Eigenspace and dimensions of linear transformation in complex plane
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Suppose V is a finite-dimensional complex vector space and T : V → V is a linear transformation which satisfies $T^{2} = T$. Prove that for any vector v $in$ V , the vector T(v) is contained in the eigenspace $E_{1}$ of eigenvalue $1$ and v −T(v) is contained in the eigenspace $E_{0}$ of eigenvalue $0$. Also prove that T is diagonalizable, and compute the dimensions of $E_{0}$ and $E_{1}$ in terms of the rank of T.
From $T^{2}-T=0$, I could get $lambda=0$ or $lambda= 1$ but I do not know how to continue.
linear-algebra linear-transformations
asked Dec 10 '18 at 13:56
AdamAdam
1
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add a comment |
$begingroup$
Suppose V is a finite-dimensional complex vector space and T : V → V is a linear transformation which satisfies $T^{2} = T$. Prove that for any vector v $in$ V , the vector T(v) is contained in the eigenspace $E_{1}$ of eigenvalue $1$ and v −T(v) is contained in the eigenspace $E_{0}$ of eigenvalue $0$. Also prove that T is diagonalizable, and compute the dimensions of $E_{0}$ and $E_{1}$ in terms of the rank of T.
From $T^{2}-T=0$, I could get $lambda=0$ or $lambda= 1$ but I do not know how to continue.
linear-algebra linear-transformations
asked Dec 10 '18 at 13:56
AdamAdam
1
$endgroup$
add a comment |
$begingroup$
Suppose V is a finite-dimensional complex vector space and T : V → V is a linear transformation which satisfies $T^{2} = T$. Prove that for any vector v $in$ V , the vector T(v) is contained in the eigenspace $E_{1}$ of eigenvalue $1$ and v −T(v) is contained in the eigenspace $E_{0}$ of eigenvalue $0$. Also prove that T is diagonalizable, and compute the dimensions of $E_{0}$ and $E_{1}$ in terms of the rank of T.
From $T^{2}-T=0$, I could get $lambda=0$ or $lambda= 1$ but I do not know how to continue.
linear-algebra linear-transformations
asked Dec 10 '18 at 13:56
AdamAdam
1
$endgroup$
Suppose V is a finite-dimensional complex vector space and T : V → V is a linear transformation which satisfies $T^{2} = T$. Prove that for any vector v $in$ V , the vector T(v) is contained in the eigenspace $E_{1}$ of eigenvalue $1$ and v −T(v) is contained in the eigenspace $E_{0}$ of eigenvalue $0$. Also prove that T is diagonalizable, and compute the dimensions of $E_{0}$ and $E_{1}$ in terms of the rank of T.
From $T^{2}-T=0$, I could get $lambda=0$ or $lambda= 1$ but I do not know how to continue.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Dec 10 '18 at 13:56
AdamAdam
1
asked Dec 10 '18 at 13:56
AdamAdam
1
asked Dec 10 '18 at 13:56
AdamAdam
1
asked Dec 10 '18 at 13:56
AdamAdam
1
asked Dec 10 '18 at 13:56
AdamAdam
1
1
add a comment |
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Hint: If $vin V$, then $v=bigl(v-T(v)bigr)+T(v)$.
answered Dec 10 '18 at 14:15
José Carlos SantosJosé Carlos Santos
156k22126227
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1 Answer
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$begingroup$
Hint: If $vin V$, then $v=bigl(v-T(v)bigr)+T(v)$.
answered Dec 10 '18 at 14:15
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
add a comment |
$begingroup$
Hint: If $vin V$, then $v=bigl(v-T(v)bigr)+T(v)$.
answered Dec 10 '18 at 14:15
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
add a comment |
$begingroup$
Hint: If $vin V$, then $v=bigl(v-T(v)bigr)+T(v)$.
answered Dec 10 '18 at 14:15
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
Hint: If $vin V$, then $v=bigl(v-T(v)bigr)+T(v)$.
answered Dec 10 '18 at 14:15
José Carlos SantosJosé Carlos Santos
156k22126227
answered Dec 10 '18 at 14:15
José Carlos SantosJosé Carlos Santos
156k22126227
answered Dec 10 '18 at 14:15
José Carlos SantosJosé Carlos Santos
156k22126227
answered Dec 10 '18 at 14:15
José Carlos SantosJosé Carlos Santos
156k22126227
156k22126227
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