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Is $ \bf a^tbc = cb^ta $ where $\bf a, b, c \in R^d$?

$\bf a^tbc =(a^tb)c = (b^ta)c = c(b^ta) = (cb^t)a$

$\bf (cb^t)$ becomes $d*d$ matrix.

First form is some scalar multiplied by $\bf c$ vector while last form is matrix transformation on $\bf a$ vector. Is this consistent? how to think of this intuitively?

Nascimento de Cos
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2 Answers2

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As I've explained here, expressions of the form $a^tbc$ are fine if read as $(a\cdot b)c$. It doesn't make sense on a literal reading because $a^tb$ is a $1\times1$ matrix, but we can safely ignore that. There is therefore no issue with $(a\cdot b)c=(b\cdot a)c=c(b\cdot a)$. These are all ways of writing a vector whose $i$th component is $\sum_ja_jb_jc_i$.

Meanwhile, $cb^t$, with the transpose on the right, is a $d\times d$ matrix as you noted, since $a,\,b,\,c$ are column vectors. We can therefore also safely write $(a\cdot b)c=(cb^t)a$, equating a vector parallel to $c$ to $a$'s image under the square matrix $cb^t$. (This image, of course, is not in general parallel to $a$; note terms in the above sum contain factors of $a_j$ rather than $a_i$, and $c^tb$'s $ij$ entry is $c_ib_j$.)

J.G.
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  • sir But thi is what i am asking. How image of vector $\vec a$ under $cb^t$ is parallel to $vec c$. kindly elaborate (but u wrote in bracket not parallel to $\vec a$)... – Nascimento de Cos Jan 24 '21 at 14:46
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    @NascimentideCos $a$ is not in general an eigenvector of $cb^t$. Is that term familiar? – J.G. Jan 24 '21 at 14:48
  • sir eigen vectors are such vectors whose images are proportional under Matrix transformation. But again 2 vectors are equal means they are linearly dependent in which case either they are same or they are parallel. I think coincidentally under matrix transformation $cb^t$, a vector became parallel to c vector(pls correct me if i am wrong. ) – Nascimento de Cos Jan 24 '21 at 14:58
  • @NascimentodeCos It's no coincidence that $cb^ta$ is parallel to the column vector $c$, or equivalently that $a^tbc^t$ is parallel to the row vector $c^t$, since $a^tb$ is a $1\times1$ matrix. – J.G. Jan 24 '21 at 15:01
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You are mixing matrix multiplication and scalar multiplication.

Let us use different symbols to see what is going on. For a scalar $\lambda\in K$ and a vector $v\in K^n$ there is the scalar multiple $\lambda\odot \mathbf v$ (typically written as $\lambda\mathbf v$). Since matrices form a vector space as well, any $n\times m$ matrix $\mathbf X$ can also be scaled by $\lambda\in K$ as $\lambda\odot \mathbf X$ (typically written as $\lambda\mathbf X$ as well).

Now for an $n\times m$ matrix $\mathbf A$ and a $m\times k$ matrix $\mathbf B$ there is the matrix product $\mathbf A\cdot \mathbf B$, a $n\times k$ matrix (typically written as $\mathbf A\mathbf B$). When this is a $1\times 1$ matrix, we may treat it as a scalar.

We may also treat vectors in $K^n$ as column vectors, that is $n\times 1$ matrices.

Using this notation, your question concerns the equation $$ (\mathbf a^t \cdot \mathbf b)\odot \mathbf c = \mathbf c \cdot \mathbf b^t \cdot \mathbf a. $$

Note that treating $\mathbf c$ as a $n\times 1$ matrix and scalars $\lambda$ as $1\times 1$ matrices, we do indeed have $$ \lambda\odot \mathbf c = \begin{pmatrix} \lambda c_1 \\ \vdots \\ \lambda c_n\end{pmatrix} = \mathbf c \cdot (\lambda) = \mathbf c \cdot \lambda. $$ Hence, $$ (\mathbf a^t \cdot \mathbf b)\odot \mathbf c = (\mathbf b^t \cdot \mathbf a)\odot \mathbf c = \mathbf c\cdot \mathbf b^t\cdot \mathbf a. $$

What you see here is that the linear map $\mathbf x \mapsto (\mathbf x^t\cdot \mathbf b)\odot \mathbf c$ is described by the matrix $\mathbf A=\mathbf c\cdot \mathbf b^t$ as $\mathbf x\mapsto \mathbf A\cdot \mathbf x$.

In general, the matrix product $\mathbf A\cdot\mathbf x$ of a matrix and a column vector is obtained as the linear combination of the columns of $\mathbf A$ with the coefficients from $\mathbf x$. Now note that the columns of $\mathbf c \mathbf b^t$ are all scalar multiples of $\mathbf c$, so any linear combination of those will be a scalar multiple of $\mathbf c$ as well. That explains why $(\mathbf c\cdot\mathbf b^t)\cdot \mathbf a$ is always a scalar multiple of $\mathbf c$.

Christoph
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