Mixing
Going from matrix form to summation for calculating the derivative
$begingroup$
I have troubles calculating the derivatives of:
$frac{dy}{dx}$ if $y= vec{x}^{T}matrix{W}vec{x}$
I think it would be easier if I go to summation form and then take it from there. However, i'm not sure on how to do that? Same thing when I have a summation to go back to matrix form. Is there an easy step by step guide for this?
summation matrix-calculus
asked Jan 31 at 8:53
Jesper.LindbergJesper.Lindberg
82
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add a comment |
$begingroup$
I have troubles calculating the derivatives of:
$frac{dy}{dx}$ if $y= vec{x}^{T}matrix{W}vec{x}$
I think it would be easier if I go to summation form and then take it from there. However, i'm not sure on how to do that? Same thing when I have a summation to go back to matrix form. Is there an easy step by step guide for this?
summation matrix-calculus
asked Jan 31 at 8:53
Jesper.LindbergJesper.Lindberg
82
$endgroup$
add a comment |
$begingroup$
I have troubles calculating the derivatives of:
$frac{dy}{dx}$ if $y= vec{x}^{T}matrix{W}vec{x}$
I think it would be easier if I go to summation form and then take it from there. However, i'm not sure on how to do that? Same thing when I have a summation to go back to matrix form. Is there an easy step by step guide for this?
summation matrix-calculus
asked Jan 31 at 8:53
Jesper.LindbergJesper.Lindberg
82
$endgroup$
I have troubles calculating the derivatives of:
$frac{dy}{dx}$ if $y= vec{x}^{T}matrix{W}vec{x}$
I think it would be easier if I go to summation form and then take it from there. However, i'm not sure on how to do that? Same thing when I have a summation to go back to matrix form. Is there an easy step by step guide for this?
summation matrix-calculus
summation matrix-calculus
asked Jan 31 at 8:53
Jesper.LindbergJesper.Lindberg
82
asked Jan 31 at 8:53
Jesper.LindbergJesper.Lindberg
82
asked Jan 31 at 8:53
Jesper.LindbergJesper.Lindberg
82
asked Jan 31 at 8:53
Jesper.LindbergJesper.Lindberg
82
asked Jan 31 at 8:53
Jesper.LindbergJesper.Lindberg
82
82
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint:
Let $mathbf{A}$ be a matrix of size $n times m$ with elements $A_{ik}$ for $i = 1, 2 dots, n$ and $k = 1, 2, dots, m$
Let $mathbf{B}$ be a matrix of size $m times p$ with elements $B_{kj}$ for $k = 1, 2 dots, m$ and $j = 1, 2, dots, p$
Then, we have
begin{align}
(mathbf{A} , mathbf{B})_{ij} ; = ; sum_{k=1}^{m}A_{ik} , B_{kj}
end{align}
answered Jan 31 at 9:01
K_inverseK_inverse
288213
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
Hint:
Let $mathbf{A}$ be a matrix of size $n times m$ with elements $A_{ik}$ for $i = 1, 2 dots, n$ and $k = 1, 2, dots, m$
Let $mathbf{B}$ be a matrix of size $m times p$ with elements $B_{kj}$ for $k = 1, 2 dots, m$ and $j = 1, 2, dots, p$
Then, we have
begin{align}
(mathbf{A} , mathbf{B})_{ij} ; = ; sum_{k=1}^{m}A_{ik} , B_{kj}
end{align}
answered Jan 31 at 9:01
K_inverseK_inverse
288213
$endgroup$
add a comment |
$begingroup$
Hint:
Let $mathbf{A}$ be a matrix of size $n times m$ with elements $A_{ik}$ for $i = 1, 2 dots, n$ and $k = 1, 2, dots, m$
Let $mathbf{B}$ be a matrix of size $m times p$ with elements $B_{kj}$ for $k = 1, 2 dots, m$ and $j = 1, 2, dots, p$
Then, we have
begin{align}
(mathbf{A} , mathbf{B})_{ij} ; = ; sum_{k=1}^{m}A_{ik} , B_{kj}
end{align}
answered Jan 31 at 9:01
K_inverseK_inverse
288213
$endgroup$
add a comment |
$begingroup$
Hint:
Let $mathbf{A}$ be a matrix of size $n times m$ with elements $A_{ik}$ for $i = 1, 2 dots, n$ and $k = 1, 2, dots, m$
Let $mathbf{B}$ be a matrix of size $m times p$ with elements $B_{kj}$ for $k = 1, 2 dots, m$ and $j = 1, 2, dots, p$
Then, we have
begin{align}
(mathbf{A} , mathbf{B})_{ij} ; = ; sum_{k=1}^{m}A_{ik} , B_{kj}
end{align}
answered Jan 31 at 9:01
K_inverseK_inverse
288213
$endgroup$
Hint:
Let $mathbf{A}$ be a matrix of size $n times m$ with elements $A_{ik}$ for $i = 1, 2 dots, n$ and $k = 1, 2, dots, m$
Let $mathbf{B}$ be a matrix of size $m times p$ with elements $B_{kj}$ for $k = 1, 2 dots, m$ and $j = 1, 2, dots, p$
Then, we have
begin{align}
(mathbf{A} , mathbf{B})_{ij} ; = ; sum_{k=1}^{m}A_{ik} , B_{kj}
end{align}
answered Jan 31 at 9:01
K_inverseK_inverse
288213
answered Jan 31 at 9:01
K_inverseK_inverse
288213
answered Jan 31 at 9:01
K_inverseK_inverse
288213
answered Jan 31 at 9:01
K_inverseK_inverse
288213
288213
add a comment |
add a comment |
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