Why row vectors equal to linear combination of basis vectors of rowspc(in this case)?

Let $\displaystyle v_1, v_2, ...,\,\,\, and \,\,\,v_m$ be the row vectors of a matrix:

$\displaystyle A = \begin{bmatrix} a_{11} & a_{12} & ... &a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ... & ... & ...\\a_{m1} & a_{m2} & ... & a_{mn} \end{bmatrix} $

Suppose the row space of $\displaystyle A$ has dimension $\displaystyle r$ and basis $\displaystyle S = \{b_1, b_2, ..., b_r\}$ where $\displaystyle b_i = (b_{i1}, b_{i2}, ..., b_{in})$.

Using this basis, you can write the row vectors of $\displaystyle A$ as(the book wrote this)

$\displaystyle v_1 = c_{11}b_1 + c_{12}b_2 + ... + c_{1r}b_r $

$\displaystyle v_2 = c_{21}b_1 + c_{22}b_2 + ... + c_{2r}b_r $

$\displaystyle ...\,\,\, ... \,\,\, ....\,\,\, $

$\displaystyle v_m = c_{m1}b_1 + c_{m2}b_2 + ... + c_{mr}b_r $

My question: why $\displaystyle v_1 = c_{11}b_1 + c_{12}b_2 + ... + c_{1r}b_r$ and $\displaystyle v_2 = c_{21}b_1 + c_{22}b_2 + ... + c_{2r}b_r$ and $\displaystyle v_m = c_{m1}b_1 + c_{m2}b_2 + ... + c_{mr}b_r$ ?

Why the book states that the linear combination of basis vectors equal to $\displaystyle v_1, v_2, ...\,\,\, and\,\,\, v_m$?

What method they used to find this? I can't see it. Sorry.

Re: Why row vectors equal to linear combination of basis vectors of rowspc(in this ca

We know that a set of vectors $\displaystyle S = \{g_1, g_2, ...,g_n\}$ in a vector space $\displaystyle V$ is called a basis for $\displaystyle V$ if the following conditions are true.

1) $\displaystyle S$ spans $\displaystyle V$

2) $\displaystyle S$ is linearly independent

I can't find the answer of my above mentioned question. All I can think of is this.

So in the above post $\displaystyle \{v_1, v_2, ..., v_r\}$ are the row vectors of matrix $\displaystyle A$ and so spans the row space of $\displaystyle A$ which is $\displaystyle row\,\,space\,\,of\,\,A \,\,\, = d_1 v_1 + d_2 v_2 + ... + d_m v_m$:

Shouldn't $\displaystyle d_1 v_1 + d_2 v_2 + ... + d_m v_m = e_1 b_1 + e_2 b_2 +... + e_r b_r$ because $\displaystyle \{b_1, b_2, ..., b_r\}$ is the basis of row space of $\displaystyle A$ and for the fact that $\displaystyle row\,\,space\,\,of\,\,A \,\,\, = e_1 b_1 + e_2 b_2 +... + e_r b_r$?

So how and why

$\displaystyle v_1 = c_{11}b_1 + c_{12}b_2 + ... + c_{1r}b_r $

$\displaystyle v_2 = c_{21}b_1 + c_{22}b_2 + ... + c_{2r}b_r $

$\displaystyle ...\,\,\, ... \,\,\, .... $

$\displaystyle v_m = c_{m1}b_1 + c_{m2}b_2 + ... + c_{mr}b_r $

?

Can anyone kindly help me understand please?

Re: Why row vectors equal to linear combination of basis vectors of rowspc(in this ca

Quote:

Originally Posted by

**x3bnm** Let [tex]

Why the book states that the linear combination of basis vectors equal to $\displaystyle v_1, v_2, ...\,\,\, and\,\,\, v_m$?

$\displaystyle b_{1}, b_{2},...,b_{r}$ are a basis for the rowspace of A, so every combination of the rows of A can be written in terms of this basis.

Each of $\displaystyle v_1, v_2, ...\,\,\, and\,\,\, v_m$ is in the row space of A, and so can be written in terms of $\displaystyle b_{1}, b_{2},...,b_{r}$.

Did you understand column spaces? Row space works the same way, except obviously related to rows rather than columns. If you did understand column space,

note that the row space of $\displaystyle A$ is equal to the column space of $\displaystyle A^{T}$, and a basis for the row space of $\displaystyle A$ is a basis for the column space of $\displaystyle A^{T}$

Re: Why row vectors equal to linear combination of basis vectors of rowspc(in this ca

Quote:

Originally Posted by

**Joanna** $\displaystyle b_{1}, b_{2},...,b_{r}$ are a basis for the rowspace of A, so every combination of the rows of A can be written in terms of this basis.

Each of $\displaystyle v_1, v_2, ...\,\,\, and\,\,\, v_m$ is in the row space of A, and so can be written in terms of $\displaystyle b_{1}, b_{2},...,b_{r}$.

Did you understand column spaces? Row space works the same way, except obviously related to rows rather than columns. If you did understand column space,

note that the row space of $\displaystyle A$ is equal to the column space of $\displaystyle A^{T}$, and a basis for the row space of $\displaystyle A$ is a basis for the column space of $\displaystyle A^{T}$

Thanks a lot for clarifying this.