Find the rank and the nullity of the following linear map from $\mathbb{R}^{4}$ to $\mathbb{R}^{3}$
$\left(x,y,z,t\right)\rightarrow(x-t,z-y,x-2y+2z-t)$
I understand how to find the rank and nullity, since the nullity is the dimension of the null space and the rank is the dimension of the column space. But I am having difficulty putting it into a matrix to solve.
And can someone also explain why we can take a linear map and essentially say that it is the "same" structurally as a the corresponding matrix. How do isomorphisms link into this idea ?
From this how would you know if it is onto ?
Best Answer
You can write the transformation as $$T(\mathbf{x})=\mathbf{A}\mathbf{x}=\begin{bmatrix}1&0&0&-1\\0&-1&1&0\\1&-2&2&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\\t\end{bmatrix}$$ where $\mathbf{A}$ is the transformation matrix. This matrix acts on a given vector $\mathbf{x}$ to give a transformed vector. If you carry out the matrix multiplication, you end up with $$\begin{bmatrix}1&0&0&-1\\0&-1&1&0\\1&-2&2&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\\t\end{bmatrix}=\begin{bmatrix}x-t\\-y+z\\x-2y+2z-t\end{bmatrix}$$