How do I prove that these vectors form a basis?
I know that by definition of basis, the vectors v1 and v2 should span the entire subspace. Therefore, if the first constant is not equal to the second constant, and if both of the constants give a linear transformation, then they must be linearly independent and therefore must form a basis. Is that the correct proof, or am I missing this? Also I don't know what the matrix of the linear transformation is.
linear-algebra
asked Mar 16 '16 at 22:41
Jeansandcofffeee
395
add a comment |
I know that by definition of basis, the vectors v1 and v2 should span the entire subspace. Therefore, if the first constant is not equal to the second constant, and if both of the constants give a linear transformation, then they must be linearly independent and therefore must form a basis. Is that the correct proof, or am I missing this? Also I don't know what the matrix of the linear transformation is.
linear-algebra
asked Mar 16 '16 at 22:41
Jeansandcofffeee
395
1
"...then they must be linearly independent". I don't see where you justify this claim. And what is "they"? You want to prove the vectors $v_1,v_2$ are linearly independent, but you only refer to constants and linear transformations in the argument leading up to that. There's no place where you have considered anything about the vectors.
– Erick Wong
Mar 16 '16 at 22:44
A standard exercize shows that two eigenvectors relative to two distinct eigenvalues must be linearly independent. If you are using this fact, then everything becomes obvious; but if you don't, then you are using some unjustified argument.
– Crostul
Mar 16 '16 at 23:22
add a comment |
I know that by definition of basis, the vectors v1 and v2 should span the entire subspace. Therefore, if the first constant is not equal to the second constant, and if both of the constants give a linear transformation, then they must be linearly independent and therefore must form a basis. Is that the correct proof, or am I missing this? Also I don't know what the matrix of the linear transformation is.
linear-algebra
asked Mar 16 '16 at 22:41
Jeansandcofffeee
395
I know that by definition of basis, the vectors v1 and v2 should span the entire subspace. Therefore, if the first constant is not equal to the second constant, and if both of the constants give a linear transformation, then they must be linearly independent and therefore must form a basis. Is that the correct proof, or am I missing this? Also I don't know what the matrix of the linear transformation is.
linear-algebra
linear-algebra
asked Mar 16 '16 at 22:41
Jeansandcofffeee
395
asked Mar 16 '16 at 22:41
Jeansandcofffeee
395
asked Mar 16 '16 at 22:41
Jeansandcofffeee
395
asked Mar 16 '16 at 22:41
Jeansandcofffeee
395
asked Mar 16 '16 at 22:41
Jeansandcofffeee
395
395
1
"...then they must be linearly independent". I don't see where you justify this claim. And what is "they"? You want to prove the vectors $v_1,v_2$ are linearly independent, but you only refer to constants and linear transformations in the argument leading up to that. There's no place where you have considered anything about the vectors.
– Erick Wong
Mar 16 '16 at 22:44
A standard exercize shows that two eigenvectors relative to two distinct eigenvalues must be linearly independent. If you are using this fact, then everything becomes obvious; but if you don't, then you are using some unjustified argument.
– Crostul
Mar 16 '16 at 23:22
add a comment |
1
"...then they must be linearly independent". I don't see where you justify this claim. And what is "they"? You want to prove the vectors $v_1,v_2$ are linearly independent, but you only refer to constants and linear transformations in the argument leading up to that. There's no place where you have considered anything about the vectors.
– Erick Wong
Mar 16 '16 at 22:44
A standard exercize shows that two eigenvectors relative to two distinct eigenvalues must be linearly independent. If you are using this fact, then everything becomes obvious; but if you don't, then you are using some unjustified argument.
– Crostul
Mar 16 '16 at 23:22
1
1
"...then they must be linearly independent". I don't see where you justify this claim. And what is "they"? You want to prove the vectors $v_1,v_2$ are linearly independent, but you only refer to constants and linear transformations in the argument leading up to that. There's no place where you have considered anything about the vectors.
– Erick Wong
Mar 16 '16 at 22:44
"...then they must be linearly independent". I don't see where you justify this claim. And what is "they"? You want to prove the vectors $v_1,v_2$ are linearly independent, but you only refer to constants and linear transformations in the argument leading up to that. There's no place where you have considered anything about the vectors.
– Erick Wong
Mar 16 '16 at 22:44
A standard exercize shows that two eigenvectors relative to two distinct eigenvalues must be linearly independent. If you are using this fact, then everything becomes obvious; but if you don't, then you are using some unjustified argument.
– Crostul
Mar 16 '16 at 23:22
A standard exercize shows that two eigenvectors relative to two distinct eigenvalues must be linearly independent. If you are using this fact, then everything becomes obvious; but if you don't, then you are using some unjustified argument.
– Crostul
Mar 16 '16 at 23:22
add a comment |
1 Answer
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Suppose that $c_1 vec v_1 + c_2 vec v_2 = 0.$ Multiply both sides by $A$ to get $c_1 lambda_1 vec v_1 + c_2 lambda_2 vec v_2 = 0.$ Multiply the first equation by $lambda_1,$ and subtract it from the the second to get $(lambda_2 - lambda_1) c_2 vec v_2 = 0.$ Because the $lambda$s are distinct and $vec v_2$ is not zero, we must have $c_2 = 0.$ By a similar argument, $c_1 = 0.$ Hence $vec v_1, vec v_2$ are linearly independent. Now, there are two of them, and the space $Bbb R^2$ has exactly two dimensions, so $vec v_1, vec v_2$ form a basis of $Bbb R^2.$
The $i$th column of the matrix of $T_A$ is $[T_A vec v_i]_{mathcal B}$ where $[ cdot ]_{mathcal B}$ denotes the coordinate matrix relative to $mathcal B.$ We compute
$$begin{align}
[T_A vec v_1]_{mathcal B} & = [A vec v_1]_{mathcal B} = [lambda_1 vec v_1]_{mathcal B} = [lambda_1 vec v_1 + 0 cdot vec v_2]_{mathcal B} = begin{bmatrix} lambda_1\ 0end{bmatrix}\\
[T_A vec v_2]_{mathcal B} & = [A vec v_2]_{mathcal B} = [lambda_2 vec v_2]_{mathcal B} = [0 cdot vec v_1 + lambda_2 vec v_2]_{mathcal B} = begin{bmatrix} 0\ lambda_2end{bmatrix}.
end{align}$$
Thus, the matrix of $T_A$ is
$$begin{bmatrix} lambda_1 & 0\ 0 & lambda_2end{bmatrix}.$$
answered Nov 24 '18 at 21:38
Maurice P
1,3901732
add a comment |
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Suppose that $c_1 vec v_1 + c_2 vec v_2 = 0.$ Multiply both sides by $A$ to get $c_1 lambda_1 vec v_1 + c_2 lambda_2 vec v_2 = 0.$ Multiply the first equation by $lambda_1,$ and subtract it from the the second to get $(lambda_2 - lambda_1) c_2 vec v_2 = 0.$ Because the $lambda$s are distinct and $vec v_2$ is not zero, we must have $c_2 = 0.$ By a similar argument, $c_1 = 0.$ Hence $vec v_1, vec v_2$ are linearly independent. Now, there are two of them, and the space $Bbb R^2$ has exactly two dimensions, so $vec v_1, vec v_2$ form a basis of $Bbb R^2.$
The $i$th column of the matrix of $T_A$ is $[T_A vec v_i]_{mathcal B}$ where $[ cdot ]_{mathcal B}$ denotes the coordinate matrix relative to $mathcal B.$ We compute
$$begin{align}
[T_A vec v_1]_{mathcal B} & = [A vec v_1]_{mathcal B} = [lambda_1 vec v_1]_{mathcal B} = [lambda_1 vec v_1 + 0 cdot vec v_2]_{mathcal B} = begin{bmatrix} lambda_1\ 0end{bmatrix}\\
[T_A vec v_2]_{mathcal B} & = [A vec v_2]_{mathcal B} = [lambda_2 vec v_2]_{mathcal B} = [0 cdot vec v_1 + lambda_2 vec v_2]_{mathcal B} = begin{bmatrix} 0\ lambda_2end{bmatrix}.
end{align}$$
Thus, the matrix of $T_A$ is
$$begin{bmatrix} lambda_1 & 0\ 0 & lambda_2end{bmatrix}.$$
answered Nov 24 '18 at 21:38
Maurice P
1,3901732
add a comment |
Suppose that $c_1 vec v_1 + c_2 vec v_2 = 0.$ Multiply both sides by $A$ to get $c_1 lambda_1 vec v_1 + c_2 lambda_2 vec v_2 = 0.$ Multiply the first equation by $lambda_1,$ and subtract it from the the second to get $(lambda_2 - lambda_1) c_2 vec v_2 = 0.$ Because the $lambda$s are distinct and $vec v_2$ is not zero, we must have $c_2 = 0.$ By a similar argument, $c_1 = 0.$ Hence $vec v_1, vec v_2$ are linearly independent. Now, there are two of them, and the space $Bbb R^2$ has exactly two dimensions, so $vec v_1, vec v_2$ form a basis of $Bbb R^2.$
The $i$th column of the matrix of $T_A$ is $[T_A vec v_i]_{mathcal B}$ where $[ cdot ]_{mathcal B}$ denotes the coordinate matrix relative to $mathcal B.$ We compute
$$begin{align}
[T_A vec v_1]_{mathcal B} & = [A vec v_1]_{mathcal B} = [lambda_1 vec v_1]_{mathcal B} = [lambda_1 vec v_1 + 0 cdot vec v_2]_{mathcal B} = begin{bmatrix} lambda_1\ 0end{bmatrix}\\
[T_A vec v_2]_{mathcal B} & = [A vec v_2]_{mathcal B} = [lambda_2 vec v_2]_{mathcal B} = [0 cdot vec v_1 + lambda_2 vec v_2]_{mathcal B} = begin{bmatrix} 0\ lambda_2end{bmatrix}.
end{align}$$
Thus, the matrix of $T_A$ is
$$begin{bmatrix} lambda_1 & 0\ 0 & lambda_2end{bmatrix}.$$
answered Nov 24 '18 at 21:38
Maurice P
1,3901732
add a comment |
Suppose that $c_1 vec v_1 + c_2 vec v_2 = 0.$ Multiply both sides by $A$ to get $c_1 lambda_1 vec v_1 + c_2 lambda_2 vec v_2 = 0.$ Multiply the first equation by $lambda_1,$ and subtract it from the the second to get $(lambda_2 - lambda_1) c_2 vec v_2 = 0.$ Because the $lambda$s are distinct and $vec v_2$ is not zero, we must have $c_2 = 0.$ By a similar argument, $c_1 = 0.$ Hence $vec v_1, vec v_2$ are linearly independent. Now, there are two of them, and the space $Bbb R^2$ has exactly two dimensions, so $vec v_1, vec v_2$ form a basis of $Bbb R^2.$
The $i$th column of the matrix of $T_A$ is $[T_A vec v_i]_{mathcal B}$ where $[ cdot ]_{mathcal B}$ denotes the coordinate matrix relative to $mathcal B.$ We compute
$$begin{align}
[T_A vec v_1]_{mathcal B} & = [A vec v_1]_{mathcal B} = [lambda_1 vec v_1]_{mathcal B} = [lambda_1 vec v_1 + 0 cdot vec v_2]_{mathcal B} = begin{bmatrix} lambda_1\ 0end{bmatrix}\\
[T_A vec v_2]_{mathcal B} & = [A vec v_2]_{mathcal B} = [lambda_2 vec v_2]_{mathcal B} = [0 cdot vec v_1 + lambda_2 vec v_2]_{mathcal B} = begin{bmatrix} 0\ lambda_2end{bmatrix}.
end{align}$$
Thus, the matrix of $T_A$ is
$$begin{bmatrix} lambda_1 & 0\ 0 & lambda_2end{bmatrix}.$$
answered Nov 24 '18 at 21:38
Maurice P
1,3901732
Suppose that $c_1 vec v_1 + c_2 vec v_2 = 0.$ Multiply both sides by $A$ to get $c_1 lambda_1 vec v_1 + c_2 lambda_2 vec v_2 = 0.$ Multiply the first equation by $lambda_1,$ and subtract it from the the second to get $(lambda_2 - lambda_1) c_2 vec v_2 = 0.$ Because the $lambda$s are distinct and $vec v_2$ is not zero, we must have $c_2 = 0.$ By a similar argument, $c_1 = 0.$ Hence $vec v_1, vec v_2$ are linearly independent. Now, there are two of them, and the space $Bbb R^2$ has exactly two dimensions, so $vec v_1, vec v_2$ form a basis of $Bbb R^2.$
The $i$th column of the matrix of $T_A$ is $[T_A vec v_i]_{mathcal B}$ where $[ cdot ]_{mathcal B}$ denotes the coordinate matrix relative to $mathcal B.$ We compute
$$begin{align}
[T_A vec v_1]_{mathcal B} & = [A vec v_1]_{mathcal B} = [lambda_1 vec v_1]_{mathcal B} = [lambda_1 vec v_1 + 0 cdot vec v_2]_{mathcal B} = begin{bmatrix} lambda_1\ 0end{bmatrix}\\
[T_A vec v_2]_{mathcal B} & = [A vec v_2]_{mathcal B} = [lambda_2 vec v_2]_{mathcal B} = [0 cdot vec v_1 + lambda_2 vec v_2]_{mathcal B} = begin{bmatrix} 0\ lambda_2end{bmatrix}.
end{align}$$
Thus, the matrix of $T_A$ is
$$begin{bmatrix} lambda_1 & 0\ 0 & lambda_2end{bmatrix}.$$
answered Nov 24 '18 at 21:38
Maurice P
1,3901732
edited Nov 25 '18 at 14:39
answered Nov 24 '18 at 21:38
Maurice P
1,3901732
answered Nov 24 '18 at 21:38
Maurice P
1,3901732
answered Nov 24 '18 at 21:38
Maurice P
1,3901732
1,3901732
add a comment |
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1
"...then they must be linearly independent". I don't see where you justify this claim. And what is "they"? You want to prove the vectors $v_1,v_2$ are linearly independent, but you only refer to constants and linear transformations in the argument leading up to that. There's no place where you have considered anything about the vectors.
– Erick Wong
Mar 16 '16 at 22:44
A standard exercize shows that two eigenvectors relative to two distinct eigenvalues must be linearly independent. If you are using this fact, then everything becomes obvious; but if you don't, then you are using some unjustified argument.
– Crostul
Mar 16 '16 at 23:22