Linear Transformations: Why w1 is a Linear Combination of v

[Amin2014]

Given w = T (v), where T is a linear transformation and w and v are vectors, why is it that we can write any coefficient of w, such as w₁ as a linear combination of the coefficients of v? i.e. w₁ = av₁ + bv₂ + cv₃

Supposably this is a consequence of the definition of linear transformations, but I don't see the connection.

 

[fresh_42]

In order to write $w=(w_1,w_2,w_3)$ and $v=(v_1,v_2,v_3)$ what do you need? And what does this mean for $T$?

 

[Amin2014]

In order to write $w=(w_1,w_2,w_3)$ and $v=(v_1,v_2,v_3)$ what do you need? And what does this mean for $T$?

We need a basis to be able to express a vector in terms of coefficents. I don't know what your second question implies.

 

[fresh_42]

$T$ is a linear transformation. What does this mean? Sure, it means that $T(\alpha u + \beta v)=\alpha T(u)+\beta T(v)$, but since we have a basis, we can express this transformation in terms of the basis, too:
We write down $T(1,0,\ldots ,0)\, , \, \ldots \, , \,T(0,\ldots,0,1)$ as column vectors with respect to the same basis we used to write $w$. The result is a number scheme which we call a matrix. Now it turns out that $w=Tv$ is exactly the matrix multiplication $(w_1,\ldots,w_m) = \left( \sum_{j=1}^n T_{ij}v_j \right)_{1\leq i \leq m}$ where $n=\dim V$ and $m=\dim W$.

So the coefficients of $v$ and $w$ need a bases to write them as a tuple of components, and the same bases allow us to write the linear transformation as a number scheme, too.

Your example had $n=m=3$ and the coefficients are $a=T_{11}, b=T_{12},c=T_{13}$.

 

[Amin2014]

$T$ is a linear transformation. What does this mean? Sure, it means that $T(\alpha u + \beta v)=\alpha T(u)+\beta T(v)$, but since we have a basis, we can express this transformation in terms of the basis, too:
We write down $T(1,0,\ldots ,0)\, , \, \ldots \, , \,T(0,\ldots,0,1)$ as column vectors with respect to the same basis we used to write $w$. The result is a number scheme which we call a matrix. Now it turns out that $w=Tv$ is exactly the matrix multiplication $(w_1,\ldots,w_m) = \left( \sum_{j=1}^n T_{ij}v_j \right)_{1\leq i \leq m}$ where $n=\dim V$ and $m=\dim W$.

So the coefficients of $v$ and $w$ need a bases to write them as a tuple of components, and the same bases allow us to write the linear transformation as a number scheme, too.

Your example had $n=m=3$ and the coefficients are $a=T_{11}, b=T_{12},c=T_{13}$.

 

[WWGD]

Given $T: V \rightarrow W$, Consider a basis $\{ v_1, v_2,...,v_n\}$ for $V$. Then $T(v)=...$

 

[Amin2014]

$T$ is a linear transformation. What does this mean? Sure, it means that $T(\alpha u + \beta v)=\alpha T(u)+\beta T(v)$, but since we have a basis, we can express this transformation in terms of the basis, too:
We write down $T(1,0,\ldots ,0)\, , \, \ldots \, , \,T(0,\ldots,0,1)$ as column vectors with respect to the same basis we used to write $w$. The result is a number scheme which we call a matrix. Now it turns out that $w=Tv$ is exactly the matrix multiplication $(w_1,\ldots,w_m) = \left( \sum_{j=1}^n T_{ij}v_j \right)_{1\leq i \leq m}$ where $n=\dim V$ and $m=\dim W$.

So the coefficients of $v$ and $w$ need a bases to write them as a tuple of components, and the same bases allow us to write the linear transformation as a number scheme, too.

Your example had $n=m=3$ and the coefficients are $a=T_{11}, b=T_{12},c=T_{13}$.

Got it!