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If $Ain M_ntimes n^mathbbC$ and $T(X)=AX$ for all $Xin M_ntimes :n^mathbbC$ then $T$ is normal iff $A$ is normal
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My try:
$underlineRightarrow :$ Assume $T$ is a normal linear transformation.
Therefore there exists a basis $B=leftX_1,...,X_nright$
for $M_ntimes :n^mathbbC$ that contains eigenvectors of $T$.
Assuming $lambda _j$ is an egienvalue of the egienvector $X_j$ then
$lambda _jX_j=T(X_j)=AX_j$
Therefore $lambda _j$ is an eigenvalue of A and $X_j$ is the corresponding egienvector of the eigenvalue $lambda _j$ . So we can conclude A has $n$ eigenvectors that form a basis for $M_ntimes :n^mathbbC$ and therefore A is normal.
the other direction can be showed in a similar manner.
In the proof I relied on a theorem that say: If $V$ is a finite vector space and $T:V to V$ is a linear transformation over $mathbbC$ then $T$ is normal iff there exists an orthonormal basis for $V$ that contains eigenvectors of $T$...
The same goes for A.
I Would like to know what you think about my proof please.
linear-algebra eigenvalues-eigenvectors
asked Jul 18 at 15:51
idan di
372110
add a comment |Â
up vote
2
down vote
favorite
My try:
$underlineRightarrow :$ Assume $T$ is a normal linear transformation.
Therefore there exists a basis $B=leftX_1,...,X_nright$
for $M_ntimes :n^mathbbC$ that contains eigenvectors of $T$.
Assuming $lambda _j$ is an egienvalue of the egienvector $X_j$ then
$lambda _jX_j=T(X_j)=AX_j$
Therefore $lambda _j$ is an eigenvalue of A and $X_j$ is the corresponding egienvector of the eigenvalue $lambda _j$ . So we can conclude A has $n$ eigenvectors that form a basis for $M_ntimes :n^mathbbC$ and therefore A is normal.
the other direction can be showed in a similar manner.
In the proof I relied on a theorem that say: If $V$ is a finite vector space and $T:V to V$ is a linear transformation over $mathbbC$ then $T$ is normal iff there exists an orthonormal basis for $V$ that contains eigenvectors of $T$...
The same goes for A.
I Would like to know what you think about my proof please.
linear-algebra eigenvalues-eigenvectors
asked Jul 18 at 15:51
idan di
372110
1
You seem to think that "normal" means "diagonalizable", but "normal" is more. It means that $A^*A = AA^*$. Hence, the question arises which scalar product is used on $M_ntimes n^mathbb C$ to define the adjoint $T^*$. I guess it's $langle X,Yrangle := operatornameTr(XY^*)$.
– amsmath
Jul 18 at 18:02
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
My try:
$underlineRightarrow :$ Assume $T$ is a normal linear transformation.
Therefore there exists a basis $B=leftX_1,...,X_nright$
for $M_ntimes :n^mathbbC$ that contains eigenvectors of $T$.
Assuming $lambda _j$ is an egienvalue of the egienvector $X_j$ then
$lambda _jX_j=T(X_j)=AX_j$
Therefore $lambda _j$ is an eigenvalue of A and $X_j$ is the corresponding egienvector of the eigenvalue $lambda _j$ . So we can conclude A has $n$ eigenvectors that form a basis for $M_ntimes :n^mathbbC$ and therefore A is normal.
the other direction can be showed in a similar manner.
In the proof I relied on a theorem that say: If $V$ is a finite vector space and $T:V to V$ is a linear transformation over $mathbbC$ then $T$ is normal iff there exists an orthonormal basis for $V$ that contains eigenvectors of $T$...
The same goes for A.
I Would like to know what you think about my proof please.
linear-algebra eigenvalues-eigenvectors
asked Jul 18 at 15:51
idan di
372110
My try:
$underlineRightarrow :$ Assume $T$ is a normal linear transformation.
Therefore there exists a basis $B=leftX_1,...,X_nright$
for $M_ntimes :n^mathbbC$ that contains eigenvectors of $T$.
Assuming $lambda _j$ is an egienvalue of the egienvector $X_j$ then
$lambda _jX_j=T(X_j)=AX_j$
Therefore $lambda _j$ is an eigenvalue of A and $X_j$ is the corresponding egienvector of the eigenvalue $lambda _j$ . So we can conclude A has $n$ eigenvectors that form a basis for $M_ntimes :n^mathbbC$ and therefore A is normal.
the other direction can be showed in a similar manner.
In the proof I relied on a theorem that say: If $V$ is a finite vector space and $T:V to V$ is a linear transformation over $mathbbC$ then $T$ is normal iff there exists an orthonormal basis for $V$ that contains eigenvectors of $T$...
The same goes for A.
I Would like to know what you think about my proof please.
linear-algebra eigenvalues-eigenvectors
asked Jul 18 at 15:51
idan di
372110
edited Jul 18 at 15:56
asked Jul 18 at 15:51
idan di
372110
asked Jul 18 at 15:51
idan di
372110
asked Jul 18 at 15:51
idan di
372110
372110
1
You seem to think that "normal" means "diagonalizable", but "normal" is more. It means that $A^*A = AA^*$. Hence, the question arises which scalar product is used on $M_ntimes n^mathbb C$ to define the adjoint $T^*$. I guess it's $langle X,Yrangle := operatornameTr(XY^*)$.
– amsmath
Jul 18 at 18:02
add a comment |Â
1
You seem to think that "normal" means "diagonalizable", but "normal" is more. It means that $A^*A = AA^*$. Hence, the question arises which scalar product is used on $M_ntimes n^mathbb C$ to define the adjoint $T^*$. I guess it's $langle X,Yrangle := operatornameTr(XY^*)$.
– amsmath
Jul 18 at 18:02
1
1
You seem to think that "normal" means "diagonalizable", but "normal" is more. It means that $A^*A = AA^*$. Hence, the question arises which scalar product is used on $M_ntimes n^mathbb C$ to define the adjoint $T^*$. I guess it's $langle X,Yrangle := operatornameTr(XY^*)$.
– amsmath
Jul 18 at 18:02
You seem to think that "normal" means "diagonalizable", but "normal" is more. It means that $A^*A = AA^*$. Hence, the question arises which scalar product is used on $M_ntimes n^mathbb C$ to define the adjoint $T^*$. I guess it's $langle X,Yrangle := operatornameTr(XY^*)$.
– amsmath
Jul 18 at 18:02
add a comment |Â
1 Answer
1
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1
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Let us use the scalar product $(X,Y) := operatornameTr(XY^*)$ on $M_ntimes n^mathbb C$. Then
$$
(T(X),Y) = (AX,Y) = operatornameTr(AXY^*) = operatornameTr(XY^*A) = operatornameTr(X(A^*Y)^*) = (X,A^*Y).
$$
Hence, $T^*(Y) = A^*Y$. Now it is really simple to show that $T^*T = TT^*$ is equivalent to $A^*A = AA^*$.
answered Jul 18 at 18:09
amsmath
1,613114
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let us use the scalar product $(X,Y) := operatornameTr(XY^*)$ on $M_ntimes n^mathbb C$. Then
$$
(T(X),Y) = (AX,Y) = operatornameTr(AXY^*) = operatornameTr(XY^*A) = operatornameTr(X(A^*Y)^*) = (X,A^*Y).
$$
Hence, $T^*(Y) = A^*Y$. Now it is really simple to show that $T^*T = TT^*$ is equivalent to $A^*A = AA^*$.
answered Jul 18 at 18:09
amsmath
1,613114
add a comment |Â
up vote
1
down vote
accepted
Let us use the scalar product $(X,Y) := operatornameTr(XY^*)$ on $M_ntimes n^mathbb C$. Then
$$
(T(X),Y) = (AX,Y) = operatornameTr(AXY^*) = operatornameTr(XY^*A) = operatornameTr(X(A^*Y)^*) = (X,A^*Y).
$$
Hence, $T^*(Y) = A^*Y$. Now it is really simple to show that $T^*T = TT^*$ is equivalent to $A^*A = AA^*$.
answered Jul 18 at 18:09
amsmath
1,613114
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let us use the scalar product $(X,Y) := operatornameTr(XY^*)$ on $M_ntimes n^mathbb C$. Then
$$
(T(X),Y) = (AX,Y) = operatornameTr(AXY^*) = operatornameTr(XY^*A) = operatornameTr(X(A^*Y)^*) = (X,A^*Y).
$$
Hence, $T^*(Y) = A^*Y$. Now it is really simple to show that $T^*T = TT^*$ is equivalent to $A^*A = AA^*$.
answered Jul 18 at 18:09
amsmath
1,613114
Let us use the scalar product $(X,Y) := operatornameTr(XY^*)$ on $M_ntimes n^mathbb C$. Then
$$
(T(X),Y) = (AX,Y) = operatornameTr(AXY^*) = operatornameTr(XY^*A) = operatornameTr(X(A^*Y)^*) = (X,A^*Y).
$$
Hence, $T^*(Y) = A^*Y$. Now it is really simple to show that $T^*T = TT^*$ is equivalent to $A^*A = AA^*$.
answered Jul 18 at 18:09
amsmath
1,613114
answered Jul 18 at 18:09
amsmath
1,613114
answered Jul 18 at 18:09
amsmath
1,613114
answered Jul 18 at 18:09
amsmath
1,613114
1,613114
add a comment |Â
add a comment |Â
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1
You seem to think that "normal" means "diagonalizable", but "normal" is more. It means that $A^*A = AA^*$. Hence, the question arises which scalar product is used on $M_ntimes n^mathbb C$ to define the adjoint $T^*$. I guess it's $langle X,Yrangle := operatornameTr(XY^*)$.
– amsmath
Jul 18 at 18:02