Mixing
Are eigenvectors a basis set for any invariant subspace?
2
$begingroup$
Let $Ain mathbb{R}^{ntimes n}$ (i.e., a real square matrix).
Assume that $v_1,dots, v_m$ ($mle n$) are eigenvectors of $A$.
Can any $d-$dimensional invariant subspace of $A$ be constructed using $d$ numbers of $v_i$'s?
linear-algebra matrices linear-transformations
asked Jan 10 at 7:00
Bashir SadeghiBashir Sadeghi
424
$endgroup$
|
show 2 more comments
2
$begingroup$
Let $Ain mathbb{R}^{ntimes n}$ (i.e., a real square matrix).
Assume that $v_1,dots, v_m$ ($mle n$) are eigenvectors of $A$.
Can any $d-$dimensional invariant subspace of $A$ be constructed using $d$ numbers of $v_i$'s?
linear-algebra matrices linear-transformations
asked Jan 10 at 7:00
Bashir SadeghiBashir Sadeghi
424
$endgroup$
1
$begingroup$
What do you mean by "constructed"?
$endgroup$
– BigbearZzz
Jan 10 at 7:13
6
$begingroup$
No, even assuming that the vectors are linearly independent and exhaust the eigenspaces. Consider a rotation of $mathbb{R}^3$ about the $z$-axis, e.g. $left(begin{array}{ccc}0 & -1& 0\1 & 0 & 0\0 & 0 & 1end{array}right)$. The only eigenvectors are multiples of $(0,0,1)$, but $W={(x,y,0)mid x,yinmathbb{R}}$ is invariant.
$endgroup$
– Arturo Magidin
Jan 10 at 7:13
$begingroup$
The point is precisely that invariant subspaces are more general than spaces generated by eigenvectors...
$endgroup$
– Arturo Magidin
Jan 10 at 7:14
$begingroup$
@BigbearZzz, by 'constructing' I mean linear combination.
$endgroup$
– Bashir Sadeghi
Jan 10 at 7:17
1
$begingroup$
However, even in the case of a diagonalizable matrix, the result is not quite right. For example, if you take the identity matrix and you take your $v_i$ to be the standard basis, then any subspace is invariant, but the only subspaces you can generate with subsets of your $v_i$ are the coordinate subspaces. E.g., with $V=mathbb{R}^2$, $v_1=(1,0)$, and $v_2=(0,1)$, you can’t get the subspace ${(x,x)mid xinmathbb{R}}$ which is invariant under the identity map.
$endgroup$
– Arturo Magidin
Jan 14 at 3:26
|
show 2 more comments
2
2
2
$begingroup$
Let $Ain mathbb{R}^{ntimes n}$ (i.e., a real square matrix).
Assume that $v_1,dots, v_m$ ($mle n$) are eigenvectors of $A$.
Can any $d-$dimensional invariant subspace of $A$ be constructed using $d$ numbers of $v_i$'s?
linear-algebra matrices linear-transformations
asked Jan 10 at 7:00
Bashir SadeghiBashir Sadeghi
424
$endgroup$
Let $Ain mathbb{R}^{ntimes n}$ (i.e., a real square matrix).
Assume that $v_1,dots, v_m$ ($mle n$) are eigenvectors of $A$.
Can any $d-$dimensional invariant subspace of $A$ be constructed using $d$ numbers of $v_i$'s?
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Jan 10 at 7:00
Bashir SadeghiBashir Sadeghi
424
asked Jan 10 at 7:00
Bashir SadeghiBashir Sadeghi
424
edited Jan 10 at 7:08
Bashir Sadeghi
asked Jan 10 at 7:00
Bashir SadeghiBashir Sadeghi
424
asked Jan 10 at 7:00
Bashir SadeghiBashir Sadeghi
424
asked Jan 10 at 7:00
Bashir SadeghiBashir Sadeghi
424
424
1
$begingroup$
What do you mean by "constructed"?
$endgroup$
– BigbearZzz
Jan 10 at 7:13
6
$begingroup$
No, even assuming that the vectors are linearly independent and exhaust the eigenspaces. Consider a rotation of $mathbb{R}^3$ about the $z$-axis, e.g. $left(begin{array}{ccc}0 & -1& 0\1 & 0 & 0\0 & 0 & 1end{array}right)$. The only eigenvectors are multiples of $(0,0,1)$, but $W={(x,y,0)mid x,yinmathbb{R}}$ is invariant.
$endgroup$
– Arturo Magidin
Jan 10 at 7:13
$begingroup$
The point is precisely that invariant subspaces are more general than spaces generated by eigenvectors...
$endgroup$
– Arturo Magidin
Jan 10 at 7:14
$begingroup$
@BigbearZzz, by 'constructing' I mean linear combination.
$endgroup$
– Bashir Sadeghi
Jan 10 at 7:17
1
$begingroup$
However, even in the case of a diagonalizable matrix, the result is not quite right. For example, if you take the identity matrix and you take your $v_i$ to be the standard basis, then any subspace is invariant, but the only subspaces you can generate with subsets of your $v_i$ are the coordinate subspaces. E.g., with $V=mathbb{R}^2$, $v_1=(1,0)$, and $v_2=(0,1)$, you can’t get the subspace ${(x,x)mid xinmathbb{R}}$ which is invariant under the identity map.
$endgroup$
– Arturo Magidin
Jan 14 at 3:26
|
show 2 more comments
1
$begingroup$
What do you mean by "constructed"?
$endgroup$
– BigbearZzz
Jan 10 at 7:13
6
$begingroup$
No, even assuming that the vectors are linearly independent and exhaust the eigenspaces. Consider a rotation of $mathbb{R}^3$ about the $z$-axis, e.g. $left(begin{array}{ccc}0 & -1& 0\1 & 0 & 0\0 & 0 & 1end{array}right)$. The only eigenvectors are multiples of $(0,0,1)$, but $W={(x,y,0)mid x,yinmathbb{R}}$ is invariant.
$endgroup$
– Arturo Magidin
Jan 10 at 7:13
$begingroup$
The point is precisely that invariant subspaces are more general than spaces generated by eigenvectors...
$endgroup$
– Arturo Magidin
Jan 10 at 7:14
$begingroup$
@BigbearZzz, by 'constructing' I mean linear combination.
$endgroup$
– Bashir Sadeghi
Jan 10 at 7:17
1
$begingroup$
However, even in the case of a diagonalizable matrix, the result is not quite right. For example, if you take the identity matrix and you take your $v_i$ to be the standard basis, then any subspace is invariant, but the only subspaces you can generate with subsets of your $v_i$ are the coordinate subspaces. E.g., with $V=mathbb{R}^2$, $v_1=(1,0)$, and $v_2=(0,1)$, you can’t get the subspace ${(x,x)mid xinmathbb{R}}$ which is invariant under the identity map.
$endgroup$
– Arturo Magidin
Jan 14 at 3:26
1
1
$begingroup$
What do you mean by "constructed"?
$endgroup$
– BigbearZzz
Jan 10 at 7:13
$begingroup$
What do you mean by "constructed"?
$endgroup$
– BigbearZzz
Jan 10 at 7:13
6
6
$begingroup$
No, even assuming that the vectors are linearly independent and exhaust the eigenspaces. Consider a rotation of $mathbb{R}^3$ about the $z$-axis, e.g. $left(begin{array}{ccc}0 & -1& 0\1 & 0 & 0\0 & 0 & 1end{array}right)$. The only eigenvectors are multiples of $(0,0,1)$, but $W={(x,y,0)mid x,yinmathbb{R}}$ is invariant.
$endgroup$
– Arturo Magidin
Jan 10 at 7:13
$begingroup$
No, even assuming that the vectors are linearly independent and exhaust the eigenspaces. Consider a rotation of $mathbb{R}^3$ about the $z$-axis, e.g. $left(begin{array}{ccc}0 & -1& 0\1 & 0 & 0\0 & 0 & 1end{array}right)$. The only eigenvectors are multiples of $(0,0,1)$, but $W={(x,y,0)mid x,yinmathbb{R}}$ is invariant.
$endgroup$
– Arturo Magidin
Jan 10 at 7:13
$begingroup$
The point is precisely that invariant subspaces are more general than spaces generated by eigenvectors...
$endgroup$
– Arturo Magidin
Jan 10 at 7:14
$begingroup$
The point is precisely that invariant subspaces are more general than spaces generated by eigenvectors...
$endgroup$
– Arturo Magidin
Jan 10 at 7:14
$begingroup$
@BigbearZzz, by 'constructing' I mean linear combination.
$endgroup$
– Bashir Sadeghi
Jan 10 at 7:17
$begingroup$
@BigbearZzz, by 'constructing' I mean linear combination.
$endgroup$
– Bashir Sadeghi
Jan 10 at 7:17
1
1
$begingroup$
However, even in the case of a diagonalizable matrix, the result is not quite right. For example, if you take the identity matrix and you take your $v_i$ to be the standard basis, then any subspace is invariant, but the only subspaces you can generate with subsets of your $v_i$ are the coordinate subspaces. E.g., with $V=mathbb{R}^2$, $v_1=(1,0)$, and $v_2=(0,1)$, you can’t get the subspace ${(x,x)mid xinmathbb{R}}$ which is invariant under the identity map.
$endgroup$
– Arturo Magidin
Jan 14 at 3:26
$begingroup$
However, even in the case of a diagonalizable matrix, the result is not quite right. For example, if you take the identity matrix and you take your $v_i$ to be the standard basis, then any subspace is invariant, but the only subspaces you can generate with subsets of your $v_i$ are the coordinate subspaces. E.g., with $V=mathbb{R}^2$, $v_1=(1,0)$, and $v_2=(0,1)$, you can’t get the subspace ${(x,x)mid xinmathbb{R}}$ which is invariant under the identity map.
$endgroup$
– Arturo Magidin
Jan 14 at 3:26
|
show 2 more comments
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1
$begingroup$
What do you mean by "constructed"?
$endgroup$
– BigbearZzz
Jan 10 at 7:13
6
$begingroup$
No, even assuming that the vectors are linearly independent and exhaust the eigenspaces. Consider a rotation of $mathbb{R}^3$ about the $z$-axis, e.g. $left(begin{array}{ccc}0 & -1& 0\1 & 0 & 0\0 & 0 & 1end{array}right)$. The only eigenvectors are multiples of $(0,0,1)$, but $W={(x,y,0)mid x,yinmathbb{R}}$ is invariant.
$endgroup$
– Arturo Magidin
Jan 10 at 7:13
$begingroup$
The point is precisely that invariant subspaces are more general than spaces generated by eigenvectors...
$endgroup$
– Arturo Magidin
Jan 10 at 7:14
$begingroup$
@BigbearZzz, by 'constructing' I mean linear combination.
$endgroup$
– Bashir Sadeghi
Jan 10 at 7:17
1
$begingroup$
However, even in the case of a diagonalizable matrix, the result is not quite right. For example, if you take the identity matrix and you take your $v_i$ to be the standard basis, then any subspace is invariant, but the only subspaces you can generate with subsets of your $v_i$ are the coordinate subspaces. E.g., with $V=mathbb{R}^2$, $v_1=(1,0)$, and $v_2=(0,1)$, you can’t get the subspace ${(x,x)mid xinmathbb{R}}$ which is invariant under the identity map.
$endgroup$
– Arturo Magidin
Jan 14 at 3:26