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I have the following equation, with one matrix and a vector.

$$\mathbf C*\vec n =0$$

with $\vec n = \{a,b,d,e\}$ and $\mathbf C$ having being a 3x4 matrix.

The issue is, I know $a$ but want to represent the system in the more common form $\mathbf Ax=b$, where $x$ is fully unknown. I though of a way to isolate $a$, but I am not sure if this reasoning is correct:

It starts by representing this multiplication as a system of equations,

$$c_{11}*a+c_{12}*b+c_{13}*d+c_{14}*e=0$$ $$c_{21}*a+c_{22}*b+c_{23}*d+c_{24}*e=0$$ $$c_{31}*a+c_{32}*b+c_{33}*d+c_{34}*e=0$$ $$c_{41}*a+c_{42}*b+c_{43}*d+c_{44}*e=0$$

we could then group the first terms as a vector and 'undo' the matrix multiplication for the other terms.

$$\vec {c_1}\,a + \mathbf C_{-1} *n_{-1} =0 $$

where $\mathbf C_{-1}$ is $ \mathbf C$ without the first column, and $n_{-1} = \{b,d,e\}$.

Is this reasoning correct?

It seems sound to me, but I am yet to see any kind of material on the internet about manipulating matrix order, which make me suspicious that might be something more wrong with it. And, if this reasoning is correct, does this technique has a name? I heard about matrix splitting and decomposition but the cases I saw focus almost entirely on square matrices.

Martin Argerami
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JMenezes
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1 Answers1

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As you noticed, if $\mathrm C=[\vec c_1\ \vec c_2\ \cdots\ \vec c_r]$ and $\vec n=[n_1\ \cdots\ n_r]^T$, then $$ \mathrm C\,\vec n=n_1\,\vec c_1+\cdots+n_r\,\vec c_r. $$ I don't recall a name being assigned to this. It is mentioned in every linear algebra book, as far as I can tell. The key word is, I guess, column space.

Martin Argerami
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  • Your guess was correct. Wikipedia's page on column space a problem very similar to mine as example. Thank you. – JMenezes Sep 06 '16 at 16:15