Mixing
Minimize Trace for non-symmetric matrix
$begingroup$
Let $A$ be a given matrix and $X$ a symmetric positive definite matrix which
shall be optimized:
min $tr (AX)$ subject to $X>0$
This optimization problem is convex if $A$ is symmetric as it is the standard form of Semidefinite Programming. But is the problem convex for non-symmetric $A$ as well?
convex-optimization
asked Oct 2 '13 at 9:08
Lyapunov123Lyapunov123
313
$endgroup$
add a comment |
$begingroup$
Let $A$ be a given matrix and $X$ a symmetric positive definite matrix which
shall be optimized:
min $tr (AX)$ subject to $X>0$
This optimization problem is convex if $A$ is symmetric as it is the standard form of Semidefinite Programming. But is the problem convex for non-symmetric $A$ as well?
convex-optimization
asked Oct 2 '13 at 9:08
Lyapunov123Lyapunov123
313
$endgroup$
add a comment |
$begingroup$
Let $A$ be a given matrix and $X$ a symmetric positive definite matrix which
shall be optimized:
min $tr (AX)$ subject to $X>0$
This optimization problem is convex if $A$ is symmetric as it is the standard form of Semidefinite Programming. But is the problem convex for non-symmetric $A$ as well?
convex-optimization
asked Oct 2 '13 at 9:08
Lyapunov123Lyapunov123
313
$endgroup$
Let $A$ be a given matrix and $X$ a symmetric positive definite matrix which
shall be optimized:
min $tr (AX)$ subject to $X>0$
This optimization problem is convex if $A$ is symmetric as it is the standard form of Semidefinite Programming. But is the problem convex for non-symmetric $A$ as well?
convex-optimization
convex-optimization
asked Oct 2 '13 at 9:08
Lyapunov123Lyapunov123
313
asked Oct 2 '13 at 9:08
Lyapunov123Lyapunov123
313
asked Oct 2 '13 at 9:08
Lyapunov123Lyapunov123
313
asked Oct 2 '13 at 9:08
Lyapunov123Lyapunov123
313
asked Oct 2 '13 at 9:08
Lyapunov123Lyapunov123
313
313
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, the program is convex for any $A$.
Let $X in S^n$ (the space of $ntimes n$ dimensional symmetric matrices). Then,
begin{align*}
textrm{trace}(AX) &= textrm{trace}left(Aleft(sum_{i=1}^nnu_i x_i x_i^Tright)right)\
&= textrm{trace}left(sum_{i=1}^nnu_i A x_i x_i^Tright)\
&= sum_{i=1}^nnu_i textrm{trace}left( A x_i x_i^Tright)\
&= sum_{i=1}^nnu_i textrm{trace}left(x_i^T A x_i right)\
&= sum_{i=1}^nnu_i left(x_i^T A x_i right)\
end{align*}
We will use the following identity: For any square matrix $A$
$$ x^TAx = frac{1}{2}x^T(A+ A^T)x$$
Proof:
begin{align*}
frac{1}{2}x^T(A+ A^T)x &= frac{1}{2}x^TAx + frac{1}{2}x^TA^Tx\
&= frac{1}{2}x^T(Ax) + frac{1}{2}(Ax)^Tx\
&= frac{1}{2}x^T(Ax) + frac{1}{2}x^T(Ax)\
&= x^TAx
end{align*}
Thus, we get
begin{align*}
textrm{trace}(AX) &= sum_{i=1}^nnu_i left(x_i^T frac{1}{2}(A+ A^T)x_i right)\
&= frac{1}{2}textrm{trace}left(sum_{i=1}^nnu_i x_i^T(A+ A^T)x_i right)\
& = frac{1}{2}textrm{trace}left((A+ A^T)sum_{i=1}^nnu_i x_ix_i^T right)\
& = frac{1}{2}textrm{trace}left((A+ A^T)Xright)
end{align*}
The term $textrm{trace}left((A+ A^T)Xright)$ is in standard form as desired.
answered Oct 2 '13 at 16:10
ThejaTheja
21317
$endgroup$
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Yes, the program is convex for any $A$.
Let $X in S^n$ (the space of $ntimes n$ dimensional symmetric matrices). Then,
begin{align*}
textrm{trace}(AX) &= textrm{trace}left(Aleft(sum_{i=1}^nnu_i x_i x_i^Tright)right)\
&= textrm{trace}left(sum_{i=1}^nnu_i A x_i x_i^Tright)\
&= sum_{i=1}^nnu_i textrm{trace}left( A x_i x_i^Tright)\
&= sum_{i=1}^nnu_i textrm{trace}left(x_i^T A x_i right)\
&= sum_{i=1}^nnu_i left(x_i^T A x_i right)\
end{align*}
We will use the following identity: For any square matrix $A$
$$ x^TAx = frac{1}{2}x^T(A+ A^T)x$$
Proof:
begin{align*}
frac{1}{2}x^T(A+ A^T)x &= frac{1}{2}x^TAx + frac{1}{2}x^TA^Tx\
&= frac{1}{2}x^T(Ax) + frac{1}{2}(Ax)^Tx\
&= frac{1}{2}x^T(Ax) + frac{1}{2}x^T(Ax)\
&= x^TAx
end{align*}
Thus, we get
begin{align*}
textrm{trace}(AX) &= sum_{i=1}^nnu_i left(x_i^T frac{1}{2}(A+ A^T)x_i right)\
&= frac{1}{2}textrm{trace}left(sum_{i=1}^nnu_i x_i^T(A+ A^T)x_i right)\
& = frac{1}{2}textrm{trace}left((A+ A^T)sum_{i=1}^nnu_i x_ix_i^T right)\
& = frac{1}{2}textrm{trace}left((A+ A^T)Xright)
end{align*}
The term $textrm{trace}left((A+ A^T)Xright)$ is in standard form as desired.
answered Oct 2 '13 at 16:10
ThejaTheja
21317
$endgroup$
add a comment |
$begingroup$
Yes, the program is convex for any $A$.
Let $X in S^n$ (the space of $ntimes n$ dimensional symmetric matrices). Then,
begin{align*}
textrm{trace}(AX) &= textrm{trace}left(Aleft(sum_{i=1}^nnu_i x_i x_i^Tright)right)\
&= textrm{trace}left(sum_{i=1}^nnu_i A x_i x_i^Tright)\
&= sum_{i=1}^nnu_i textrm{trace}left( A x_i x_i^Tright)\
&= sum_{i=1}^nnu_i textrm{trace}left(x_i^T A x_i right)\
&= sum_{i=1}^nnu_i left(x_i^T A x_i right)\
end{align*}
We will use the following identity: For any square matrix $A$
$$ x^TAx = frac{1}{2}x^T(A+ A^T)x$$
Proof:
begin{align*}
frac{1}{2}x^T(A+ A^T)x &= frac{1}{2}x^TAx + frac{1}{2}x^TA^Tx\
&= frac{1}{2}x^T(Ax) + frac{1}{2}(Ax)^Tx\
&= frac{1}{2}x^T(Ax) + frac{1}{2}x^T(Ax)\
&= x^TAx
end{align*}
Thus, we get
begin{align*}
textrm{trace}(AX) &= sum_{i=1}^nnu_i left(x_i^T frac{1}{2}(A+ A^T)x_i right)\
&= frac{1}{2}textrm{trace}left(sum_{i=1}^nnu_i x_i^T(A+ A^T)x_i right)\
& = frac{1}{2}textrm{trace}left((A+ A^T)sum_{i=1}^nnu_i x_ix_i^T right)\
& = frac{1}{2}textrm{trace}left((A+ A^T)Xright)
end{align*}
The term $textrm{trace}left((A+ A^T)Xright)$ is in standard form as desired.
answered Oct 2 '13 at 16:10
ThejaTheja
21317
$endgroup$
add a comment |
$begingroup$
Yes, the program is convex for any $A$.
Let $X in S^n$ (the space of $ntimes n$ dimensional symmetric matrices). Then,
begin{align*}
textrm{trace}(AX) &= textrm{trace}left(Aleft(sum_{i=1}^nnu_i x_i x_i^Tright)right)\
&= textrm{trace}left(sum_{i=1}^nnu_i A x_i x_i^Tright)\
&= sum_{i=1}^nnu_i textrm{trace}left( A x_i x_i^Tright)\
&= sum_{i=1}^nnu_i textrm{trace}left(x_i^T A x_i right)\
&= sum_{i=1}^nnu_i left(x_i^T A x_i right)\
end{align*}
We will use the following identity: For any square matrix $A$
$$ x^TAx = frac{1}{2}x^T(A+ A^T)x$$
Proof:
begin{align*}
frac{1}{2}x^T(A+ A^T)x &= frac{1}{2}x^TAx + frac{1}{2}x^TA^Tx\
&= frac{1}{2}x^T(Ax) + frac{1}{2}(Ax)^Tx\
&= frac{1}{2}x^T(Ax) + frac{1}{2}x^T(Ax)\
&= x^TAx
end{align*}
Thus, we get
begin{align*}
textrm{trace}(AX) &= sum_{i=1}^nnu_i left(x_i^T frac{1}{2}(A+ A^T)x_i right)\
&= frac{1}{2}textrm{trace}left(sum_{i=1}^nnu_i x_i^T(A+ A^T)x_i right)\
& = frac{1}{2}textrm{trace}left((A+ A^T)sum_{i=1}^nnu_i x_ix_i^T right)\
& = frac{1}{2}textrm{trace}left((A+ A^T)Xright)
end{align*}
The term $textrm{trace}left((A+ A^T)Xright)$ is in standard form as desired.
answered Oct 2 '13 at 16:10
ThejaTheja
21317
$endgroup$
Yes, the program is convex for any $A$.
Let $X in S^n$ (the space of $ntimes n$ dimensional symmetric matrices). Then,
begin{align*}
textrm{trace}(AX) &= textrm{trace}left(Aleft(sum_{i=1}^nnu_i x_i x_i^Tright)right)\
&= textrm{trace}left(sum_{i=1}^nnu_i A x_i x_i^Tright)\
&= sum_{i=1}^nnu_i textrm{trace}left( A x_i x_i^Tright)\
&= sum_{i=1}^nnu_i textrm{trace}left(x_i^T A x_i right)\
&= sum_{i=1}^nnu_i left(x_i^T A x_i right)\
end{align*}
We will use the following identity: For any square matrix $A$
$$ x^TAx = frac{1}{2}x^T(A+ A^T)x$$
Proof:
begin{align*}
frac{1}{2}x^T(A+ A^T)x &= frac{1}{2}x^TAx + frac{1}{2}x^TA^Tx\
&= frac{1}{2}x^T(Ax) + frac{1}{2}(Ax)^Tx\
&= frac{1}{2}x^T(Ax) + frac{1}{2}x^T(Ax)\
&= x^TAx
end{align*}
Thus, we get
begin{align*}
textrm{trace}(AX) &= sum_{i=1}^nnu_i left(x_i^T frac{1}{2}(A+ A^T)x_i right)\
&= frac{1}{2}textrm{trace}left(sum_{i=1}^nnu_i x_i^T(A+ A^T)x_i right)\
& = frac{1}{2}textrm{trace}left((A+ A^T)sum_{i=1}^nnu_i x_ix_i^T right)\
& = frac{1}{2}textrm{trace}left((A+ A^T)Xright)
end{align*}
The term $textrm{trace}left((A+ A^T)Xright)$ is in standard form as desired.
answered Oct 2 '13 at 16:10
ThejaTheja
21317
answered Oct 2 '13 at 16:10
ThejaTheja
21317
answered Oct 2 '13 at 16:10
ThejaTheja
21317
answered Oct 2 '13 at 16:10
ThejaTheja
21317
21317
add a comment |
add a comment |
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