Multiplying only diagonal elements of a matrix
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I'm dealing with the following problem:
Suppose that I gave a upper diagonal matrix A of the form:
$A= begin{bmatrix} a_{11} & a_{12}&dots &a_{1n}\
0 & a_{22}&dots &a_{2n} \
vdots & vdots&dots &vdots\
0 & 0&dots &a_{nn} end{bmatrix}$
And a matrix $A'$ that is $A$ with only the diagonal elements multiplied by values $u_1,u_2,dots,u_n$, but the upper diagonal elements stay intact, that is:
$A' = begin{bmatrix} u_1 a_{11} & a_{12}&dots &a_{1n}\
0 & u_2 a_{22}&dots &a_{2n} \
vdots & vdots&dots &vdots\
0 & 0&dots &u_n a_{nn} end{bmatrix}$
How can I express $A'$ in terms of $A$?
Is there a way to express it in the form $A' =UA$? But then, what is the form of the matrix U?
linear-algebra matrices matrix-calculus matrix-decomposition
asked Nov 22 at 17:32
Arthur T
1819
|
show 2 more comments
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0
down vote
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I'm dealing with the following problem:
Suppose that I gave a upper diagonal matrix A of the form:
$A= begin{bmatrix} a_{11} & a_{12}&dots &a_{1n}\
0 & a_{22}&dots &a_{2n} \
vdots & vdots&dots &vdots\
0 & 0&dots &a_{nn} end{bmatrix}$
And a matrix $A'$ that is $A$ with only the diagonal elements multiplied by values $u_1,u_2,dots,u_n$, but the upper diagonal elements stay intact, that is:
$A' = begin{bmatrix} u_1 a_{11} & a_{12}&dots &a_{1n}\
0 & u_2 a_{22}&dots &a_{2n} \
vdots & vdots&dots &vdots\
0 & 0&dots &u_n a_{nn} end{bmatrix}$
How can I express $A'$ in terms of $A$?
Is there a way to express it in the form $A' =UA$? But then, what is the form of the matrix U?
linear-algebra matrices matrix-calculus matrix-decomposition
asked Nov 22 at 17:32
Arthur T
1819
$A' = begin{bmatrix} u_{1} & 0&dots &0\ 0 & u_{2}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &u_{n} end{bmatrix}begin{bmatrix} a_{11} & 0&dots &0\ 0 & a_{22}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &a_{nn} end{bmatrix}+begin{bmatrix} 0 & a_{12}&dots &a_{1n}\ 0 & 0&dots &a_{2n} \ vdots & vdots&dots &vdots\ 0 & 0&dots &0 end{bmatrix}$. You got to isolate the diagonal elements and then multiply I guess.
– Yadati Kiran
Nov 22 at 17:45
Just calculate $U=A'A^{-1}$
– Widawensen
Nov 22 at 17:51
1
How about $$eqalign{ A' &= A + {rm Diag}(u!-!1){,rm Diag}(A) cr U &= A'A^{-1} = I + {rm Diag}(u!-!1){,rm Diag}(A)A^{-1} cr }$$
– greg
Nov 22 at 18:01
@Widawensen I need something that does not involve $A'$
– Arthur T
Nov 22 at 18:08
1
@ArthurT Anyway, the result is diag$(u_1,u_2,u_3)$ + some nilpotent matrix with more complicated entries.
– Widawensen
Nov 22 at 18:19
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm dealing with the following problem:
Suppose that I gave a upper diagonal matrix A of the form:
$A= begin{bmatrix} a_{11} & a_{12}&dots &a_{1n}\
0 & a_{22}&dots &a_{2n} \
vdots & vdots&dots &vdots\
0 & 0&dots &a_{nn} end{bmatrix}$
And a matrix $A'$ that is $A$ with only the diagonal elements multiplied by values $u_1,u_2,dots,u_n$, but the upper diagonal elements stay intact, that is:
$A' = begin{bmatrix} u_1 a_{11} & a_{12}&dots &a_{1n}\
0 & u_2 a_{22}&dots &a_{2n} \
vdots & vdots&dots &vdots\
0 & 0&dots &u_n a_{nn} end{bmatrix}$
How can I express $A'$ in terms of $A$?
Is there a way to express it in the form $A' =UA$? But then, what is the form of the matrix U?
linear-algebra matrices matrix-calculus matrix-decomposition
asked Nov 22 at 17:32
Arthur T
1819
I'm dealing with the following problem:
Suppose that I gave a upper diagonal matrix A of the form:
$A= begin{bmatrix} a_{11} & a_{12}&dots &a_{1n}\
0 & a_{22}&dots &a_{2n} \
vdots & vdots&dots &vdots\
0 & 0&dots &a_{nn} end{bmatrix}$
And a matrix $A'$ that is $A$ with only the diagonal elements multiplied by values $u_1,u_2,dots,u_n$, but the upper diagonal elements stay intact, that is:
$A' = begin{bmatrix} u_1 a_{11} & a_{12}&dots &a_{1n}\
0 & u_2 a_{22}&dots &a_{2n} \
vdots & vdots&dots &vdots\
0 & 0&dots &u_n a_{nn} end{bmatrix}$
How can I express $A'$ in terms of $A$?
Is there a way to express it in the form $A' =UA$? But then, what is the form of the matrix U?
linear-algebra matrices matrix-calculus matrix-decomposition
linear-algebra matrices matrix-calculus matrix-decomposition
asked Nov 22 at 17:32
Arthur T
1819
asked Nov 22 at 17:32
Arthur T
1819
asked Nov 22 at 17:32
Arthur T
1819
asked Nov 22 at 17:32
Arthur T
1819
asked Nov 22 at 17:32
Arthur T
1819
1819
$A' = begin{bmatrix} u_{1} & 0&dots &0\ 0 & u_{2}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &u_{n} end{bmatrix}begin{bmatrix} a_{11} & 0&dots &0\ 0 & a_{22}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &a_{nn} end{bmatrix}+begin{bmatrix} 0 & a_{12}&dots &a_{1n}\ 0 & 0&dots &a_{2n} \ vdots & vdots&dots &vdots\ 0 & 0&dots &0 end{bmatrix}$. You got to isolate the diagonal elements and then multiply I guess.
– Yadati Kiran
Nov 22 at 17:45
Just calculate $U=A'A^{-1}$
– Widawensen
Nov 22 at 17:51
1
How about $$eqalign{ A' &= A + {rm Diag}(u!-!1){,rm Diag}(A) cr U &= A'A^{-1} = I + {rm Diag}(u!-!1){,rm Diag}(A)A^{-1} cr }$$
– greg
Nov 22 at 18:01
@Widawensen I need something that does not involve $A'$
– Arthur T
Nov 22 at 18:08
1
@ArthurT Anyway, the result is diag$(u_1,u_2,u_3)$ + some nilpotent matrix with more complicated entries.
– Widawensen
Nov 22 at 18:19
|
show 2 more comments
$A' = begin{bmatrix} u_{1} & 0&dots &0\ 0 & u_{2}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &u_{n} end{bmatrix}begin{bmatrix} a_{11} & 0&dots &0\ 0 & a_{22}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &a_{nn} end{bmatrix}+begin{bmatrix} 0 & a_{12}&dots &a_{1n}\ 0 & 0&dots &a_{2n} \ vdots & vdots&dots &vdots\ 0 & 0&dots &0 end{bmatrix}$. You got to isolate the diagonal elements and then multiply I guess.
– Yadati Kiran
Nov 22 at 17:45
Just calculate $U=A'A^{-1}$
– Widawensen
Nov 22 at 17:51
1
How about $$eqalign{ A' &= A + {rm Diag}(u!-!1){,rm Diag}(A) cr U &= A'A^{-1} = I + {rm Diag}(u!-!1){,rm Diag}(A)A^{-1} cr }$$
– greg
Nov 22 at 18:01
@Widawensen I need something that does not involve $A'$
– Arthur T
Nov 22 at 18:08
1
@ArthurT Anyway, the result is diag$(u_1,u_2,u_3)$ + some nilpotent matrix with more complicated entries.
– Widawensen
Nov 22 at 18:19
$A' = begin{bmatrix} u_{1} & 0&dots &0\ 0 & u_{2}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &u_{n} end{bmatrix}begin{bmatrix} a_{11} & 0&dots &0\ 0 & a_{22}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &a_{nn} end{bmatrix}+begin{bmatrix} 0 & a_{12}&dots &a_{1n}\ 0 & 0&dots &a_{2n} \ vdots & vdots&dots &vdots\ 0 & 0&dots &0 end{bmatrix}$. You got to isolate the diagonal elements and then multiply I guess.
– Yadati Kiran
Nov 22 at 17:45
$A' = begin{bmatrix} u_{1} & 0&dots &0\ 0 & u_{2}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &u_{n} end{bmatrix}begin{bmatrix} a_{11} & 0&dots &0\ 0 & a_{22}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &a_{nn} end{bmatrix}+begin{bmatrix} 0 & a_{12}&dots &a_{1n}\ 0 & 0&dots &a_{2n} \ vdots & vdots&dots &vdots\ 0 & 0&dots &0 end{bmatrix}$. You got to isolate the diagonal elements and then multiply I guess.
– Yadati Kiran
Nov 22 at 17:45
Just calculate $U=A'A^{-1}$
– Widawensen
Nov 22 at 17:51
Just calculate $U=A'A^{-1}$
– Widawensen
Nov 22 at 17:51
1
1
How about $$eqalign{ A' &= A + {rm Diag}(u!-!1){,rm Diag}(A) cr U &= A'A^{-1} = I + {rm Diag}(u!-!1){,rm Diag}(A)A^{-1} cr }$$
– greg
Nov 22 at 18:01
How about $$eqalign{ A' &= A + {rm Diag}(u!-!1){,rm Diag}(A) cr U &= A'A^{-1} = I + {rm Diag}(u!-!1){,rm Diag}(A)A^{-1} cr }$$
– greg
Nov 22 at 18:01
@Widawensen I need something that does not involve $A'$
– Arthur T
Nov 22 at 18:08
@Widawensen I need something that does not involve $A'$
– Arthur T
Nov 22 at 18:08
1
1
@ArthurT Anyway, the result is diag$(u_1,u_2,u_3)$ + some nilpotent matrix with more complicated entries.
– Widawensen
Nov 22 at 18:19
@ArthurT Anyway, the result is diag$(u_1,u_2,u_3)$ + some nilpotent matrix with more complicated entries.
– Widawensen
Nov 22 at 18:19
|
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$A' = begin{bmatrix} u_{1} & 0&dots &0\ 0 & u_{2}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &u_{n} end{bmatrix}begin{bmatrix} a_{11} & 0&dots &0\ 0 & a_{22}&dots &0 \ vdots & vdots&dots &vdots\ 0 & 0&dots &a_{nn} end{bmatrix}+begin{bmatrix} 0 & a_{12}&dots &a_{1n}\ 0 & 0&dots &a_{2n} \ vdots & vdots&dots &vdots\ 0 & 0&dots &0 end{bmatrix}$. You got to isolate the diagonal elements and then multiply I guess.
– Yadati Kiran
Nov 22 at 17:45
Just calculate $U=A'A^{-1}$
– Widawensen
Nov 22 at 17:51
1
How about $$eqalign{ A' &= A + {rm Diag}(u!-!1){,rm Diag}(A) cr U &= A'A^{-1} = I + {rm Diag}(u!-!1){,rm Diag}(A)A^{-1} cr }$$
– greg
Nov 22 at 18:01
@Widawensen I need something that does not involve $A'$
– Arthur T
Nov 22 at 18:08
1
@ArthurT Anyway, the result is diag$(u_1,u_2,u_3)$ + some nilpotent matrix with more complicated entries.
– Widawensen
Nov 22 at 18:19