Mixing
Can i prove that this matrix is PSD?
$begingroup$
I have matrix $A in mathbb{R}^{N times N}$, such that $A(i,j)=trace(B_iCB_j), forall ij$.
Matrices $B_i$ and C are PSD and symmetric with positive entries. Can I prove that $A$ is PSD too?
In fact, I've tried different random matrices for $C$ and $B_i$ with different dimensions and $A$ was always PSD.
linear-algebra matrices positive-semidefinite
asked Jan 18 at 14:23
BabakBabak
354111
$endgroup$
add a comment |
$begingroup$
I have matrix $A in mathbb{R}^{N times N}$, such that $A(i,j)=trace(B_iCB_j), forall ij$.
Matrices $B_i$ and C are PSD and symmetric with positive entries. Can I prove that $A$ is PSD too?
In fact, I've tried different random matrices for $C$ and $B_i$ with different dimensions and $A$ was always PSD.
linear-algebra matrices positive-semidefinite
asked Jan 18 at 14:23
BabakBabak
354111
$endgroup$
add a comment |
$begingroup$
I have matrix $A in mathbb{R}^{N times N}$, such that $A(i,j)=trace(B_iCB_j), forall ij$.
Matrices $B_i$ and C are PSD and symmetric with positive entries. Can I prove that $A$ is PSD too?
In fact, I've tried different random matrices for $C$ and $B_i$ with different dimensions and $A$ was always PSD.
linear-algebra matrices positive-semidefinite
asked Jan 18 at 14:23
BabakBabak
354111
$endgroup$
I have matrix $A in mathbb{R}^{N times N}$, such that $A(i,j)=trace(B_iCB_j), forall ij$.
Matrices $B_i$ and C are PSD and symmetric with positive entries. Can I prove that $A$ is PSD too?
In fact, I've tried different random matrices for $C$ and $B_i$ with different dimensions and $A$ was always PSD.
linear-algebra matrices positive-semidefinite
linear-algebra matrices positive-semidefinite
asked Jan 18 at 14:23
BabakBabak
354111
asked Jan 18 at 14:23
BabakBabak
354111
edited Jan 18 at 14:36
Babak
asked Jan 18 at 14:23
BabakBabak
354111
asked Jan 18 at 14:23
BabakBabak
354111
asked Jan 18 at 14:23
BabakBabak
354111
354111
add a comment |
add a comment |
1 Answer
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For any $x in mathbb{R}^n$ I will prove that $x^TAx geq 0$. Let $x in mathbb{R}^n$, then
$$begin{align}x^TAx &= sum_{i,j} x_i x_j text{tr}(B_iCB_j) \
&= text{tr}(sum_{i,j} x_i x_j B_iCB_j) \
&= text{tr}(sum_{i} sum_j (x_i B_i)C (x_j B_j)) \
&= text{tr}((sum_{i} x_i B_i)C(sum_j x_j B_j)) \ end{align}$$
Let $D = sum_{i} x_i B_i$, then this expression equals $text{tr}(DCD)$. Since $D$ is symmetric, $DCD=D^TCD$, which is positive semidefinite, so the trace is indeed nonnegative.
answered Jan 18 at 14:41
LinAlgLinAlg
9,8791521
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$begingroup$
Note that I never used that the entries of the matrices are positive.
$endgroup$
– LinAlg
Jan 18 at 14:42
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For any $x in mathbb{R}^n$ I will prove that $x^TAx geq 0$. Let $x in mathbb{R}^n$, then
$$begin{align}x^TAx &= sum_{i,j} x_i x_j text{tr}(B_iCB_j) \
&= text{tr}(sum_{i,j} x_i x_j B_iCB_j) \
&= text{tr}(sum_{i} sum_j (x_i B_i)C (x_j B_j)) \
&= text{tr}((sum_{i} x_i B_i)C(sum_j x_j B_j)) \ end{align}$$
Let $D = sum_{i} x_i B_i$, then this expression equals $text{tr}(DCD)$. Since $D$ is symmetric, $DCD=D^TCD$, which is positive semidefinite, so the trace is indeed nonnegative.
answered Jan 18 at 14:41
LinAlgLinAlg
9,8791521
$endgroup$
$begingroup$
Note that I never used that the entries of the matrices are positive.
$endgroup$
– LinAlg
Jan 18 at 14:42
add a comment |
$begingroup$
For any $x in mathbb{R}^n$ I will prove that $x^TAx geq 0$. Let $x in mathbb{R}^n$, then
$$begin{align}x^TAx &= sum_{i,j} x_i x_j text{tr}(B_iCB_j) \
&= text{tr}(sum_{i,j} x_i x_j B_iCB_j) \
&= text{tr}(sum_{i} sum_j (x_i B_i)C (x_j B_j)) \
&= text{tr}((sum_{i} x_i B_i)C(sum_j x_j B_j)) \ end{align}$$
Let $D = sum_{i} x_i B_i$, then this expression equals $text{tr}(DCD)$. Since $D$ is symmetric, $DCD=D^TCD$, which is positive semidefinite, so the trace is indeed nonnegative.
answered Jan 18 at 14:41
LinAlgLinAlg
9,8791521
$endgroup$
$begingroup$
Note that I never used that the entries of the matrices are positive.
$endgroup$
– LinAlg
Jan 18 at 14:42
add a comment |
$begingroup$
For any $x in mathbb{R}^n$ I will prove that $x^TAx geq 0$. Let $x in mathbb{R}^n$, then
$$begin{align}x^TAx &= sum_{i,j} x_i x_j text{tr}(B_iCB_j) \
&= text{tr}(sum_{i,j} x_i x_j B_iCB_j) \
&= text{tr}(sum_{i} sum_j (x_i B_i)C (x_j B_j)) \
&= text{tr}((sum_{i} x_i B_i)C(sum_j x_j B_j)) \ end{align}$$
Let $D = sum_{i} x_i B_i$, then this expression equals $text{tr}(DCD)$. Since $D$ is symmetric, $DCD=D^TCD$, which is positive semidefinite, so the trace is indeed nonnegative.
answered Jan 18 at 14:41
LinAlgLinAlg
9,8791521
$endgroup$
For any $x in mathbb{R}^n$ I will prove that $x^TAx geq 0$. Let $x in mathbb{R}^n$, then
$$begin{align}x^TAx &= sum_{i,j} x_i x_j text{tr}(B_iCB_j) \
&= text{tr}(sum_{i,j} x_i x_j B_iCB_j) \
&= text{tr}(sum_{i} sum_j (x_i B_i)C (x_j B_j)) \
&= text{tr}((sum_{i} x_i B_i)C(sum_j x_j B_j)) \ end{align}$$
Let $D = sum_{i} x_i B_i$, then this expression equals $text{tr}(DCD)$. Since $D$ is symmetric, $DCD=D^TCD$, which is positive semidefinite, so the trace is indeed nonnegative.
answered Jan 18 at 14:41
LinAlgLinAlg
9,8791521
answered Jan 18 at 14:41
LinAlgLinAlg
9,8791521
answered Jan 18 at 14:41
LinAlgLinAlg
9,8791521
answered Jan 18 at 14:41
LinAlgLinAlg
9,8791521
9,8791521
$begingroup$
Note that I never used that the entries of the matrices are positive.
$endgroup$
– LinAlg
Jan 18 at 14:42
add a comment |
$begingroup$
Note that I never used that the entries of the matrices are positive.
$endgroup$
– LinAlg
Jan 18 at 14:42
$begingroup$
Note that I never used that the entries of the matrices are positive.
$endgroup$
– LinAlg
Jan 18 at 14:42
$begingroup$
Note that I never used that the entries of the matrices are positive.
$endgroup$
– LinAlg
Jan 18 at 14:42
add a comment |
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