Eigenspace with dimension equal to the domain
$begingroup$
Is it possible to have a linear transformation $T:Vto V$ ($V$ a vector space over field $mathbb{C}$, with dimension $n$) such that there is an eigenvalue lambda with an associated eigenspace which has a dimension $n$?
I am currently solving a problem and in the problem the Eigenspace is denoted as $W$ and the domain is denoted as $V$. Im wondering whether $W$ is a subset or a proper subset of $V$.
linear-algebra
edited Dec 28 '18 at 19:42
A. Goodier
3,59651427
asked Dec 28 '18 at 19:38
Sam.SSam.S
719
$endgroup$
add a comment |
$begingroup$
Is it possible to have a linear transformation $T:Vto V$ ($V$ a vector space over field $mathbb{C}$, with dimension $n$) such that there is an eigenvalue lambda with an associated eigenspace which has a dimension $n$?
I am currently solving a problem and in the problem the Eigenspace is denoted as $W$ and the domain is denoted as $V$. Im wondering whether $W$ is a subset or a proper subset of $V$.
linear-algebra
edited Dec 28 '18 at 19:42
A. Goodier
3,59651427
asked Dec 28 '18 at 19:38
Sam.SSam.S
719
$endgroup$
3
$begingroup$
Look at the identity operator.
$endgroup$
– David Mitra
Dec 28 '18 at 19:41
$begingroup$
Thank you. So W is a subset of V.
$endgroup$
– Sam.S
Dec 28 '18 at 19:43
add a comment |
$begingroup$
Is it possible to have a linear transformation $T:Vto V$ ($V$ a vector space over field $mathbb{C}$, with dimension $n$) such that there is an eigenvalue lambda with an associated eigenspace which has a dimension $n$?
I am currently solving a problem and in the problem the Eigenspace is denoted as $W$ and the domain is denoted as $V$. Im wondering whether $W$ is a subset or a proper subset of $V$.
linear-algebra
edited Dec 28 '18 at 19:42
A. Goodier
3,59651427
asked Dec 28 '18 at 19:38
Sam.SSam.S
719
$endgroup$
Is it possible to have a linear transformation $T:Vto V$ ($V$ a vector space over field $mathbb{C}$, with dimension $n$) such that there is an eigenvalue lambda with an associated eigenspace which has a dimension $n$?
I am currently solving a problem and in the problem the Eigenspace is denoted as $W$ and the domain is denoted as $V$. Im wondering whether $W$ is a subset or a proper subset of $V$.
linear-algebra
linear-algebra
edited Dec 28 '18 at 19:42
A. Goodier
3,59651427
asked Dec 28 '18 at 19:38
Sam.SSam.S
719
edited Dec 28 '18 at 19:42
A. Goodier
3,59651427
asked Dec 28 '18 at 19:38
Sam.SSam.S
719
edited Dec 28 '18 at 19:42
A. Goodier
3,59651427
edited Dec 28 '18 at 19:42
A. Goodier
3,59651427
edited Dec 28 '18 at 19:42
A. Goodier
3,59651427
3,59651427
asked Dec 28 '18 at 19:38
Sam.SSam.S
719
asked Dec 28 '18 at 19:38
Sam.SSam.S
719
asked Dec 28 '18 at 19:38
Sam.SSam.S
719
719
3
$begingroup$
Look at the identity operator.
$endgroup$
– David Mitra
Dec 28 '18 at 19:41
$begingroup$
Thank you. So W is a subset of V.
$endgroup$
– Sam.S
Dec 28 '18 at 19:43
add a comment |
3
$begingroup$
Look at the identity operator.
$endgroup$
– David Mitra
Dec 28 '18 at 19:41
$begingroup$
Thank you. So W is a subset of V.
$endgroup$
– Sam.S
Dec 28 '18 at 19:43
3
3
$begingroup$
Look at the identity operator.
$endgroup$
– David Mitra
Dec 28 '18 at 19:41
$begingroup$
Look at the identity operator.
$endgroup$
– David Mitra
Dec 28 '18 at 19:41
$begingroup$
Thank you. So W is a subset of V.
$endgroup$
– Sam.S
Dec 28 '18 at 19:43
$begingroup$
Thank you. So W is a subset of V.
$endgroup$
– Sam.S
Dec 28 '18 at 19:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can do this for any $lambda$. Fix $lambdainmathbb C$ and consider the linear operator $T:Vto V$ which sends $xmapsto lambda x$. This operator has one eigenvalue, $lambda$, whose corresponding eigenspace is $V$, which has dimension $n$.
The only way the eigenspace $W$ can have dimension $n$ is if it is all of $V$.
answered Dec 28 '18 at 19:44
DaveDave
9,22211033
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We can do this for any $lambda$. Fix $lambdainmathbb C$ and consider the linear operator $T:Vto V$ which sends $xmapsto lambda x$. This operator has one eigenvalue, $lambda$, whose corresponding eigenspace is $V$, which has dimension $n$.
The only way the eigenspace $W$ can have dimension $n$ is if it is all of $V$.
answered Dec 28 '18 at 19:44
DaveDave
9,22211033
$endgroup$
add a comment |
$begingroup$
We can do this for any $lambda$. Fix $lambdainmathbb C$ and consider the linear operator $T:Vto V$ which sends $xmapsto lambda x$. This operator has one eigenvalue, $lambda$, whose corresponding eigenspace is $V$, which has dimension $n$.
The only way the eigenspace $W$ can have dimension $n$ is if it is all of $V$.
answered Dec 28 '18 at 19:44
DaveDave
9,22211033
$endgroup$
add a comment |
$begingroup$
We can do this for any $lambda$. Fix $lambdainmathbb C$ and consider the linear operator $T:Vto V$ which sends $xmapsto lambda x$. This operator has one eigenvalue, $lambda$, whose corresponding eigenspace is $V$, which has dimension $n$.
The only way the eigenspace $W$ can have dimension $n$ is if it is all of $V$.
answered Dec 28 '18 at 19:44
DaveDave
9,22211033
$endgroup$
We can do this for any $lambda$. Fix $lambdainmathbb C$ and consider the linear operator $T:Vto V$ which sends $xmapsto lambda x$. This operator has one eigenvalue, $lambda$, whose corresponding eigenspace is $V$, which has dimension $n$.
The only way the eigenspace $W$ can have dimension $n$ is if it is all of $V$.
answered Dec 28 '18 at 19:44
DaveDave
9,22211033
answered Dec 28 '18 at 19:44
DaveDave
9,22211033
answered Dec 28 '18 at 19:44
DaveDave
9,22211033
answered Dec 28 '18 at 19:44
DaveDave
9,22211033
9,22211033
add a comment |
add a comment |
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3
$begingroup$
Look at the identity operator.
$endgroup$
– David Mitra
Dec 28 '18 at 19:41
$begingroup$
Thank you. So W is a subset of V.
$endgroup$
– Sam.S
Dec 28 '18 at 19:43