Mixing
Finding values for which a linear transformation $L: mathbb{R}^3 rightarrow mathbb{R}^3 $ satisfies $v_1, v_2...
$begingroup$
I am working on an old exam question as preparation for my exam this coming week. The question states that $v_1 = (1, a, a^2), v_2 = (a^2, 1, a), v_3 = (a, a^2, 1)$.
Part A asked to find the values of $a$ for which ${v_1, v_2, v_3 }$ is a basis of $mathbb{R}$. I did this by setting up $v_1, v_2, v_3$ in a matrix, finding the determinant and seeing for which values of $a$ it is equal to zero. I found that ${v_1, v_2, v_3 }$ is a basis of $mathbb{R}$ if $a neq 1$.
Part B asks: for which values of $a$ is $L: mathbb{R}^3 rightarrow mathbb{R}^3$ a linear map so that
$$v_1, v_2 in ker(L) text{ and } L(v_3) = (1, 1, 1)$$
I do not really have an idea of where to get started with this problem. I know that $L(v_1) = 0$ and $L(v_2) =0$, but I'm not sure where to go with the problem from there.
Part C asks for the eigenvalues and eigenvectors of $L$, for the values of $a$ given in part B. I assume this is straightforward if I know what the linear map is.
linear-algebra linear-transformations
asked Jan 18 at 15:08
Elliot SElliot S
657
$endgroup$
add a comment |
$begingroup$
I am working on an old exam question as preparation for my exam this coming week. The question states that $v_1 = (1, a, a^2), v_2 = (a^2, 1, a), v_3 = (a, a^2, 1)$.
Part A asked to find the values of $a$ for which ${v_1, v_2, v_3 }$ is a basis of $mathbb{R}$. I did this by setting up $v_1, v_2, v_3$ in a matrix, finding the determinant and seeing for which values of $a$ it is equal to zero. I found that ${v_1, v_2, v_3 }$ is a basis of $mathbb{R}$ if $a neq 1$.
Part B asks: for which values of $a$ is $L: mathbb{R}^3 rightarrow mathbb{R}^3$ a linear map so that
$$v_1, v_2 in ker(L) text{ and } L(v_3) = (1, 1, 1)$$
I do not really have an idea of where to get started with this problem. I know that $L(v_1) = 0$ and $L(v_2) =0$, but I'm not sure where to go with the problem from there.
Part C asks for the eigenvalues and eigenvectors of $L$, for the values of $a$ given in part B. I assume this is straightforward if I know what the linear map is.
linear-algebra linear-transformations
asked Jan 18 at 15:08
Elliot SElliot S
657
$endgroup$
$begingroup$
For part C, you should be able to spot two eigenvectors of $L$ and their associated eigenvalues without constructing $L$ explicitly.
$endgroup$
– amd
Jan 19 at 1:23
add a comment |
$begingroup$
I am working on an old exam question as preparation for my exam this coming week. The question states that $v_1 = (1, a, a^2), v_2 = (a^2, 1, a), v_3 = (a, a^2, 1)$.
Part A asked to find the values of $a$ for which ${v_1, v_2, v_3 }$ is a basis of $mathbb{R}$. I did this by setting up $v_1, v_2, v_3$ in a matrix, finding the determinant and seeing for which values of $a$ it is equal to zero. I found that ${v_1, v_2, v_3 }$ is a basis of $mathbb{R}$ if $a neq 1$.
Part B asks: for which values of $a$ is $L: mathbb{R}^3 rightarrow mathbb{R}^3$ a linear map so that
$$v_1, v_2 in ker(L) text{ and } L(v_3) = (1, 1, 1)$$
I do not really have an idea of where to get started with this problem. I know that $L(v_1) = 0$ and $L(v_2) =0$, but I'm not sure where to go with the problem from there.
Part C asks for the eigenvalues and eigenvectors of $L$, for the values of $a$ given in part B. I assume this is straightforward if I know what the linear map is.
linear-algebra linear-transformations
asked Jan 18 at 15:08
Elliot SElliot S
657
$endgroup$
I am working on an old exam question as preparation for my exam this coming week. The question states that $v_1 = (1, a, a^2), v_2 = (a^2, 1, a), v_3 = (a, a^2, 1)$.
Part A asked to find the values of $a$ for which ${v_1, v_2, v_3 }$ is a basis of $mathbb{R}$. I did this by setting up $v_1, v_2, v_3$ in a matrix, finding the determinant and seeing for which values of $a$ it is equal to zero. I found that ${v_1, v_2, v_3 }$ is a basis of $mathbb{R}$ if $a neq 1$.
Part B asks: for which values of $a$ is $L: mathbb{R}^3 rightarrow mathbb{R}^3$ a linear map so that
$$v_1, v_2 in ker(L) text{ and } L(v_3) = (1, 1, 1)$$
I do not really have an idea of where to get started with this problem. I know that $L(v_1) = 0$ and $L(v_2) =0$, but I'm not sure where to go with the problem from there.
Part C asks for the eigenvalues and eigenvectors of $L$, for the values of $a$ given in part B. I assume this is straightforward if I know what the linear map is.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Jan 18 at 15:08
Elliot SElliot S
657
asked Jan 18 at 15:08
Elliot SElliot S
657
asked Jan 18 at 15:08
Elliot SElliot S
657
asked Jan 18 at 15:08
Elliot SElliot S
657
asked Jan 18 at 15:08
Elliot SElliot S
657
657
$begingroup$
For part C, you should be able to spot two eigenvectors of $L$ and their associated eigenvalues without constructing $L$ explicitly.
$endgroup$
– amd
Jan 19 at 1:23
add a comment |
$begingroup$
For part C, you should be able to spot two eigenvectors of $L$ and their associated eigenvalues without constructing $L$ explicitly.
$endgroup$
– amd
Jan 19 at 1:23
$begingroup$
For part C, you should be able to spot two eigenvectors of $L$ and their associated eigenvalues without constructing $L$ explicitly.
$endgroup$
– amd
Jan 19 at 1:23
$begingroup$
For part C, you should be able to spot two eigenvectors of $L$ and their associated eigenvalues without constructing $L$ explicitly.
$endgroup$
– amd
Jan 19 at 1:23
add a comment |
2 Answers
2
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$begingroup$
If $aneq1$, then ${v_1,v_2,v_3}$ is a basis, and therefore there is one and only one linear map $Lcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that $L(v_1)=(0,0,0)$, that $L(v_2)=(0,0,0)$, and that $L(v_3)=(1,1,1)$.
Otherwise, $v_1=v_2=v_3=(1,1,1)$. So, no such linear map $L$ exists, since $L(1,1,1)$ cannot be equal to $(0,0,0)$ and to $(1,1,1)$ simultaneously.
answered Jan 18 at 15:15
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 7:57
add a comment |
$begingroup$
Part B : The information you provided allows you to explicitly write the Matrix for the linear transformation L in the basis $B=[v_1, v_2, v_3].$
$$ [L]_B =begin{pmatrix}
0 & 0 & 1 \
0 & 0 & 1 \
0 & 0 & 1
end{pmatrix} $$
Part C: You can remember that if a non zero vector is in the kernel of a linear map that also means that such vector is an eigenvector for the eigenvalue 0. So you already have two eigenvector $ (v_1,v_2) $ which are linearly independent. Once you have made these observations you can see if there are other eigenvalues and finish the exercise.
answered Jan 18 at 16:39
NassoumoNassoumo
812
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add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
If $aneq1$, then ${v_1,v_2,v_3}$ is a basis, and therefore there is one and only one linear map $Lcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that $L(v_1)=(0,0,0)$, that $L(v_2)=(0,0,0)$, and that $L(v_3)=(1,1,1)$.
Otherwise, $v_1=v_2=v_3=(1,1,1)$. So, no such linear map $L$ exists, since $L(1,1,1)$ cannot be equal to $(0,0,0)$ and to $(1,1,1)$ simultaneously.
answered Jan 18 at 15:15
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 7:57
add a comment |
$begingroup$
If $aneq1$, then ${v_1,v_2,v_3}$ is a basis, and therefore there is one and only one linear map $Lcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that $L(v_1)=(0,0,0)$, that $L(v_2)=(0,0,0)$, and that $L(v_3)=(1,1,1)$.
Otherwise, $v_1=v_2=v_3=(1,1,1)$. So, no such linear map $L$ exists, since $L(1,1,1)$ cannot be equal to $(0,0,0)$ and to $(1,1,1)$ simultaneously.
answered Jan 18 at 15:15
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 7:57
add a comment |
$begingroup$
If $aneq1$, then ${v_1,v_2,v_3}$ is a basis, and therefore there is one and only one linear map $Lcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that $L(v_1)=(0,0,0)$, that $L(v_2)=(0,0,0)$, and that $L(v_3)=(1,1,1)$.
Otherwise, $v_1=v_2=v_3=(1,1,1)$. So, no such linear map $L$ exists, since $L(1,1,1)$ cannot be equal to $(0,0,0)$ and to $(1,1,1)$ simultaneously.
answered Jan 18 at 15:15
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
If $aneq1$, then ${v_1,v_2,v_3}$ is a basis, and therefore there is one and only one linear map $Lcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that $L(v_1)=(0,0,0)$, that $L(v_2)=(0,0,0)$, and that $L(v_3)=(1,1,1)$.
Otherwise, $v_1=v_2=v_3=(1,1,1)$. So, no such linear map $L$ exists, since $L(1,1,1)$ cannot be equal to $(0,0,0)$ and to $(1,1,1)$ simultaneously.
answered Jan 18 at 15:15
José Carlos SantosJosé Carlos Santos
164k22131234
edited Jan 19 at 7:57
answered Jan 18 at 15:15
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 18 at 15:15
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 18 at 15:15
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 7:57
add a comment |
$begingroup$
Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 7:57
$begingroup$
Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 7:57
$begingroup$
Oops! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 7:57
add a comment |
$begingroup$
Part B : The information you provided allows you to explicitly write the Matrix for the linear transformation L in the basis $B=[v_1, v_2, v_3].$
$$ [L]_B =begin{pmatrix}
0 & 0 & 1 \
0 & 0 & 1 \
0 & 0 & 1
end{pmatrix} $$
Part C: You can remember that if a non zero vector is in the kernel of a linear map that also means that such vector is an eigenvector for the eigenvalue 0. So you already have two eigenvector $ (v_1,v_2) $ which are linearly independent. Once you have made these observations you can see if there are other eigenvalues and finish the exercise.
answered Jan 18 at 16:39
NassoumoNassoumo
812
$endgroup$
add a comment |
$begingroup$
Part B : The information you provided allows you to explicitly write the Matrix for the linear transformation L in the basis $B=[v_1, v_2, v_3].$
$$ [L]_B =begin{pmatrix}
0 & 0 & 1 \
0 & 0 & 1 \
0 & 0 & 1
end{pmatrix} $$
Part C: You can remember that if a non zero vector is in the kernel of a linear map that also means that such vector is an eigenvector for the eigenvalue 0. So you already have two eigenvector $ (v_1,v_2) $ which are linearly independent. Once you have made these observations you can see if there are other eigenvalues and finish the exercise.
answered Jan 18 at 16:39
NassoumoNassoumo
812
$endgroup$
add a comment |
$begingroup$
Part B : The information you provided allows you to explicitly write the Matrix for the linear transformation L in the basis $B=[v_1, v_2, v_3].$
$$ [L]_B =begin{pmatrix}
0 & 0 & 1 \
0 & 0 & 1 \
0 & 0 & 1
end{pmatrix} $$
Part C: You can remember that if a non zero vector is in the kernel of a linear map that also means that such vector is an eigenvector for the eigenvalue 0. So you already have two eigenvector $ (v_1,v_2) $ which are linearly independent. Once you have made these observations you can see if there are other eigenvalues and finish the exercise.
answered Jan 18 at 16:39
NassoumoNassoumo
812
$endgroup$
Part B : The information you provided allows you to explicitly write the Matrix for the linear transformation L in the basis $B=[v_1, v_2, v_3].$
$$ [L]_B =begin{pmatrix}
0 & 0 & 1 \
0 & 0 & 1 \
0 & 0 & 1
end{pmatrix} $$
Part C: You can remember that if a non zero vector is in the kernel of a linear map that also means that such vector is an eigenvector for the eigenvalue 0. So you already have two eigenvector $ (v_1,v_2) $ which are linearly independent. Once you have made these observations you can see if there are other eigenvalues and finish the exercise.
answered Jan 18 at 16:39
NassoumoNassoumo
812
answered Jan 18 at 16:39
NassoumoNassoumo
812
answered Jan 18 at 16:39
NassoumoNassoumo
812
answered Jan 18 at 16:39
NassoumoNassoumo
812
812
add a comment |
add a comment |
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$begingroup$
For part C, you should be able to spot two eigenvectors of $L$ and their associated eigenvalues without constructing $L$ explicitly.
$endgroup$
– amd
Jan 19 at 1:23