Linear transformation $ R^3 to R^3$

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Let $T:R^3 to R^3$ be the linear transformation of projection onto $x_1x_2$-plane. What is the linear transformation one obtains when you compose $T$ with itself?



I think $x_1x_2$ means projecting onto a two dimensional $XY$ plane. But what does it mean to compose $T$ with itself? is it the composition of $T$? Then I assume the composition of a linear transformation should be $R^3$ as well, since $T$ is not specified here, I am a little confused whether $T$'s composition is itself?



Any explanation would be appreciated thanks.




matrices linear-transformations projection




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    You have $T(x_1, x_2, x_3) = (x_1, x_2, 0)$. Then, you would then apply $T$ again: $T(x_1, x_2, 0) = $?.
    – Joe Johnson 126
    Jul 24 at 11:08











  • What is your definition of aprojection? There are many projections onto the $x$-$y$ plane. Which one are you talking about?
    – amd
    Jul 24 at 19:08














up vote
1
down vote

favorite
1












Let $T:R^3 to R^3$ be the linear transformation of projection onto $x_1x_2$-plane. What is the linear transformation one obtains when you compose $T$ with itself?



I think $x_1x_2$ means projecting onto a two dimensional $XY$ plane. But what does it mean to compose $T$ with itself? is it the composition of $T$? Then I assume the composition of a linear transformation should be $R^3$ as well, since $T$ is not specified here, I am a little confused whether $T$'s composition is itself?



Any explanation would be appreciated thanks.




matrices linear-transformations projection




share|cite|improve this question

















  • 3




    You have $T(x_1, x_2, x_3) = (x_1, x_2, 0)$. Then, you would then apply $T$ again: $T(x_1, x_2, 0) = $?.
    – Joe Johnson 126
    Jul 24 at 11:08











  • What is your definition of aprojection? There are many projections onto the $x$-$y$ plane. Which one are you talking about?
    – amd
    Jul 24 at 19:08












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
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Let $T:R^3 to R^3$ be the linear transformation of projection onto $x_1x_2$-plane. What is the linear transformation one obtains when you compose $T$ with itself?



I think $x_1x_2$ means projecting onto a two dimensional $XY$ plane. But what does it mean to compose $T$ with itself? is it the composition of $T$? Then I assume the composition of a linear transformation should be $R^3$ as well, since $T$ is not specified here, I am a little confused whether $T$'s composition is itself?



Any explanation would be appreciated thanks.




matrices linear-transformations projection




share|cite|improve this question













Let $T:R^3 to R^3$ be the linear transformation of projection onto $x_1x_2$-plane. What is the linear transformation one obtains when you compose $T$ with itself?



I think $x_1x_2$ means projecting onto a two dimensional $XY$ plane. But what does it mean to compose $T$ with itself? is it the composition of $T$? Then I assume the composition of a linear transformation should be $R^3$ as well, since $T$ is not specified here, I am a little confused whether $T$'s composition is itself?



Any explanation would be appreciated thanks.





matrices linear-transformations projection





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edited Jul 24 at 11:22









Javi

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asked Jul 24 at 11:04









james black

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  • 3




    You have $T(x_1, x_2, x_3) = (x_1, x_2, 0)$. Then, you would then apply $T$ again: $T(x_1, x_2, 0) = $?.
    – Joe Johnson 126
    Jul 24 at 11:08











  • What is your definition of aprojection? There are many projections onto the $x$-$y$ plane. Which one are you talking about?
    – amd
    Jul 24 at 19:08












  • 3




    You have $T(x_1, x_2, x_3) = (x_1, x_2, 0)$. Then, you would then apply $T$ again: $T(x_1, x_2, 0) = $?.
    – Joe Johnson 126
    Jul 24 at 11:08











  • What is your definition of aprojection? There are many projections onto the $x$-$y$ plane. Which one are you talking about?
    – amd
    Jul 24 at 19:08







3




3




You have $T(x_1, x_2, x_3) = (x_1, x_2, 0)$. Then, you would then apply $T$ again: $T(x_1, x_2, 0) = $?.
– Joe Johnson 126
Jul 24 at 11:08





You have $T(x_1, x_2, x_3) = (x_1, x_2, 0)$. Then, you would then apply $T$ again: $T(x_1, x_2, 0) = $?.
– Joe Johnson 126
Jul 24 at 11:08













What is your definition of aprojection? There are many projections onto the $x$-$y$ plane. Which one are you talking about?
– amd
Jul 24 at 19:08




What is your definition of aprojection? There are many projections onto the $x$-$y$ plane. Which one are you talking about?
– amd
Jul 24 at 19:08










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If $T(x_1,x_2,x_3)=(x_1,x_2,0)$ then $(T circ T)(x_1,x_2,x_3)=T(x_1,x_2,0)=(x_1,x_2,0)$, hence $T circ T=T$.






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  • got it thank you because for matrices, A(BC)=(AB)C, right?
    – james black
    Jul 24 at 11:19










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If $T(x_1,x_2,x_3)=(x_1,x_2,0)$ then $(T circ T)(x_1,x_2,x_3)=T(x_1,x_2,0)=(x_1,x_2,0)$, hence $T circ T=T$.






share|cite|improve this answer





















  • got it thank you because for matrices, A(BC)=(AB)C, right?
    – james black
    Jul 24 at 11:19














up vote
4
down vote













If $T(x_1,x_2,x_3)=(x_1,x_2,0)$ then $(T circ T)(x_1,x_2,x_3)=T(x_1,x_2,0)=(x_1,x_2,0)$, hence $T circ T=T$.






share|cite|improve this answer





















  • got it thank you because for matrices, A(BC)=(AB)C, right?
    – james black
    Jul 24 at 11:19












up vote
4
down vote










up vote
4
down vote









If $T(x_1,x_2,x_3)=(x_1,x_2,0)$ then $(T circ T)(x_1,x_2,x_3)=T(x_1,x_2,0)=(x_1,x_2,0)$, hence $T circ T=T$.






share|cite|improve this answer













If $T(x_1,x_2,x_3)=(x_1,x_2,0)$ then $(T circ T)(x_1,x_2,x_3)=T(x_1,x_2,0)=(x_1,x_2,0)$, hence $T circ T=T$.







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answered Jul 24 at 11:08









Fred

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  • got it thank you because for matrices, A(BC)=(AB)C, right?
    – james black
    Jul 24 at 11:19
















  • got it thank you because for matrices, A(BC)=(AB)C, right?
    – james black
    Jul 24 at 11:19















got it thank you because for matrices, A(BC)=(AB)C, right?
– james black
Jul 24 at 11:19




got it thank you because for matrices, A(BC)=(AB)C, right?
– james black
Jul 24 at 11:19












 

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