Binding eigenvalue properties with normal operators

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Let $N$ be a normal transformation over $Bbb C$.

A) True/False:$$N = N^* LeftarrowRightarrow lambda in Bbb R$$
(where $lambda$ is any eigenvalue of $N$)

B) True/False:$$N^* = N^-1 LeftarrowRightarrow |lambda|=1$$
(where $lambda$ is any eigenvalue of $N$)

---

In both cases, I am able to prove the direction $Rightarrow$ quite easily.
but in the other direction i am having a struggle (perhaps its false?).

for ex. statement B: I thought of showing that since $N$ is normal, then it is diagonalizable, therefore $N^*$ is also diagonalizable with the conjugate eigenvalues on its diagonal, therefore for any $lambda = (a+bi)$ then $NN^*$ has $(a+bi)(a-bi)$ on its diagonal, which is the identity matrix (since $|lambda|=1$) and therefore $N$ is unitary, but i'm not sure my transitions are valid.

Anyway, any help much appreciated on both statements!

linear-algebra eigenvalues-eigenvectors adjoint-operators

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asked Jul 30 at 12:20

Limitless

986

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up vote
2
down vote

favorite

Let $N$ be a normal transformation over $Bbb C$.

A) True/False:$$N = N^* LeftarrowRightarrow lambda in Bbb R$$
(where $lambda$ is any eigenvalue of $N$)

B) True/False:$$N^* = N^-1 LeftarrowRightarrow |lambda|=1$$
(where $lambda$ is any eigenvalue of $N$)

---

In both cases, I am able to prove the direction $Rightarrow$ quite easily.
but in the other direction i am having a struggle (perhaps its false?).

for ex. statement B: I thought of showing that since $N$ is normal, then it is diagonalizable, therefore $N^*$ is also diagonalizable with the conjugate eigenvalues on its diagonal, therefore for any $lambda = (a+bi)$ then $NN^*$ has $(a+bi)(a-bi)$ on its diagonal, which is the identity matrix (since $|lambda|=1$) and therefore $N$ is unitary, but i'm not sure my transitions are valid.

Anyway, any help much appreciated on both statements!

linear-algebra eigenvalues-eigenvectors adjoint-operators

share|cite|improve this question

asked Jul 30 at 12:20

Limitless

986

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $N$ be a normal transformation over $Bbb C$.

A) True/False:$$N = N^* LeftarrowRightarrow lambda in Bbb R$$
(where $lambda$ is any eigenvalue of $N$)

B) True/False:$$N^* = N^-1 LeftarrowRightarrow |lambda|=1$$
(where $lambda$ is any eigenvalue of $N$)

---

In both cases, I am able to prove the direction $Rightarrow$ quite easily.
but in the other direction i am having a struggle (perhaps its false?).

for ex. statement B: I thought of showing that since $N$ is normal, then it is diagonalizable, therefore $N^*$ is also diagonalizable with the conjugate eigenvalues on its diagonal, therefore for any $lambda = (a+bi)$ then $NN^*$ has $(a+bi)(a-bi)$ on its diagonal, which is the identity matrix (since $|lambda|=1$) and therefore $N$ is unitary, but i'm not sure my transitions are valid.

Anyway, any help much appreciated on both statements!

linear-algebra eigenvalues-eigenvectors adjoint-operators

share|cite|improve this question

asked Jul 30 at 12:20

Limitless

986

Let $N$ be a normal transformation over $Bbb C$.

A) True/False:$$N = N^* LeftarrowRightarrow lambda in Bbb R$$
(where $lambda$ is any eigenvalue of $N$)

B) True/False:$$N^* = N^-1 LeftarrowRightarrow |lambda|=1$$
(where $lambda$ is any eigenvalue of $N$)

---

In both cases, I am able to prove the direction $Rightarrow$ quite easily.
but in the other direction i am having a struggle (perhaps its false?).

for ex. statement B: I thought of showing that since $N$ is normal, then it is diagonalizable, therefore $N^*$ is also diagonalizable with the conjugate eigenvalues on its diagonal, therefore for any $lambda = (a+bi)$ then $NN^*$ has $(a+bi)(a-bi)$ on its diagonal, which is the identity matrix (since $|lambda|=1$) and therefore $N$ is unitary, but i'm not sure my transitions are valid.

Anyway, any help much appreciated on both statements!

linear-algebra eigenvalues-eigenvectors adjoint-operators

share|cite|improve this question

asked Jul 30 at 12:20

Limitless

986

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 12:20

Limitless

986

asked Jul 30 at 12:20

Limitless

986

asked Jul 30 at 12:20

Limitless

986

986

add a comment | 

add a comment | 

2 Answers
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Since $N$ is a normal $n times n$ - matrix, there is a unitary matrix $U$ such that $N=UDU^*$ , where $D=diag (lambda_1,..., lambda_n)$ and the $ lambda_k$ are the eigenvalues of $N$.

A) if all $ lambda_k$ are real, then $D^*=D$, hence $N^*=UD^*U^*=UDU^*=N$.

B) if for all $ lambda_k$ we have $|lambda_k|=1$, then $overlinelambda_k= frac1lambda_k$, thus $N^*=UD^*U^*=UD^-1U^*=N^-1$.

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answered Jul 30 at 12:40

Fred

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A): Since $mathcal N$ is normal, there is a eigenbasis $(e_j)$ of the vector space. Since $(mathcal N - mathcal N^*)e_j = (lambda - lambda) e_j = 0$ for all $j = 1, ldots, n$ [assuming the vector space $V$ has dimension $n < infty$]. Hence
$$
mathcal Nx = mathcal N^*x
$$
for all $x in V$, i.e. $mathcal N = mathcal N^*$.

B) If $|lambda| = 1$, then applied the operator $mathcal Nmathcal N^*$ to $e_j$ gives
$$
mathcal N mathcal N^* e_j = mathcal N (overline lambda e_j) = lambda overline lambda e_j = e_jquad [j = 1, ldots, n].
$$
Thus $mathcal N mathcal N^* = mathcal I$ [the identical mapping].

P.S. the symbol $iff$ is available as "iff" .

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edited Jul 30 at 12:53

answered Jul 30 at 12:44

xbh

1,0257

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Since $N$ is a normal $n times n$ - matrix, there is a unitary matrix $U$ such that $N=UDU^*$ , where $D=diag (lambda_1,..., lambda_n)$ and the $ lambda_k$ are the eigenvalues of $N$.

A) if all $ lambda_k$ are real, then $D^*=D$, hence $N^*=UD^*U^*=UDU^*=N$.

B) if for all $ lambda_k$ we have $|lambda_k|=1$, then $overlinelambda_k= frac1lambda_k$, thus $N^*=UD^*U^*=UD^-1U^*=N^-1$.

share|cite|improve this answer

answered Jul 30 at 12:40

Fred

37k1237

add a comment | 

up vote
2
down vote



accepted

Since $N$ is a normal $n times n$ - matrix, there is a unitary matrix $U$ such that $N=UDU^*$ , where $D=diag (lambda_1,..., lambda_n)$ and the $ lambda_k$ are the eigenvalues of $N$.

A) if all $ lambda_k$ are real, then $D^*=D$, hence $N^*=UD^*U^*=UDU^*=N$.

B) if for all $ lambda_k$ we have $|lambda_k|=1$, then $overlinelambda_k= frac1lambda_k$, thus $N^*=UD^*U^*=UD^-1U^*=N^-1$.

share|cite|improve this answer

answered Jul 30 at 12:40

Fred

37k1237

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Since $N$ is a normal $n times n$ - matrix, there is a unitary matrix $U$ such that $N=UDU^*$ , where $D=diag (lambda_1,..., lambda_n)$ and the $ lambda_k$ are the eigenvalues of $N$.

A) if all $ lambda_k$ are real, then $D^*=D$, hence $N^*=UD^*U^*=UDU^*=N$.

B) if for all $ lambda_k$ we have $|lambda_k|=1$, then $overlinelambda_k= frac1lambda_k$, thus $N^*=UD^*U^*=UD^-1U^*=N^-1$.

share|cite|improve this answer

answered Jul 30 at 12:40

Fred

37k1237

Since $N$ is a normal $n times n$ - matrix, there is a unitary matrix $U$ such that $N=UDU^*$ , where $D=diag (lambda_1,..., lambda_n)$ and the $ lambda_k$ are the eigenvalues of $N$.

A) if all $ lambda_k$ are real, then $D^*=D$, hence $N^*=UD^*U^*=UDU^*=N$.

B) if for all $ lambda_k$ we have $|lambda_k|=1$, then $overlinelambda_k= frac1lambda_k$, thus $N^*=UD^*U^*=UD^-1U^*=N^-1$.

share|cite|improve this answer

answered Jul 30 at 12:40

Fred

37k1237

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 12:40

Fred

37k1237

answered Jul 30 at 12:40

Fred

37k1237

answered Jul 30 at 12:40

Fred

37k1237

37k1237

add a comment | 

add a comment | 

up vote
1
down vote

A): Since $mathcal N$ is normal, there is a eigenbasis $(e_j)$ of the vector space. Since $(mathcal N - mathcal N^*)e_j = (lambda - lambda) e_j = 0$ for all $j = 1, ldots, n$ [assuming the vector space $V$ has dimension $n < infty$]. Hence
$$
mathcal Nx = mathcal N^*x
$$
for all $x in V$, i.e. $mathcal N = mathcal N^*$.

B) If $|lambda| = 1$, then applied the operator $mathcal Nmathcal N^*$ to $e_j$ gives
$$
mathcal N mathcal N^* e_j = mathcal N (overline lambda e_j) = lambda overline lambda e_j = e_jquad [j = 1, ldots, n].
$$
Thus $mathcal N mathcal N^* = mathcal I$ [the identical mapping].

P.S. the symbol $iff$ is available as "iff" .

share|cite|improve this answer

edited Jul 30 at 12:53

answered Jul 30 at 12:44

xbh

1,0257

add a comment | 

up vote
1
down vote

A): Since $mathcal N$ is normal, there is a eigenbasis $(e_j)$ of the vector space. Since $(mathcal N - mathcal N^*)e_j = (lambda - lambda) e_j = 0$ for all $j = 1, ldots, n$ [assuming the vector space $V$ has dimension $n < infty$]. Hence
$$
mathcal Nx = mathcal N^*x
$$
for all $x in V$, i.e. $mathcal N = mathcal N^*$.

B) If $|lambda| = 1$, then applied the operator $mathcal Nmathcal N^*$ to $e_j$ gives
$$
mathcal N mathcal N^* e_j = mathcal N (overline lambda e_j) = lambda overline lambda e_j = e_jquad [j = 1, ldots, n].
$$
Thus $mathcal N mathcal N^* = mathcal I$ [the identical mapping].

P.S. the symbol $iff$ is available as "iff" .

share|cite|improve this answer

edited Jul 30 at 12:53

answered Jul 30 at 12:44

xbh

1,0257

add a comment | 

up vote
1
down vote

up vote
1
down vote

A): Since $mathcal N$ is normal, there is a eigenbasis $(e_j)$ of the vector space. Since $(mathcal N - mathcal N^*)e_j = (lambda - lambda) e_j = 0$ for all $j = 1, ldots, n$ [assuming the vector space $V$ has dimension $n < infty$]. Hence
$$
mathcal Nx = mathcal N^*x
$$
for all $x in V$, i.e. $mathcal N = mathcal N^*$.

B) If $|lambda| = 1$, then applied the operator $mathcal Nmathcal N^*$ to $e_j$ gives
$$
mathcal N mathcal N^* e_j = mathcal N (overline lambda e_j) = lambda overline lambda e_j = e_jquad [j = 1, ldots, n].
$$
Thus $mathcal N mathcal N^* = mathcal I$ [the identical mapping].

P.S. the symbol $iff$ is available as "iff" .

share|cite|improve this answer

edited Jul 30 at 12:53

answered Jul 30 at 12:44

xbh

1,0257

A): Since $mathcal N$ is normal, there is a eigenbasis $(e_j)$ of the vector space. Since $(mathcal N - mathcal N^*)e_j = (lambda - lambda) e_j = 0$ for all $j = 1, ldots, n$ [assuming the vector space $V$ has dimension $n < infty$]. Hence
$$
mathcal Nx = mathcal N^*x
$$
for all $x in V$, i.e. $mathcal N = mathcal N^*$.

B) If $|lambda| = 1$, then applied the operator $mathcal Nmathcal N^*$ to $e_j$ gives
$$
mathcal N mathcal N^* e_j = mathcal N (overline lambda e_j) = lambda overline lambda e_j = e_jquad [j = 1, ldots, n].
$$
Thus $mathcal N mathcal N^* = mathcal I$ [the identical mapping].

P.S. the symbol $iff$ is available as "iff" .

share|cite|improve this answer

edited Jul 30 at 12:53

answered Jul 30 at 12:44

xbh

1,0257

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 12:53

edited Jul 30 at 12:53

edited Jul 30 at 12:53

answered Jul 30 at 12:44

xbh

1,0257

answered Jul 30 at 12:44

xbh

1,0257

answered Jul 30 at 12:44

xbh

1,0257

1,0257

add a comment | 

add a comment | 

 

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