Solutions to equations involving linear transformations

[Mr Davis 97]

I have learned that for matrix theory, for $A \vec{x} = \vec{b}$, if there exists a particular solution $p$, then every solution looks like $p+k$, where $k \in \ker A$.

Can someone help me find material on this online, but only in the context of general linear transformations? For example, I want something explaining that in general solutions to $T (\vec{x}) = \vec{b}$ looks like a translation of the kernel.

 

[mathwonk]

do you know the definition of a linear transformation? If so try using it to prove that if T is linear and T(v) = b = T(w), then v-w is in the kernel of T. Then ask yourself how this relates to your question.

 

[fresh_42]

Another helpful thought is to visualize it geometrically. Solve $\begin{bmatrix}2&-1\\ 0 & 0\end{bmatrix} \vec{x}=\begin{bmatrix}-3 \\ 0 \end{bmatrix}$ which can be drawn on a piece of paper, and think about what it means for the question: what is $\vec{b}$, what $\vec{p}$ and what the kernel? Even three dimensional examples can easily be drawn, although it'll be a bit more work to do; on the other hand, there will be more dimensions for the kernel available.

 

[Mark44]

I have learned that for matrix theory, for $A \vec{x} = \vec{b}$, if there exists a particular solution $p$, then every solution looks like $p+k$, where $k \in \ker A$.

Can someone help me find material on this online, but only in the context of general linear transformations? For example, I want something explaining that in general solutions to $T (\vec{x}) = \vec{b}$ looks like a translation of the kernel.

Let $\vec x = \vec p + \vec k$, where $\vec p$ is a solution to $T(\vec p) = \vec b$, and $\vec k$ is any vector in the kernel of T. Then $T(\vec x) = T(\vec p + \vec k) = T(\vec p) + T(\vec k) = \vec b + \vec 0 = \vec b$, by the linearity of linear transformations. The matrix form of the equation follows immediately.