Showing $T^{-1} = T^*$ where T is matrix of normalized eigenvectors of Hermitian matrix
$begingroup$
Let A be an nxn Hermitian matrix with n distinct eigenvalues and T be the matrix whose columns are normalized eigenvectors of A. I want to show that $T^{-1} = T^*$.
I know that the eigenvectors of A are mutually orthogonal, and since they are normalized, then they are orthonormal. Where do I proceed in showing that $T^{-1} = T^*$? Thanks.
linear-algebra
asked Dec 5 '18 at 2:57
appleapple
445
$endgroup$
add a comment |
$begingroup$
Let A be an nxn Hermitian matrix with n distinct eigenvalues and T be the matrix whose columns are normalized eigenvectors of A. I want to show that $T^{-1} = T^*$.
I know that the eigenvectors of A are mutually orthogonal, and since they are normalized, then they are orthonormal. Where do I proceed in showing that $T^{-1} = T^*$? Thanks.
linear-algebra
asked Dec 5 '18 at 2:57
appleapple
445
$endgroup$
$begingroup$
Compute $TT^*$. Write $T = [t_1 t_2 cdots t_n]$.
$endgroup$
– xbh
Dec 5 '18 at 3:01
add a comment |
$begingroup$
Let A be an nxn Hermitian matrix with n distinct eigenvalues and T be the matrix whose columns are normalized eigenvectors of A. I want to show that $T^{-1} = T^*$.
I know that the eigenvectors of A are mutually orthogonal, and since they are normalized, then they are orthonormal. Where do I proceed in showing that $T^{-1} = T^*$? Thanks.
linear-algebra
asked Dec 5 '18 at 2:57
appleapple
445
$endgroup$
Let A be an nxn Hermitian matrix with n distinct eigenvalues and T be the matrix whose columns are normalized eigenvectors of A. I want to show that $T^{-1} = T^*$.
I know that the eigenvectors of A are mutually orthogonal, and since they are normalized, then they are orthonormal. Where do I proceed in showing that $T^{-1} = T^*$? Thanks.
linear-algebra
linear-algebra
asked Dec 5 '18 at 2:57
appleapple
445
asked Dec 5 '18 at 2:57
appleapple
445
asked Dec 5 '18 at 2:57
appleapple
445
asked Dec 5 '18 at 2:57
appleapple
445
asked Dec 5 '18 at 2:57
appleapple
445
445
$begingroup$
Compute $TT^*$. Write $T = [t_1 t_2 cdots t_n]$.
$endgroup$
– xbh
Dec 5 '18 at 3:01
add a comment |
$begingroup$
Compute $TT^*$. Write $T = [t_1 t_2 cdots t_n]$.
$endgroup$
– xbh
Dec 5 '18 at 3:01
$begingroup$
Compute $TT^*$. Write $T = [t_1 t_2 cdots t_n]$.
$endgroup$
– xbh
Dec 5 '18 at 3:01
$begingroup$
Compute $TT^*$. Write $T = [t_1 t_2 cdots t_n]$.
$endgroup$
– xbh
Dec 5 '18 at 3:01
add a comment |
1 Answer
1
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$begingroup$
Proof.
Let $T = begin{bmatrix}t_1 & cdots & t_n end{bmatrix}$, where $t_j$'s are $ntimes 1$ column matrices. Since $A$ is Hermitian with $n$ distinct eigenvalues, and $t_j$'s are normalized, ${t_j}_1^n$ satisfies $t_j^* t_k = delta_{j,k}$, thus
$$
T^* T = [t_j^* t_k]_{(j,k)=(1,1)}^{(n,n)} = [delta_{j,k}] = I.
$$
answered Dec 5 '18 at 3:45
xbhxbh
6,1251522
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Proof.
Let $T = begin{bmatrix}t_1 & cdots & t_n end{bmatrix}$, where $t_j$'s are $ntimes 1$ column matrices. Since $A$ is Hermitian with $n$ distinct eigenvalues, and $t_j$'s are normalized, ${t_j}_1^n$ satisfies $t_j^* t_k = delta_{j,k}$, thus
$$
T^* T = [t_j^* t_k]_{(j,k)=(1,1)}^{(n,n)} = [delta_{j,k}] = I.
$$
answered Dec 5 '18 at 3:45
xbhxbh
6,1251522
$endgroup$
add a comment |
$begingroup$
Proof.
Let $T = begin{bmatrix}t_1 & cdots & t_n end{bmatrix}$, where $t_j$'s are $ntimes 1$ column matrices. Since $A$ is Hermitian with $n$ distinct eigenvalues, and $t_j$'s are normalized, ${t_j}_1^n$ satisfies $t_j^* t_k = delta_{j,k}$, thus
$$
T^* T = [t_j^* t_k]_{(j,k)=(1,1)}^{(n,n)} = [delta_{j,k}] = I.
$$
answered Dec 5 '18 at 3:45
xbhxbh
6,1251522
$endgroup$
add a comment |
$begingroup$
Proof.
Let $T = begin{bmatrix}t_1 & cdots & t_n end{bmatrix}$, where $t_j$'s are $ntimes 1$ column matrices. Since $A$ is Hermitian with $n$ distinct eigenvalues, and $t_j$'s are normalized, ${t_j}_1^n$ satisfies $t_j^* t_k = delta_{j,k}$, thus
$$
T^* T = [t_j^* t_k]_{(j,k)=(1,1)}^{(n,n)} = [delta_{j,k}] = I.
$$
answered Dec 5 '18 at 3:45
xbhxbh
6,1251522
$endgroup$
Proof.
Let $T = begin{bmatrix}t_1 & cdots & t_n end{bmatrix}$, where $t_j$'s are $ntimes 1$ column matrices. Since $A$ is Hermitian with $n$ distinct eigenvalues, and $t_j$'s are normalized, ${t_j}_1^n$ satisfies $t_j^* t_k = delta_{j,k}$, thus
$$
T^* T = [t_j^* t_k]_{(j,k)=(1,1)}^{(n,n)} = [delta_{j,k}] = I.
$$
answered Dec 5 '18 at 3:45
xbhxbh
6,1251522
answered Dec 5 '18 at 3:45
xbhxbh
6,1251522
answered Dec 5 '18 at 3:45
xbhxbh
6,1251522
answered Dec 5 '18 at 3:45
xbhxbh
6,1251522
6,1251522
add a comment |
add a comment |
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$begingroup$
Compute $TT^*$. Write $T = [t_1 t_2 cdots t_n]$.
$endgroup$
– xbh
Dec 5 '18 at 3:01