What is the kernel and range of T?












0














Define the linear transformation T by T($vec x$) = A$vec x$. Find (1) the kernel of T and (2) the range of T



$A = begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$




1) Finding the Kernel(T)




$begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$ => RREF $begin{bmatrix}1&0&2\0&1&.5end{bmatrix}$



$begin{bmatrix}x_1\x_2\x_3end{bmatrix}$ = $begin{bmatrix}-2x_3\-.5x_3\tend{bmatrix}$ =
$begin{bmatrix}-2t\-.5t\tend{bmatrix}$ =
t$begin{bmatrix}-2\-.5\1end{bmatrix}$



My Answer - Kernel(T) = {t(-2, -.5, -1)}, t is any real number




2) Finding the Range(T)




$begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$ => RREF $begin{bmatrix}1&0&2\0&1&.5end{bmatrix}$



My Answer - Basis for range(T) = {(1.0), (0,1)}



This is what I got for the kernel and range, however the answer words it in an odd way. It says "Kernel of A is null space of A which is span{(4,1,-2)} and the Range of A is $R^2$.



Is my answer equivalent with the book's answer? Did I do anything wrong?






linear-algebra







share|cite|improve this question


















  • 1




    The span of $(4, 1, -2)$ is the same as the span of $(-2, -1/2, 1)$, which is what you found (minus your typo in writing down the final answer from the line before it). You found a basis for the range which has dimension $2$, so the range is all of $mathbb{R}^2$.
    – T. Bongers
    Nov 30 at 0:55










  • I would be a bit pedantic and say that although saying that the basis for the range of $T$ being ${(1,0),(0,1)}$ (or their transposes if you prefer) will clearly imply that the range of $T$ is all of $Bbb R^2$, the question did ask you for what the range of $T$ was, it did not ask you what a basis for the range of $T$ was. It is like answering "The 45th president of the united states" rather than "Donald Trump" when asked the question of "Who is the current president of the united states?"
    – JMoravitz
    Nov 30 at 0:59






  • 1




    Just to check your understanding: are you saying that this is the basis for the range of $T$ because those are the first two columns of the RREF or because the RREF tells you that the rank of the matrix is 2 and so the range is all of $mathbb R^2$?
    – amd
    Nov 30 at 0:59










  • I am saying this is the basis for the range of T because those are the first two columns of the RREF that have a 1 as the first non-zero number.
    – Evan Kim
    Nov 30 at 1:03






  • 1




    That’s wrong, though in this case it happens to give you a correct answer. The pivot columns of the RREF generally do not themselves form a basis for the range (column space), but indicate which of the columns of the original matrix do. E.g., the RREF of $pmatrix{1&1\1&1}$ is $pmatrix{1&1\0&0}$, but $(1,0)^T$ is obviously not a basis for its column space.
    – amd
    Nov 30 at 22:45


















0














Define the linear transformation T by T($vec x$) = A$vec x$. Find (1) the kernel of T and (2) the range of T



$A = begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$




1) Finding the Kernel(T)




$begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$ => RREF $begin{bmatrix}1&0&2\0&1&.5end{bmatrix}$



$begin{bmatrix}x_1\x_2\x_3end{bmatrix}$ = $begin{bmatrix}-2x_3\-.5x_3\tend{bmatrix}$ =
$begin{bmatrix}-2t\-.5t\tend{bmatrix}$ =
t$begin{bmatrix}-2\-.5\1end{bmatrix}$



My Answer - Kernel(T) = {t(-2, -.5, -1)}, t is any real number




2) Finding the Range(T)




$begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$ => RREF $begin{bmatrix}1&0&2\0&1&.5end{bmatrix}$



My Answer - Basis for range(T) = {(1.0), (0,1)}



This is what I got for the kernel and range, however the answer words it in an odd way. It says "Kernel of A is null space of A which is span{(4,1,-2)} and the Range of A is $R^2$.



Is my answer equivalent with the book's answer? Did I do anything wrong?






linear-algebra







share|cite|improve this question


















  • 1




    The span of $(4, 1, -2)$ is the same as the span of $(-2, -1/2, 1)$, which is what you found (minus your typo in writing down the final answer from the line before it). You found a basis for the range which has dimension $2$, so the range is all of $mathbb{R}^2$.
    – T. Bongers
    Nov 30 at 0:55










  • I would be a bit pedantic and say that although saying that the basis for the range of $T$ being ${(1,0),(0,1)}$ (or their transposes if you prefer) will clearly imply that the range of $T$ is all of $Bbb R^2$, the question did ask you for what the range of $T$ was, it did not ask you what a basis for the range of $T$ was. It is like answering "The 45th president of the united states" rather than "Donald Trump" when asked the question of "Who is the current president of the united states?"
    – JMoravitz
    Nov 30 at 0:59






  • 1




    Just to check your understanding: are you saying that this is the basis for the range of $T$ because those are the first two columns of the RREF or because the RREF tells you that the rank of the matrix is 2 and so the range is all of $mathbb R^2$?
    – amd
    Nov 30 at 0:59










  • I am saying this is the basis for the range of T because those are the first two columns of the RREF that have a 1 as the first non-zero number.
    – Evan Kim
    Nov 30 at 1:03






  • 1




    That’s wrong, though in this case it happens to give you a correct answer. The pivot columns of the RREF generally do not themselves form a basis for the range (column space), but indicate which of the columns of the original matrix do. E.g., the RREF of $pmatrix{1&1\1&1}$ is $pmatrix{1&1\0&0}$, but $(1,0)^T$ is obviously not a basis for its column space.
    – amd
    Nov 30 at 22:45
















0












0








0







Define the linear transformation T by T($vec x$) = A$vec x$. Find (1) the kernel of T and (2) the range of T



$A = begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$




1) Finding the Kernel(T)




$begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$ => RREF $begin{bmatrix}1&0&2\0&1&.5end{bmatrix}$



$begin{bmatrix}x_1\x_2\x_3end{bmatrix}$ = $begin{bmatrix}-2x_3\-.5x_3\tend{bmatrix}$ =
$begin{bmatrix}-2t\-.5t\tend{bmatrix}$ =
t$begin{bmatrix}-2\-.5\1end{bmatrix}$



My Answer - Kernel(T) = {t(-2, -.5, -1)}, t is any real number




2) Finding the Range(T)




$begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$ => RREF $begin{bmatrix}1&0&2\0&1&.5end{bmatrix}$



My Answer - Basis for range(T) = {(1.0), (0,1)}



This is what I got for the kernel and range, however the answer words it in an odd way. It says "Kernel of A is null space of A which is span{(4,1,-2)} and the Range of A is $R^2$.



Is my answer equivalent with the book's answer? Did I do anything wrong?






linear-algebra







share|cite|improve this question













Define the linear transformation T by T($vec x$) = A$vec x$. Find (1) the kernel of T and (2) the range of T



$A = begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$




1) Finding the Kernel(T)




$begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$ => RREF $begin{bmatrix}1&0&2\0&1&.5end{bmatrix}$



$begin{bmatrix}x_1\x_2\x_3end{bmatrix}$ = $begin{bmatrix}-2x_3\-.5x_3\tend{bmatrix}$ =
$begin{bmatrix}-2t\-.5t\tend{bmatrix}$ =
t$begin{bmatrix}-2\-.5\1end{bmatrix}$



My Answer - Kernel(T) = {t(-2, -.5, -1)}, t is any real number




2) Finding the Range(T)




$begin{bmatrix}1&-2&1\0&2&1end{bmatrix}$ => RREF $begin{bmatrix}1&0&2\0&1&.5end{bmatrix}$



My Answer - Basis for range(T) = {(1.0), (0,1)}



This is what I got for the kernel and range, however the answer words it in an odd way. It says "Kernel of A is null space of A which is span{(4,1,-2)} and the Range of A is $R^2$.



Is my answer equivalent with the book's answer? Did I do anything wrong?






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 at 0:49









Evan Kim

828




828








  • 1




    The span of $(4, 1, -2)$ is the same as the span of $(-2, -1/2, 1)$, which is what you found (minus your typo in writing down the final answer from the line before it). You found a basis for the range which has dimension $2$, so the range is all of $mathbb{R}^2$.
    – T. Bongers
    Nov 30 at 0:55










  • I would be a bit pedantic and say that although saying that the basis for the range of $T$ being ${(1,0),(0,1)}$ (or their transposes if you prefer) will clearly imply that the range of $T$ is all of $Bbb R^2$, the question did ask you for what the range of $T$ was, it did not ask you what a basis for the range of $T$ was. It is like answering "The 45th president of the united states" rather than "Donald Trump" when asked the question of "Who is the current president of the united states?"
    – JMoravitz
    Nov 30 at 0:59






  • 1




    Just to check your understanding: are you saying that this is the basis for the range of $T$ because those are the first two columns of the RREF or because the RREF tells you that the rank of the matrix is 2 and so the range is all of $mathbb R^2$?
    – amd
    Nov 30 at 0:59










  • I am saying this is the basis for the range of T because those are the first two columns of the RREF that have a 1 as the first non-zero number.
    – Evan Kim
    Nov 30 at 1:03






  • 1




    That’s wrong, though in this case it happens to give you a correct answer. The pivot columns of the RREF generally do not themselves form a basis for the range (column space), but indicate which of the columns of the original matrix do. E.g., the RREF of $pmatrix{1&1\1&1}$ is $pmatrix{1&1\0&0}$, but $(1,0)^T$ is obviously not a basis for its column space.
    – amd
    Nov 30 at 22:45
















  • 1




    The span of $(4, 1, -2)$ is the same as the span of $(-2, -1/2, 1)$, which is what you found (minus your typo in writing down the final answer from the line before it). You found a basis for the range which has dimension $2$, so the range is all of $mathbb{R}^2$.
    – T. Bongers
    Nov 30 at 0:55










  • I would be a bit pedantic and say that although saying that the basis for the range of $T$ being ${(1,0),(0,1)}$ (or their transposes if you prefer) will clearly imply that the range of $T$ is all of $Bbb R^2$, the question did ask you for what the range of $T$ was, it did not ask you what a basis for the range of $T$ was. It is like answering "The 45th president of the united states" rather than "Donald Trump" when asked the question of "Who is the current president of the united states?"
    – JMoravitz
    Nov 30 at 0:59






  • 1




    Just to check your understanding: are you saying that this is the basis for the range of $T$ because those are the first two columns of the RREF or because the RREF tells you that the rank of the matrix is 2 and so the range is all of $mathbb R^2$?
    – amd
    Nov 30 at 0:59










  • I am saying this is the basis for the range of T because those are the first two columns of the RREF that have a 1 as the first non-zero number.
    – Evan Kim
    Nov 30 at 1:03






  • 1




    That’s wrong, though in this case it happens to give you a correct answer. The pivot columns of the RREF generally do not themselves form a basis for the range (column space), but indicate which of the columns of the original matrix do. E.g., the RREF of $pmatrix{1&1\1&1}$ is $pmatrix{1&1\0&0}$, but $(1,0)^T$ is obviously not a basis for its column space.
    – amd
    Nov 30 at 22:45










1




1




The span of $(4, 1, -2)$ is the same as the span of $(-2, -1/2, 1)$, which is what you found (minus your typo in writing down the final answer from the line before it). You found a basis for the range which has dimension $2$, so the range is all of $mathbb{R}^2$.
– T. Bongers
Nov 30 at 0:55




The span of $(4, 1, -2)$ is the same as the span of $(-2, -1/2, 1)$, which is what you found (minus your typo in writing down the final answer from the line before it). You found a basis for the range which has dimension $2$, so the range is all of $mathbb{R}^2$.
– T. Bongers
Nov 30 at 0:55












I would be a bit pedantic and say that although saying that the basis for the range of $T$ being ${(1,0),(0,1)}$ (or their transposes if you prefer) will clearly imply that the range of $T$ is all of $Bbb R^2$, the question did ask you for what the range of $T$ was, it did not ask you what a basis for the range of $T$ was. It is like answering "The 45th president of the united states" rather than "Donald Trump" when asked the question of "Who is the current president of the united states?"
– JMoravitz
Nov 30 at 0:59




I would be a bit pedantic and say that although saying that the basis for the range of $T$ being ${(1,0),(0,1)}$ (or their transposes if you prefer) will clearly imply that the range of $T$ is all of $Bbb R^2$, the question did ask you for what the range of $T$ was, it did not ask you what a basis for the range of $T$ was. It is like answering "The 45th president of the united states" rather than "Donald Trump" when asked the question of "Who is the current president of the united states?"
– JMoravitz
Nov 30 at 0:59




1




1




Just to check your understanding: are you saying that this is the basis for the range of $T$ because those are the first two columns of the RREF or because the RREF tells you that the rank of the matrix is 2 and so the range is all of $mathbb R^2$?
– amd
Nov 30 at 0:59




Just to check your understanding: are you saying that this is the basis for the range of $T$ because those are the first two columns of the RREF or because the RREF tells you that the rank of the matrix is 2 and so the range is all of $mathbb R^2$?
– amd
Nov 30 at 0:59












I am saying this is the basis for the range of T because those are the first two columns of the RREF that have a 1 as the first non-zero number.
– Evan Kim
Nov 30 at 1:03




I am saying this is the basis for the range of T because those are the first two columns of the RREF that have a 1 as the first non-zero number.
– Evan Kim
Nov 30 at 1:03




1




1




That’s wrong, though in this case it happens to give you a correct answer. The pivot columns of the RREF generally do not themselves form a basis for the range (column space), but indicate which of the columns of the original matrix do. E.g., the RREF of $pmatrix{1&1\1&1}$ is $pmatrix{1&1\0&0}$, but $(1,0)^T$ is obviously not a basis for its column space.
– amd
Nov 30 at 22:45






That’s wrong, though in this case it happens to give you a correct answer. The pivot columns of the RREF generally do not themselves form a basis for the range (column space), but indicate which of the columns of the original matrix do. E.g., the RREF of $pmatrix{1&1\1&1}$ is $pmatrix{1&1\0&0}$, but $(1,0)^T$ is obviously not a basis for its column space.
– amd
Nov 30 at 22:45












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Your answers match those of the book; they're just written differently. Note that if you set $t = -2$, you will see that the set of vectors spanned by $begin{bmatrix}-2 \ -0.5 \ 1end{bmatrix}$ is the same as that spanned by $begin{bmatrix} 4 \ 1 \ -2end{bmatrix}$. Similarly, $left{begin{bmatrix} 1 \ 0 end{bmatrix}, begin{bmatrix} 0 \ 1end{bmatrix} right}$ is a basis for $mathbb{R}^2$.






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    Your answers match those of the book; they're just written differently. Note that if you set $t = -2$, you will see that the set of vectors spanned by $begin{bmatrix}-2 \ -0.5 \ 1end{bmatrix}$ is the same as that spanned by $begin{bmatrix} 4 \ 1 \ -2end{bmatrix}$. Similarly, $left{begin{bmatrix} 1 \ 0 end{bmatrix}, begin{bmatrix} 0 \ 1end{bmatrix} right}$ is a basis for $mathbb{R}^2$.






    share|cite|improve this answer


























      1














      Your answers match those of the book; they're just written differently. Note that if you set $t = -2$, you will see that the set of vectors spanned by $begin{bmatrix}-2 \ -0.5 \ 1end{bmatrix}$ is the same as that spanned by $begin{bmatrix} 4 \ 1 \ -2end{bmatrix}$. Similarly, $left{begin{bmatrix} 1 \ 0 end{bmatrix}, begin{bmatrix} 0 \ 1end{bmatrix} right}$ is a basis for $mathbb{R}^2$.






      share|cite|improve this answer
























        1












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        1






        Your answers match those of the book; they're just written differently. Note that if you set $t = -2$, you will see that the set of vectors spanned by $begin{bmatrix}-2 \ -0.5 \ 1end{bmatrix}$ is the same as that spanned by $begin{bmatrix} 4 \ 1 \ -2end{bmatrix}$. Similarly, $left{begin{bmatrix} 1 \ 0 end{bmatrix}, begin{bmatrix} 0 \ 1end{bmatrix} right}$ is a basis for $mathbb{R}^2$.






        share|cite|improve this answer












        Your answers match those of the book; they're just written differently. Note that if you set $t = -2$, you will see that the set of vectors spanned by $begin{bmatrix}-2 \ -0.5 \ 1end{bmatrix}$ is the same as that spanned by $begin{bmatrix} 4 \ 1 \ -2end{bmatrix}$. Similarly, $left{begin{bmatrix} 1 \ 0 end{bmatrix}, begin{bmatrix} 0 \ 1end{bmatrix} right}$ is a basis for $mathbb{R}^2$.







        share|cite|improve this answer












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        answered Nov 30 at 0:56









        platty

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