exist, $(AB)^{-1}=B^{-1}A^{-1}$,
so also:
$(P^{-1}AP)^{-1}=P^{-1}A^{-1}P$
and note that:
$(P^{-1}MP)^{2}=(P^{-1}MP)(P^{-1}MP)=P^{-1}M^2P$
answered Dec 3 '18 at 15:02
Emilio NovatiEmilio Novati
52.1k43474
$endgroup$
$begingroup$
But $(P^{-1}BP)(P^{-1}AP)^{-2}=P^{-1}BP^{-1}A^{-2}P^{2}$.
$endgroup$
– weilam06
Dec 3 '18 at 15:06
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No it doesn't. $(P^{-1}XP)^k=P^{-1}X^k P$.
$endgroup$
– ancientmathematician
Dec 3 '18 at 15:07
$begingroup$
Oops, I know what's my mistake here. $(P^{-1}AP)^{k}=(P^{-1}AP)(P^{-1}AP)cdots(P^{-1}AP)=P^{-1}A^{k}P$ works here.
$endgroup$
– weilam06
Dec 3 '18 at 15:08
$begingroup$
I added to my hint another hint....:)
$endgroup$
– Emilio Novati
Dec 3 '18 at 15:11
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Hint:
remember that, if the inverse exist, $(AB)^{-1}=B^{-1}A^{-1}$,
so also:
$(P^{-1}AP)^{-1}=P^{-1}A^{-1}P$
and note that:
$(P^{-1}MP)^{2}=(P^{-1}MP)(P^{-1}MP)=P^{-1}M^2P$
answered Dec 3 '18 at 15:02
Emilio NovatiEmilio Novati
52.1k43474
$endgroup$
$begingroup$
But $(P^{-1}BP)(P^{-1}AP)^{-2}=P^{-1}BP^{-1}A^{-2}P^{2}$.
$endgroup$
– weilam06
Dec 3 '18 at 15:06
$begingroup$
No it doesn't. $(P^{-1}XP)^k=P^{-1}X^k P$.
$endgroup$
– ancientmathematician
Dec 3 '18 at 15:07
$begingroup$
Oops, I know what's my mistake here. $(P^{-1}AP)^{k}=(P^{-1}AP)(P^{-1}AP)cdots(P^{-1}AP)=P^{-1}A^{k}P$ works here.
$endgroup$
– weilam06
Dec 3 '18 at 15:08
$begingroup$
I added to my hint another hint....:)
$endgroup$
– Emilio Novati
Dec 3 '18 at 15:11
add a comment |
$begingroup$
Hint:
remember that, if the inverse exist, $(AB)^{-1}=B^{-1}A^{-1}$,
so also:
$(P^{-1}AP)^{-1}=P^{-1}A^{-1}P$
and note that:
$(P^{-1}MP)^{2}=(P^{-1}MP)(P^{-1}MP)=P^{-1}M^2P$
answered Dec 3 '18 at 15:02
Emilio NovatiEmilio Novati
52.1k43474
$endgroup$
$begingroup$
But $(P^{-1}BP)(P^{-1}AP)^{-2}=P^{-1}BP^{-1}A^{-2}P^{2}$.
$endgroup$
– weilam06
Dec 3 '18 at 15:06
$begingroup$
No it doesn't. $(P^{-1}XP)^k=P^{-1}X^k P$.
$endgroup$
– ancientmathematician
Dec 3 '18 at 15:07
$begingroup$
Oops, I know what's my mistake here. $(P^{-1}AP)^{k}=(P^{-1}AP)(P^{-1}AP)cdots(P^{-1}AP)=P^{-1}A^{k}P$ works here.
$endgroup$
– weilam06
Dec 3 '18 at 15:08
$begingroup$
I added to my hint another hint....:)
$endgroup$
– Emilio Novati
Dec 3 '18 at 15:11
add a comment |
$begingroup$
Hint:
remember that, if the inverse exist, $(AB)^{-1}=B^{-1}A^{-1}$,
so also:
$(P^{-1}AP)^{-1}=P^{-1}A^{-1}P$
and note that:
$(P^{-1}MP)^{2}=(P^{-1}MP)(P^{-1}MP)=P^{-1}M^2P$
answered Dec 3 '18 at 15:02
Emilio NovatiEmilio Novati
52.1k43474
$endgroup$
Hint:
remember that, if the inverse exist, $(AB)^{-1}=B^{-1}A^{-1}$,
so also:
$(P^{-1}AP)^{-1}=P^{-1}A^{-1}P$
and note that:
$(P^{-1}MP)^{2}=(P^{-1}MP)(P^{-1}MP)=P^{-1}M^2P$
answered Dec 3 '18 at 15:02
Emilio NovatiEmilio Novati
52.1k43474
edited Dec 3 '18 at 15:10
answered Dec 3 '18 at 15:02
Emilio NovatiEmilio Novati
52.1k43474
answered Dec 3 '18 at 15:02
Emilio NovatiEmilio Novati
52.1k43474
answered Dec 3 '18 at 15:02
Emilio NovatiEmilio Novati
52.1k43474
52.1k43474
$begingroup$
But $(P^{-1}BP)(P^{-1}AP)^{-2}=P^{-1}BP^{-1}A^{-2}P^{2}$.
$endgroup$
– weilam06
Dec 3 '18 at 15:06
$begingroup$
No it doesn't. $(P^{-1}XP)^k=P^{-1}X^k P$.
$endgroup$
– ancientmathematician
Dec 3 '18 at 15:07
$begingroup$
Oops, I know what's my mistake here. $(P^{-1}AP)^{k}=(P^{-1}AP)(P^{-1}AP)cdots(P^{-1}AP)=P^{-1}A^{k}P$ works here.
$endgroup$
– weilam06
Dec 3 '18 at 15:08
$begingroup$
I added to my hint another hint....:)
$endgroup$
– Emilio Novati
Dec 3 '18 at 15:11
add a comment |
$begingroup$
But $(P^{-1}BP)(P^{-1}AP)^{-2}=P^{-1}BP^{-1}A^{-2}P^{2}$.
$endgroup$
– weilam06
Dec 3 '18 at 15:06
$begingroup$
No it doesn't. $(P^{-1}XP)^k=P^{-1}X^k P$.
$endgroup$
– ancientmathematician
Dec 3 '18 at 15:07
$begingroup$
Oops, I know what's my mistake here. $(P^{-1}AP)^{k}=(P^{-1}AP)(P^{-1}AP)cdots(P^{-1}AP)=P^{-1}A^{k}P$ works here.
$endgroup$
– weilam06
Dec 3 '18 at 15:08
$begingroup$
I added to my hint another hint....:)
$endgroup$
– Emilio Novati
Dec 3 '18 at 15:11
$begingroup$
But $(P^{-1}BP)(P^{-1}AP)^{-2}=P^{-1}BP^{-1}A^{-2}P^{2}$.
$endgroup$
– weilam06
Dec 3 '18 at 15:06
$begingroup$
But $(P^{-1}BP)(P^{-1}AP)^{-2}=P^{-1}BP^{-1}A^{-2}P^{2}$.
$endgroup$
– weilam06
Dec 3 '18 at 15:06
$begingroup$
No it doesn't. $(P^{-1}XP)^k=P^{-1}X^k P$.
$endgroup$
– ancientmathematician
Dec 3 '18 at 15:07
$begingroup$
No it doesn't. $(P^{-1}XP)^k=P^{-1}X^k P$.
$endgroup$
– ancientmathematician
Dec 3 '18 at 15:07
$begingroup$
Oops, I know what's my mistake here. $(P^{-1}AP)^{k}=(P^{-1}AP)(P^{-1}AP)cdots(P^{-1}AP)=P^{-1}A^{k}P$ works here.
$endgroup$
– weilam06
Dec 3 '18 at 15:08
$begingroup$
Oops, I know what's my mistake here. $(P^{-1}AP)^{k}=(P^{-1}AP)(P^{-1}AP)cdots(P^{-1}AP)=P^{-1}A^{k}P$ works here.
$endgroup$
– weilam06
Dec 3 '18 at 15:08
$begingroup$
I added to my hint another hint....:)
$endgroup$
– Emilio Novati
Dec 3 '18 at 15:11
$begingroup$
I added to my hint another hint....:)
$endgroup$
– Emilio Novati
Dec 3 '18 at 15:11
add a comment |
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$begingroup$
Note that for any matrices $A$ and $B$ you have $(P^{-1} A P) (P^{-1} B P) = P^{-1} A (P P^{-1}) B P = P^{-1} (AB) P$; i.e., it doesn't matter if you change the basis of matrices before or after multiplying them. Does this help?
$endgroup$
– Connor Harris
Dec 3 '18 at 14:39
$begingroup$
Nope, I am struggling in the inverse linear transformation of $varphi$ and its matrix representation on second basis of $V$.
$endgroup$
– weilam06
Dec 3 '18 at 14:56