What is the vectorized form of this expression?












0












$begingroup$


I want to figure out a general formula for vectorizing computations.
For example, if I am given matrices, A,B such that A is $m times q$, B is $n times q$. And I want to compute the matrix $X$, where each component of $X$ is
$$
X_{ij} = sum_{r = 1}^q A_{ir} B_{jr}
$$
Note that:



$i = 1 ldots m$



$j = 1 ldots n$



This one is simple. By vectorizing, I mean we get rid of the summation and the individual component computations. So for this example, the vectorized form is $X = A*B^T$.



But what if instead we add another multplication term, $c$ that is $qtimes1$
$$
X_{ij} = sum_{r = 1}^q A_{ir} * B_{jr} * c_r
$$



How would we vectorize this? If $q$ is small, say q = 2, we could simply expand out the terms like $X = A*B^T*c[1] + A*B^T*c[2]$.
But this is obviously not feasible if q is large. How would you vectorize something like this?



Is there a general procedure to follow to a problem like this, or is the vectorization very problem-specific that no general procedure can be followed?






matrices vectors vectorization







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Let $$C = begin{bmatrix}c_1 & 0 & dots & 0\0 & c_2 & dots & 0 \vdots & vdots &ddots& vdots\0&0&dots&c_qend{bmatrix}$$ Then $$X = ACB^T$$
    $endgroup$
    – Paul Sinclair
    Jan 30 at 3:20
















0












$begingroup$


I want to figure out a general formula for vectorizing computations.
For example, if I am given matrices, A,B such that A is $m times q$, B is $n times q$. And I want to compute the matrix $X$, where each component of $X$ is
$$
X_{ij} = sum_{r = 1}^q A_{ir} B_{jr}
$$
Note that:



$i = 1 ldots m$



$j = 1 ldots n$



This one is simple. By vectorizing, I mean we get rid of the summation and the individual component computations. So for this example, the vectorized form is $X = A*B^T$.



But what if instead we add another multplication term, $c$ that is $qtimes1$
$$
X_{ij} = sum_{r = 1}^q A_{ir} * B_{jr} * c_r
$$



How would we vectorize this? If $q$ is small, say q = 2, we could simply expand out the terms like $X = A*B^T*c[1] + A*B^T*c[2]$.
But this is obviously not feasible if q is large. How would you vectorize something like this?



Is there a general procedure to follow to a problem like this, or is the vectorization very problem-specific that no general procedure can be followed?






matrices vectors vectorization







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Let $$C = begin{bmatrix}c_1 & 0 & dots & 0\0 & c_2 & dots & 0 \vdots & vdots &ddots& vdots\0&0&dots&c_qend{bmatrix}$$ Then $$X = ACB^T$$
    $endgroup$
    – Paul Sinclair
    Jan 30 at 3:20














0












0








0





$begingroup$


I want to figure out a general formula for vectorizing computations.
For example, if I am given matrices, A,B such that A is $m times q$, B is $n times q$. And I want to compute the matrix $X$, where each component of $X$ is
$$
X_{ij} = sum_{r = 1}^q A_{ir} B_{jr}
$$
Note that:



$i = 1 ldots m$



$j = 1 ldots n$



This one is simple. By vectorizing, I mean we get rid of the summation and the individual component computations. So for this example, the vectorized form is $X = A*B^T$.



But what if instead we add another multplication term, $c$ that is $qtimes1$
$$
X_{ij} = sum_{r = 1}^q A_{ir} * B_{jr} * c_r
$$



How would we vectorize this? If $q$ is small, say q = 2, we could simply expand out the terms like $X = A*B^T*c[1] + A*B^T*c[2]$.
But this is obviously not feasible if q is large. How would you vectorize something like this?



Is there a general procedure to follow to a problem like this, or is the vectorization very problem-specific that no general procedure can be followed?






matrices vectors vectorization







share|cite|improve this question









$endgroup$




I want to figure out a general formula for vectorizing computations.
For example, if I am given matrices, A,B such that A is $m times q$, B is $n times q$. And I want to compute the matrix $X$, where each component of $X$ is
$$
X_{ij} = sum_{r = 1}^q A_{ir} B_{jr}
$$
Note that:



$i = 1 ldots m$



$j = 1 ldots n$



This one is simple. By vectorizing, I mean we get rid of the summation and the individual component computations. So for this example, the vectorized form is $X = A*B^T$.



But what if instead we add another multplication term, $c$ that is $qtimes1$
$$
X_{ij} = sum_{r = 1}^q A_{ir} * B_{jr} * c_r
$$



How would we vectorize this? If $q$ is small, say q = 2, we could simply expand out the terms like $X = A*B^T*c[1] + A*B^T*c[2]$.
But this is obviously not feasible if q is large. How would you vectorize something like this?



Is there a general procedure to follow to a problem like this, or is the vectorization very problem-specific that no general procedure can be followed?






matrices vectors vectorization




matrices vectors vectorization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 29 at 19:12









IamanonIamanon

1328




1328








  • 1




    $begingroup$
    Let $$C = begin{bmatrix}c_1 & 0 & dots & 0\0 & c_2 & dots & 0 \vdots & vdots &ddots& vdots\0&0&dots&c_qend{bmatrix}$$ Then $$X = ACB^T$$
    $endgroup$
    – Paul Sinclair
    Jan 30 at 3:20














  • 1




    $begingroup$
    Let $$C = begin{bmatrix}c_1 & 0 & dots & 0\0 & c_2 & dots & 0 \vdots & vdots &ddots& vdots\0&0&dots&c_qend{bmatrix}$$ Then $$X = ACB^T$$
    $endgroup$
    – Paul Sinclair
    Jan 30 at 3:20








1




1




$begingroup$
Let $$C = begin{bmatrix}c_1 & 0 & dots & 0\0 & c_2 & dots & 0 \vdots & vdots &ddots& vdots\0&0&dots&c_qend{bmatrix}$$ Then $$X = ACB^T$$
$endgroup$
– Paul Sinclair
Jan 30 at 3:20




$begingroup$
Let $$C = begin{bmatrix}c_1 & 0 & dots & 0\0 & c_2 & dots & 0 \vdots & vdots &ddots& vdots\0&0&dots&c_qend{bmatrix}$$ Then $$X = ACB^T$$
$endgroup$
– Paul Sinclair
Jan 30 at 3:20










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