Finding linear transformation from matrix
$begingroup$
I tried solving this question by taking given matrix $[A]_{B_{1}}$ = $[T]_{B_{2}}$. The answer that I am getting is $(35,-10)$. Tried solving several times and still reaching at the same answer. Which is the right way to solve this question?
linear-algebra matrices linear-transformations change-of-basis
asked Jan 1 at 9:45
user46697user46697
215211
$endgroup$
add a comment |
$begingroup$
I tried solving this question by taking given matrix $[A]_{B_{1}}$ = $[T]_{B_{2}}$. The answer that I am getting is $(35,-10)$. Tried solving several times and still reaching at the same answer. Which is the right way to solve this question?
linear-algebra matrices linear-transformations change-of-basis
asked Jan 1 at 9:45
user46697user46697
215211
$endgroup$
add a comment |
$begingroup$
I tried solving this question by taking given matrix $[A]_{B_{1}}$ = $[T]_{B_{2}}$. The answer that I am getting is $(35,-10)$. Tried solving several times and still reaching at the same answer. Which is the right way to solve this question?
linear-algebra matrices linear-transformations change-of-basis
asked Jan 1 at 9:45
user46697user46697
215211
$endgroup$
I tried solving this question by taking given matrix $[A]_{B_{1}}$ = $[T]_{B_{2}}$. The answer that I am getting is $(35,-10)$. Tried solving several times and still reaching at the same answer. Which is the right way to solve this question?
linear-algebra matrices linear-transformations change-of-basis
linear-algebra matrices linear-transformations change-of-basis
asked Jan 1 at 9:45
user46697user46697
215211
asked Jan 1 at 9:45
user46697user46697
215211
asked Jan 1 at 9:45
user46697user46697
215211
asked Jan 1 at 9:45
user46697user46697
215211
asked Jan 1 at 9:45
user46697user46697
215211
215211
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First, you write $(5,5)$ as a linear combination of $(1,2)$ and $(2,-1)$. It turns out that $(5,5)=3(1,2)+(2,-1)$. So, now you know thatbegin{align}T(5,5)&=3T(1,2)+T(2,-1)\&=3(4,2)+(3,-4)\&=(15,2).end{align}
answered Jan 1 at 9:53
José Carlos SantosJosé Carlos Santos
164k22132235
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add a comment |
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Hey that Matrix $[T]_{mathcal{B_1}}^{mathcal{B_2}}$ will take input in Basis $mathcal{B_1}$ and spit the output in Basis $mathcal{B_2}$ .Therefore you cannot feed the input in our standard basis.
Convert $(5,5)$ into basis $mathcal{B_1}$ and then feed.
W.K.T $(5,5)=3(1,2)+1(2,-1)$ There fore putting in the matrix $(3,1)$ gives us $(15,2)$ as output convert this into our standard basis which is $mathcal{B_2}$.
Therefore Option D is correct
answered Jan 1 at 10:01
SundarNarasimhanSundarNarasimhan
234
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add a comment |
$begingroup$
Write $;(5,5);$ as a linear combination of $;B_1;$ , apply $;T;$ and then this last write as a lin. comb. of $;B_2;$ (though this last is the usual, canonical basis so you won't have to do anything!):
$$binom55=3binom21+1binom2{-1}implies left[binom 55right]_{B_1}=binom31$$
so now
$$Tleft[binom 55right]_{B_1}=begin{pmatrix}4&3\2&!-4end{pmatrix}binom31=binom{15}2$$
answered Jan 1 at 9:54
DonAntonioDonAntonio
179k1494230
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, you write $(5,5)$ as a linear combination of $(1,2)$ and $(2,-1)$. It turns out that $(5,5)=3(1,2)+(2,-1)$. So, now you know thatbegin{align}T(5,5)&=3T(1,2)+T(2,-1)\&=3(4,2)+(3,-4)\&=(15,2).end{align}
answered Jan 1 at 9:53
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
add a comment |
$begingroup$
First, you write $(5,5)$ as a linear combination of $(1,2)$ and $(2,-1)$. It turns out that $(5,5)=3(1,2)+(2,-1)$. So, now you know thatbegin{align}T(5,5)&=3T(1,2)+T(2,-1)\&=3(4,2)+(3,-4)\&=(15,2).end{align}
answered Jan 1 at 9:53
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
add a comment |
$begingroup$
First, you write $(5,5)$ as a linear combination of $(1,2)$ and $(2,-1)$. It turns out that $(5,5)=3(1,2)+(2,-1)$. So, now you know thatbegin{align}T(5,5)&=3T(1,2)+T(2,-1)\&=3(4,2)+(3,-4)\&=(15,2).end{align}
answered Jan 1 at 9:53
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
First, you write $(5,5)$ as a linear combination of $(1,2)$ and $(2,-1)$. It turns out that $(5,5)=3(1,2)+(2,-1)$. So, now you know thatbegin{align}T(5,5)&=3T(1,2)+T(2,-1)\&=3(4,2)+(3,-4)\&=(15,2).end{align}
answered Jan 1 at 9:53
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 1 at 9:53
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 1 at 9:53
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 1 at 9:53
José Carlos SantosJosé Carlos Santos
164k22132235
164k22132235
add a comment |
add a comment |
$begingroup$
Hey that Matrix $[T]_{mathcal{B_1}}^{mathcal{B_2}}$ will take input in Basis $mathcal{B_1}$ and spit the output in Basis $mathcal{B_2}$ .Therefore you cannot feed the input in our standard basis.
Convert $(5,5)$ into basis $mathcal{B_1}$ and then feed.
W.K.T $(5,5)=3(1,2)+1(2,-1)$ There fore putting in the matrix $(3,1)$ gives us $(15,2)$ as output convert this into our standard basis which is $mathcal{B_2}$.
Therefore Option D is correct
answered Jan 1 at 10:01
SundarNarasimhanSundarNarasimhan
234
$endgroup$
add a comment |
$begingroup$
Hey that Matrix $[T]_{mathcal{B_1}}^{mathcal{B_2}}$ will take input in Basis $mathcal{B_1}$ and spit the output in Basis $mathcal{B_2}$ .Therefore you cannot feed the input in our standard basis.
Convert $(5,5)$ into basis $mathcal{B_1}$ and then feed.
W.K.T $(5,5)=3(1,2)+1(2,-1)$ There fore putting in the matrix $(3,1)$ gives us $(15,2)$ as output convert this into our standard basis which is $mathcal{B_2}$.
Therefore Option D is correct
answered Jan 1 at 10:01
SundarNarasimhanSundarNarasimhan
234
$endgroup$
add a comment |
$begingroup$
Hey that Matrix $[T]_{mathcal{B_1}}^{mathcal{B_2}}$ will take input in Basis $mathcal{B_1}$ and spit the output in Basis $mathcal{B_2}$ .Therefore you cannot feed the input in our standard basis.
Convert $(5,5)$ into basis $mathcal{B_1}$ and then feed.
W.K.T $(5,5)=3(1,2)+1(2,-1)$ There fore putting in the matrix $(3,1)$ gives us $(15,2)$ as output convert this into our standard basis which is $mathcal{B_2}$.
Therefore Option D is correct
answered Jan 1 at 10:01
SundarNarasimhanSundarNarasimhan
234
$endgroup$
Hey that Matrix $[T]_{mathcal{B_1}}^{mathcal{B_2}}$ will take input in Basis $mathcal{B_1}$ and spit the output in Basis $mathcal{B_2}$ .Therefore you cannot feed the input in our standard basis.
Convert $(5,5)$ into basis $mathcal{B_1}$ and then feed.
W.K.T $(5,5)=3(1,2)+1(2,-1)$ There fore putting in the matrix $(3,1)$ gives us $(15,2)$ as output convert this into our standard basis which is $mathcal{B_2}$.
Therefore Option D is correct
answered Jan 1 at 10:01
SundarNarasimhanSundarNarasimhan
234
answered Jan 1 at 10:01
SundarNarasimhanSundarNarasimhan
234
answered Jan 1 at 10:01
SundarNarasimhanSundarNarasimhan
234
answered Jan 1 at 10:01
SundarNarasimhanSundarNarasimhan
234
234
add a comment |
add a comment |
$begingroup$
Write $;(5,5);$ as a linear combination of $;B_1;$ , apply $;T;$ and then this last write as a lin. comb. of $;B_2;$ (though this last is the usual, canonical basis so you won't have to do anything!):
$$binom55=3binom21+1binom2{-1}implies left[binom 55right]_{B_1}=binom31$$
so now
$$Tleft[binom 55right]_{B_1}=begin{pmatrix}4&3\2&!-4end{pmatrix}binom31=binom{15}2$$
answered Jan 1 at 9:54
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Write $;(5,5);$ as a linear combination of $;B_1;$ , apply $;T;$ and then this last write as a lin. comb. of $;B_2;$ (though this last is the usual, canonical basis so you won't have to do anything!):
$$binom55=3binom21+1binom2{-1}implies left[binom 55right]_{B_1}=binom31$$
so now
$$Tleft[binom 55right]_{B_1}=begin{pmatrix}4&3\2&!-4end{pmatrix}binom31=binom{15}2$$
answered Jan 1 at 9:54
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Write $;(5,5);$ as a linear combination of $;B_1;$ , apply $;T;$ and then this last write as a lin. comb. of $;B_2;$ (though this last is the usual, canonical basis so you won't have to do anything!):
$$binom55=3binom21+1binom2{-1}implies left[binom 55right]_{B_1}=binom31$$
so now
$$Tleft[binom 55right]_{B_1}=begin{pmatrix}4&3\2&!-4end{pmatrix}binom31=binom{15}2$$
answered Jan 1 at 9:54
DonAntonioDonAntonio
179k1494230
$endgroup$
Write $;(5,5);$ as a linear combination of $;B_1;$ , apply $;T;$ and then this last write as a lin. comb. of $;B_2;$ (though this last is the usual, canonical basis so you won't have to do anything!):
$$binom55=3binom21+1binom2{-1}implies left[binom 55right]_{B_1}=binom31$$
so now
$$Tleft[binom 55right]_{B_1}=begin{pmatrix}4&3\2&!-4end{pmatrix}binom31=binom{15}2$$
answered Jan 1 at 9:54
DonAntonioDonAntonio
179k1494230
edited Jan 1 at 15:23
answered Jan 1 at 9:54
DonAntonioDonAntonio
179k1494230
answered Jan 1 at 9:54
DonAntonioDonAntonio
179k1494230
answered Jan 1 at 9:54
DonAntonioDonAntonio
179k1494230
179k1494230
add a comment |
add a comment |
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