Mixing
Two Different Statements about Schur's Decomposition Theorem (Linear Algebra)
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My book: Linear Algebra Done Right 3rd Edition by Sheldon Axler states the Schur's Theorem as follows.
Schur's Theorem. Suppose $V$ is a finite-dimensional complex vector space and $T$ is a linear operator on $V$. Then $T$ has an upper-triangular matrix with respect to some orthonormal basis of $V$.
In other words, for any $n times n$ complex-valued matrix, there is an orthonormal basis in which that matrix is upper-triangular.
But many other textbooks state the Schur's Decomposition as follows.
Schur's Deocmposition Theorem. Suppose $V$ is a finite-dimensional complex vector space and $T$ is a linear operator on $V$. Let $B$ be a basis of $V$. Then there exist an unitary matrix $U$ such that $$ mathcal{M}(T,B) = U^ast A U$$
where $A$ is an upper triangular matrix.
Here, $mathcal{M}(T)$ denote matrix representation of $T$ respect to basis $B$.
What is the relationship between those two? Are they equivalent? If yes, how can I prove it?
linear-algebra
asked Jan 13 at 6:14
asdfasdfsdfsdf14asdfasdfsdfsdf14
133
$endgroup$
|
show 2 more comments
$begingroup$
My book: Linear Algebra Done Right 3rd Edition by Sheldon Axler states the Schur's Theorem as follows.
Schur's Theorem. Suppose $V$ is a finite-dimensional complex vector space and $T$ is a linear operator on $V$. Then $T$ has an upper-triangular matrix with respect to some orthonormal basis of $V$.
In other words, for any $n times n$ complex-valued matrix, there is an orthonormal basis in which that matrix is upper-triangular.
But many other textbooks state the Schur's Decomposition as follows.
Schur's Deocmposition Theorem. Suppose $V$ is a finite-dimensional complex vector space and $T$ is a linear operator on $V$. Let $B$ be a basis of $V$. Then there exist an unitary matrix $U$ such that $$ mathcal{M}(T,B) = U^ast A U$$
where $A$ is an upper triangular matrix.
Here, $mathcal{M}(T)$ denote matrix representation of $T$ respect to basis $B$.
What is the relationship between those two? Are they equivalent? If yes, how can I prove it?
linear-algebra
asked Jan 13 at 6:14
asdfasdfsdfsdf14asdfasdfsdfsdf14
133
$endgroup$
$begingroup$
The orthonormal basis in the first version of the theorem consists of the columns of $U^*$ in the second theorem. The $A$ in the second theorem is the upper-triangular matrix with respect to that basis.
$endgroup$
– angryavian
Jan 13 at 6:19
$begingroup$
@angryavian So you mean that $B$ is considered as standard basis?
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:43
$begingroup$
Yes, sorry, that's what I meant.
$endgroup$
– angryavian
Jan 13 at 6:48
$begingroup$
@xbh Oh.. so we can not freely choose basis $B$, rather it is determined by the theorem. Right? Many examples in the Internet starts from matrix, not from operator, so I was confused.
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:55
$begingroup$
Sorry, my mistake. I read the statement wrong. Ignore me.
$endgroup$
– xbh
Jan 13 at 6:58
|
show 2 more comments
$begingroup$
My book: Linear Algebra Done Right 3rd Edition by Sheldon Axler states the Schur's Theorem as follows.
Schur's Theorem. Suppose $V$ is a finite-dimensional complex vector space and $T$ is a linear operator on $V$. Then $T$ has an upper-triangular matrix with respect to some orthonormal basis of $V$.
In other words, for any $n times n$ complex-valued matrix, there is an orthonormal basis in which that matrix is upper-triangular.
But many other textbooks state the Schur's Decomposition as follows.
Schur's Deocmposition Theorem. Suppose $V$ is a finite-dimensional complex vector space and $T$ is a linear operator on $V$. Let $B$ be a basis of $V$. Then there exist an unitary matrix $U$ such that $$ mathcal{M}(T,B) = U^ast A U$$
where $A$ is an upper triangular matrix.
Here, $mathcal{M}(T)$ denote matrix representation of $T$ respect to basis $B$.
What is the relationship between those two? Are they equivalent? If yes, how can I prove it?
linear-algebra
asked Jan 13 at 6:14
asdfasdfsdfsdf14asdfasdfsdfsdf14
133
$endgroup$
My book: Linear Algebra Done Right 3rd Edition by Sheldon Axler states the Schur's Theorem as follows.
Schur's Theorem. Suppose $V$ is a finite-dimensional complex vector space and $T$ is a linear operator on $V$. Then $T$ has an upper-triangular matrix with respect to some orthonormal basis of $V$.
In other words, for any $n times n$ complex-valued matrix, there is an orthonormal basis in which that matrix is upper-triangular.
But many other textbooks state the Schur's Decomposition as follows.
Schur's Deocmposition Theorem. Suppose $V$ is a finite-dimensional complex vector space and $T$ is a linear operator on $V$. Let $B$ be a basis of $V$. Then there exist an unitary matrix $U$ such that $$ mathcal{M}(T,B) = U^ast A U$$
where $A$ is an upper triangular matrix.
Here, $mathcal{M}(T)$ denote matrix representation of $T$ respect to basis $B$.
What is the relationship between those two? Are they equivalent? If yes, how can I prove it?
linear-algebra
linear-algebra
asked Jan 13 at 6:14
asdfasdfsdfsdf14asdfasdfsdfsdf14
133
asked Jan 13 at 6:14
asdfasdfsdfsdf14asdfasdfsdfsdf14
133
asked Jan 13 at 6:14
asdfasdfsdfsdf14asdfasdfsdfsdf14
133
asked Jan 13 at 6:14
asdfasdfsdfsdf14asdfasdfsdfsdf14
133
asked Jan 13 at 6:14
asdfasdfsdfsdf14asdfasdfsdfsdf14
133
133
$begingroup$
The orthonormal basis in the first version of the theorem consists of the columns of $U^*$ in the second theorem. The $A$ in the second theorem is the upper-triangular matrix with respect to that basis.
$endgroup$
– angryavian
Jan 13 at 6:19
$begingroup$
@angryavian So you mean that $B$ is considered as standard basis?
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:43
$begingroup$
Yes, sorry, that's what I meant.
$endgroup$
– angryavian
Jan 13 at 6:48
$begingroup$
@xbh Oh.. so we can not freely choose basis $B$, rather it is determined by the theorem. Right? Many examples in the Internet starts from matrix, not from operator, so I was confused.
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:55
$begingroup$
Sorry, my mistake. I read the statement wrong. Ignore me.
$endgroup$
– xbh
Jan 13 at 6:58
|
show 2 more comments
$begingroup$
The orthonormal basis in the first version of the theorem consists of the columns of $U^*$ in the second theorem. The $A$ in the second theorem is the upper-triangular matrix with respect to that basis.
$endgroup$
– angryavian
Jan 13 at 6:19
$begingroup$
@angryavian So you mean that $B$ is considered as standard basis?
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:43
$begingroup$
Yes, sorry, that's what I meant.
$endgroup$
– angryavian
Jan 13 at 6:48
$begingroup$
@xbh Oh.. so we can not freely choose basis $B$, rather it is determined by the theorem. Right? Many examples in the Internet starts from matrix, not from operator, so I was confused.
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:55
$begingroup$
Sorry, my mistake. I read the statement wrong. Ignore me.
$endgroup$
– xbh
Jan 13 at 6:58
$begingroup$
The orthonormal basis in the first version of the theorem consists of the columns of $U^*$ in the second theorem. The $A$ in the second theorem is the upper-triangular matrix with respect to that basis.
$endgroup$
– angryavian
Jan 13 at 6:19
$begingroup$
The orthonormal basis in the first version of the theorem consists of the columns of $U^*$ in the second theorem. The $A$ in the second theorem is the upper-triangular matrix with respect to that basis.
$endgroup$
– angryavian
Jan 13 at 6:19
$begingroup$
@angryavian So you mean that $B$ is considered as standard basis?
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:43
$begingroup$
@angryavian So you mean that $B$ is considered as standard basis?
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:43
$begingroup$
Yes, sorry, that's what I meant.
$endgroup$
– angryavian
Jan 13 at 6:48
$begingroup$
Yes, sorry, that's what I meant.
$endgroup$
– angryavian
Jan 13 at 6:48
$begingroup$
@xbh Oh.. so we can not freely choose basis $B$, rather it is determined by the theorem. Right? Many examples in the Internet starts from matrix, not from operator, so I was confused.
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:55
$begingroup$
@xbh Oh.. so we can not freely choose basis $B$, rather it is determined by the theorem. Right? Many examples in the Internet starts from matrix, not from operator, so I was confused.
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:55
$begingroup$
Sorry, my mistake. I read the statement wrong. Ignore me.
$endgroup$
– xbh
Jan 13 at 6:58
$begingroup$
Sorry, my mistake. I read the statement wrong. Ignore me.
$endgroup$
– xbh
Jan 13 at 6:58
|
show 2 more comments
1 Answer
1
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oldest
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$begingroup$
Proof of the equivalency
$(2) implies (1)$
Let $mathcal E' = mathcal B U^*$, then $T (mathcal B) = mathcal B U^* A U $, or $T (mathcal E' U) = mathcal E' U U^* A U = mathcal E' A U $, hence $T (mathcal E') = mathcal E' A$, i.e. $mathcal M(T, mathcal E') = A$ is upper triangular. Now apply Gram-Schmidt orthogonalization process to $mathcal E'$ to obtain an orthonormal basis $mathcal E$, then $mathcal M(T, mathcal E)$ is still upper triangular, hence the statement.
$(1) implies (2)$
From now on, the bold-italic letters represent matrices, the regular letters are mappings, and the calligraphic letters are basis. Also $mathbb C^n cong mathrm M_{n,1}(mathbb C)$ is the collection of $n times 1$ complex matrices.
Let $boldsymbol S = mathcal M(T, mathcal B)$, on $W = mathbb C^n$, define linear operator $S colon boldsymbol x mapsto boldsymbol {Sx}$ where $boldsymbol x in mathbb C^n$, then apply $(1)$ to the operator $S$, there exists some ONB $mathcal F$ of $mathbb C^n$ that $mathcal M(S, mathcal F) = boldsymbol A$ is upper triangular. Let $mathcal E$ be the standard basis of $mathbb C^n$, and let $mathcal F = mathcal E boldsymbol U$, then $boldsymbol U$ is unitary. Write $boldsymbol U^*$ instead of $boldsymbol U$, i.e. $mathcal F = mathcal E boldsymbol U^*$ and $S(mathcal F) = mathcal F boldsymbol A$ becomes $S (mathcal E boldsymbol U^*) = mathcal E boldsymbol U^* boldsymbol A$, hence $S (mathcal E) = mathcal E (boldsymbol U^*boldsymbol {AU})$. But $S (mathcal E) = boldsymbol S$, hence $boldsymbol S = boldsymbol {U}^* boldsymbol {AU}$. Hence the statement.
answered Jan 13 at 9:05
xbhxbh
6,1351522
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add a comment |
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$begingroup$
Proof of the equivalency
$(2) implies (1)$
Let $mathcal E' = mathcal B U^*$, then $T (mathcal B) = mathcal B U^* A U $, or $T (mathcal E' U) = mathcal E' U U^* A U = mathcal E' A U $, hence $T (mathcal E') = mathcal E' A$, i.e. $mathcal M(T, mathcal E') = A$ is upper triangular. Now apply Gram-Schmidt orthogonalization process to $mathcal E'$ to obtain an orthonormal basis $mathcal E$, then $mathcal M(T, mathcal E)$ is still upper triangular, hence the statement.
$(1) implies (2)$
From now on, the bold-italic letters represent matrices, the regular letters are mappings, and the calligraphic letters are basis. Also $mathbb C^n cong mathrm M_{n,1}(mathbb C)$ is the collection of $n times 1$ complex matrices.
Let $boldsymbol S = mathcal M(T, mathcal B)$, on $W = mathbb C^n$, define linear operator $S colon boldsymbol x mapsto boldsymbol {Sx}$ where $boldsymbol x in mathbb C^n$, then apply $(1)$ to the operator $S$, there exists some ONB $mathcal F$ of $mathbb C^n$ that $mathcal M(S, mathcal F) = boldsymbol A$ is upper triangular. Let $mathcal E$ be the standard basis of $mathbb C^n$, and let $mathcal F = mathcal E boldsymbol U$, then $boldsymbol U$ is unitary. Write $boldsymbol U^*$ instead of $boldsymbol U$, i.e. $mathcal F = mathcal E boldsymbol U^*$ and $S(mathcal F) = mathcal F boldsymbol A$ becomes $S (mathcal E boldsymbol U^*) = mathcal E boldsymbol U^* boldsymbol A$, hence $S (mathcal E) = mathcal E (boldsymbol U^*boldsymbol {AU})$. But $S (mathcal E) = boldsymbol S$, hence $boldsymbol S = boldsymbol {U}^* boldsymbol {AU}$. Hence the statement.
answered Jan 13 at 9:05
xbhxbh
6,1351522
$endgroup$
add a comment |
$begingroup$
Proof of the equivalency
$(2) implies (1)$
Let $mathcal E' = mathcal B U^*$, then $T (mathcal B) = mathcal B U^* A U $, or $T (mathcal E' U) = mathcal E' U U^* A U = mathcal E' A U $, hence $T (mathcal E') = mathcal E' A$, i.e. $mathcal M(T, mathcal E') = A$ is upper triangular. Now apply Gram-Schmidt orthogonalization process to $mathcal E'$ to obtain an orthonormal basis $mathcal E$, then $mathcal M(T, mathcal E)$ is still upper triangular, hence the statement.
$(1) implies (2)$
From now on, the bold-italic letters represent matrices, the regular letters are mappings, and the calligraphic letters are basis. Also $mathbb C^n cong mathrm M_{n,1}(mathbb C)$ is the collection of $n times 1$ complex matrices.
Let $boldsymbol S = mathcal M(T, mathcal B)$, on $W = mathbb C^n$, define linear operator $S colon boldsymbol x mapsto boldsymbol {Sx}$ where $boldsymbol x in mathbb C^n$, then apply $(1)$ to the operator $S$, there exists some ONB $mathcal F$ of $mathbb C^n$ that $mathcal M(S, mathcal F) = boldsymbol A$ is upper triangular. Let $mathcal E$ be the standard basis of $mathbb C^n$, and let $mathcal F = mathcal E boldsymbol U$, then $boldsymbol U$ is unitary. Write $boldsymbol U^*$ instead of $boldsymbol U$, i.e. $mathcal F = mathcal E boldsymbol U^*$ and $S(mathcal F) = mathcal F boldsymbol A$ becomes $S (mathcal E boldsymbol U^*) = mathcal E boldsymbol U^* boldsymbol A$, hence $S (mathcal E) = mathcal E (boldsymbol U^*boldsymbol {AU})$. But $S (mathcal E) = boldsymbol S$, hence $boldsymbol S = boldsymbol {U}^* boldsymbol {AU}$. Hence the statement.
answered Jan 13 at 9:05
xbhxbh
6,1351522
$endgroup$
add a comment |
$begingroup$
Proof of the equivalency
$(2) implies (1)$
Let $mathcal E' = mathcal B U^*$, then $T (mathcal B) = mathcal B U^* A U $, or $T (mathcal E' U) = mathcal E' U U^* A U = mathcal E' A U $, hence $T (mathcal E') = mathcal E' A$, i.e. $mathcal M(T, mathcal E') = A$ is upper triangular. Now apply Gram-Schmidt orthogonalization process to $mathcal E'$ to obtain an orthonormal basis $mathcal E$, then $mathcal M(T, mathcal E)$ is still upper triangular, hence the statement.
$(1) implies (2)$
From now on, the bold-italic letters represent matrices, the regular letters are mappings, and the calligraphic letters are basis. Also $mathbb C^n cong mathrm M_{n,1}(mathbb C)$ is the collection of $n times 1$ complex matrices.
Let $boldsymbol S = mathcal M(T, mathcal B)$, on $W = mathbb C^n$, define linear operator $S colon boldsymbol x mapsto boldsymbol {Sx}$ where $boldsymbol x in mathbb C^n$, then apply $(1)$ to the operator $S$, there exists some ONB $mathcal F$ of $mathbb C^n$ that $mathcal M(S, mathcal F) = boldsymbol A$ is upper triangular. Let $mathcal E$ be the standard basis of $mathbb C^n$, and let $mathcal F = mathcal E boldsymbol U$, then $boldsymbol U$ is unitary. Write $boldsymbol U^*$ instead of $boldsymbol U$, i.e. $mathcal F = mathcal E boldsymbol U^*$ and $S(mathcal F) = mathcal F boldsymbol A$ becomes $S (mathcal E boldsymbol U^*) = mathcal E boldsymbol U^* boldsymbol A$, hence $S (mathcal E) = mathcal E (boldsymbol U^*boldsymbol {AU})$. But $S (mathcal E) = boldsymbol S$, hence $boldsymbol S = boldsymbol {U}^* boldsymbol {AU}$. Hence the statement.
answered Jan 13 at 9:05
xbhxbh
6,1351522
$endgroup$
Proof of the equivalency
$(2) implies (1)$
Let $mathcal E' = mathcal B U^*$, then $T (mathcal B) = mathcal B U^* A U $, or $T (mathcal E' U) = mathcal E' U U^* A U = mathcal E' A U $, hence $T (mathcal E') = mathcal E' A$, i.e. $mathcal M(T, mathcal E') = A$ is upper triangular. Now apply Gram-Schmidt orthogonalization process to $mathcal E'$ to obtain an orthonormal basis $mathcal E$, then $mathcal M(T, mathcal E)$ is still upper triangular, hence the statement.
$(1) implies (2)$
From now on, the bold-italic letters represent matrices, the regular letters are mappings, and the calligraphic letters are basis. Also $mathbb C^n cong mathrm M_{n,1}(mathbb C)$ is the collection of $n times 1$ complex matrices.
Let $boldsymbol S = mathcal M(T, mathcal B)$, on $W = mathbb C^n$, define linear operator $S colon boldsymbol x mapsto boldsymbol {Sx}$ where $boldsymbol x in mathbb C^n$, then apply $(1)$ to the operator $S$, there exists some ONB $mathcal F$ of $mathbb C^n$ that $mathcal M(S, mathcal F) = boldsymbol A$ is upper triangular. Let $mathcal E$ be the standard basis of $mathbb C^n$, and let $mathcal F = mathcal E boldsymbol U$, then $boldsymbol U$ is unitary. Write $boldsymbol U^*$ instead of $boldsymbol U$, i.e. $mathcal F = mathcal E boldsymbol U^*$ and $S(mathcal F) = mathcal F boldsymbol A$ becomes $S (mathcal E boldsymbol U^*) = mathcal E boldsymbol U^* boldsymbol A$, hence $S (mathcal E) = mathcal E (boldsymbol U^*boldsymbol {AU})$. But $S (mathcal E) = boldsymbol S$, hence $boldsymbol S = boldsymbol {U}^* boldsymbol {AU}$. Hence the statement.
answered Jan 13 at 9:05
xbhxbh
6,1351522
answered Jan 13 at 9:05
xbhxbh
6,1351522
answered Jan 13 at 9:05
xbhxbh
6,1351522
answered Jan 13 at 9:05
xbhxbh
6,1351522
6,1351522
add a comment |
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$begingroup$
The orthonormal basis in the first version of the theorem consists of the columns of $U^*$ in the second theorem. The $A$ in the second theorem is the upper-triangular matrix with respect to that basis.
$endgroup$
– angryavian
Jan 13 at 6:19
$begingroup$
@angryavian So you mean that $B$ is considered as standard basis?
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:43
$begingroup$
Yes, sorry, that's what I meant.
$endgroup$
– angryavian
Jan 13 at 6:48
$begingroup$
@xbh Oh.. so we can not freely choose basis $B$, rather it is determined by the theorem. Right? Many examples in the Internet starts from matrix, not from operator, so I was confused.
$endgroup$
– asdfasdfsdfsdf14
Jan 13 at 6:55
$begingroup$
Sorry, my mistake. I read the statement wrong. Ignore me.
$endgroup$
– xbh
Jan 13 at 6:58