Prove normal matrices are RPN (Range perpendicular to Null space) Matrices?
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For a normal matrix $A$, $A^* A = A A^*$ where $A^*$ is the conjugate transpose of $A$
Prove that for normal matrices
$$C(A) perp N(A)$$
linear-algebra
asked Nov 27 at 4:08
Nagabhushan S N
255114
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up vote
1
down vote
favorite
For a normal matrix $A$, $A^* A = A A^*$ where $A^*$ is the conjugate transpose of $A$
Prove that for normal matrices
$$C(A) perp N(A)$$
linear-algebra
asked Nov 27 at 4:08
Nagabhushan S N
255114
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For a normal matrix $A$, $A^* A = A A^*$ where $A^*$ is the conjugate transpose of $A$
Prove that for normal matrices
$$C(A) perp N(A)$$
linear-algebra
asked Nov 27 at 4:08
Nagabhushan S N
255114
For a normal matrix $A$, $A^* A = A A^*$ where $A^*$ is the conjugate transpose of $A$
Prove that for normal matrices
$$C(A) perp N(A)$$
linear-algebra
linear-algebra
asked Nov 27 at 4:08
Nagabhushan S N
255114
asked Nov 27 at 4:08
Nagabhushan S N
255114
asked Nov 27 at 4:08
Nagabhushan S N
255114
asked Nov 27 at 4:08
Nagabhushan S N
255114
asked Nov 27 at 4:08
Nagabhushan S N
255114
255114
add a comment |
add a comment |
1 Answer
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I will use the language of linear transformations. We have $|A^{*}x|^{2}=langle A^{*}x, A^{*}xrangle =langle AA^{*}x, xrangle =langle A^{*}Ax, xrangle ==langle Ax, Axrangle =|Ax|^{2}$ which shows that for any $x$ in the null space of $A$ we also have $A^{*}x=0$. Now let $x$ be in the null space of $A$ and consider any vector $Ay$ in the range of $A$. We have $langle Ay, xrangle =langle y, A^{*}xrangle =langle y, 0rangle =0$.
answered Nov 27 at 6:08
Kavi Rama Murthy
46.3k31854
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I will use the language of linear transformations. We have $|A^{*}x|^{2}=langle A^{*}x, A^{*}xrangle =langle AA^{*}x, xrangle =langle A^{*}Ax, xrangle ==langle Ax, Axrangle =|Ax|^{2}$ which shows that for any $x$ in the null space of $A$ we also have $A^{*}x=0$. Now let $x$ be in the null space of $A$ and consider any vector $Ay$ in the range of $A$. We have $langle Ay, xrangle =langle y, A^{*}xrangle =langle y, 0rangle =0$.
answered Nov 27 at 6:08
Kavi Rama Murthy
46.3k31854
add a comment |
up vote
2
down vote
accepted
I will use the language of linear transformations. We have $|A^{*}x|^{2}=langle A^{*}x, A^{*}xrangle =langle AA^{*}x, xrangle =langle A^{*}Ax, xrangle ==langle Ax, Axrangle =|Ax|^{2}$ which shows that for any $x$ in the null space of $A$ we also have $A^{*}x=0$. Now let $x$ be in the null space of $A$ and consider any vector $Ay$ in the range of $A$. We have $langle Ay, xrangle =langle y, A^{*}xrangle =langle y, 0rangle =0$.
answered Nov 27 at 6:08
Kavi Rama Murthy
46.3k31854
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I will use the language of linear transformations. We have $|A^{*}x|^{2}=langle A^{*}x, A^{*}xrangle =langle AA^{*}x, xrangle =langle A^{*}Ax, xrangle ==langle Ax, Axrangle =|Ax|^{2}$ which shows that for any $x$ in the null space of $A$ we also have $A^{*}x=0$. Now let $x$ be in the null space of $A$ and consider any vector $Ay$ in the range of $A$. We have $langle Ay, xrangle =langle y, A^{*}xrangle =langle y, 0rangle =0$.
answered Nov 27 at 6:08
Kavi Rama Murthy
46.3k31854
I will use the language of linear transformations. We have $|A^{*}x|^{2}=langle A^{*}x, A^{*}xrangle =langle AA^{*}x, xrangle =langle A^{*}Ax, xrangle ==langle Ax, Axrangle =|Ax|^{2}$ which shows that for any $x$ in the null space of $A$ we also have $A^{*}x=0$. Now let $x$ be in the null space of $A$ and consider any vector $Ay$ in the range of $A$. We have $langle Ay, xrangle =langle y, A^{*}xrangle =langle y, 0rangle =0$.
answered Nov 27 at 6:08
Kavi Rama Murthy
46.3k31854
answered Nov 27 at 6:08
Kavi Rama Murthy
46.3k31854
answered Nov 27 at 6:08
Kavi Rama Murthy
46.3k31854
answered Nov 27 at 6:08
Kavi Rama Murthy
46.3k31854
46.3k31854
add a comment |
add a comment |
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