Image and kernel of a matrix transformation

After a long night of studying I finally figured out the answer to these. The previous answers on transformation were all good, but I have the outlined steps on how to find $\mathrm{im}(T)$ and $\ker(T)$.

$$A = \left(\begin{array}{crc} 1 & 2 & 2 & -5 & 6\\ -1 & -2 & -1 & 1 & -1\\ 4 & 8 & 5 & -8 & 9\\ 3 & 6 & 1 & 5 & -7 \end{array}\right)$$

(1) Find $\mathrm{im}(T)$

$\mathrm{im}(T)$ is the same thing as column space or $C(A)$. The first step to getting that is to take the Transpose of $A$.

$$ A^T = \left(\begin{array}{crc} 1 & -1 & 4 & 3 \\ 2 & -2 & 8 & 6 \\ 2 & -1 & 5 & 1 \\ -5 & 1 & -8 & 5 \\ 6 & -1 & 9 & -7 \end{array}\right)$$

once that's done the next step is to reduce $A^T$ to Reduced Row Echelon Form

$$ \mathrm{rref}(A^T) = \left(\begin{array}{crc} 1 & 0 & 1 & -2 \\ 0 & 1 & -3 & -5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)$$

now on this step I honestly don't know the reasons behind it, but the thext thing you do is take the rows and that's your answer. so that:

$$\mathrm{im}(T)\ = \begin{align*} \operatorname{span}\left\{\left(\begin{array}{crc} 1 \\ 0 \\ 1 \\ -2 \end{array}\right), \left(\begin{array}{crc} 0 \\ 1 \\ -3 \\ -5 \end{array}\right)\right\} \end{align*}$$

(2) Find $\ker(T)$

$\ker(T)$ ends up being the same as the null space of matrix, and we find it by first taking the Reduced Row Echelon Form of A

$$ \mathrm{rref}(A) = \left(\begin{array}{crc} 1 & 2 & 0 & 3 & -4\\ 0 & 0 & 1 & -4 & 5\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$$

we then use that to solve for the values of $\mathbb R^5$ so that we get

$$\begin{align*} \left(\begin{array}{crc} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{array}\right) = r\left(\begin{array}{crc} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right) + s\left(\begin{array}{crc} -3 \\ 0 \\ 4 \\ 1 \\ 0 \end{array}\right) + t\left(\begin{array}{crc} 4 \\ 0 \\ -5 \\ 0 \\ 1 \end{array}\right) \end{align*}$$

from that we arrange the vectors and get our answer the vectors and that gives us our answer

$$\begin{align*} \ker(T) = \operatorname{span}\left\{\left(\begin{array}{crc} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right), \left(\begin{array}{crc} -3 \\ 0 \\ 4 \\ 1 \\ 0 \end{array}\right), \left(\begin{array}{crc} 4 \\ 0 \\ -5 \\ 0 \\ 1 \end{array}\right)\right\} \end{align*}$$

and that's that.

What they mean by the transformation $T$ is the transformation which is induced by multiplication by $A$. You can verify that matrix multiplication is in fact a linear mapping, and in our particular case we have the linear mapping $T:\ \mathbf{x}\mapsto A\mathbf{x}$.

The image is then defined as the set of all outputs of the linear mapping. That is $$\operatorname{Im}(T) = \left\{\mathbf{y}\in \mathbb{R}^4\ \big|\ \mathbf{y} = A\mathbf{x}\ \text{such that}\ \mathbf{x}\in\mathbb{R}^5 \right\}$$ If you play around with the mapping a little bit then you should find that the image is in fact a very familiar subspace associated with the matrix $A$ (take a look at how the mapping $T$ acts on the standard basis).

The kernel is correspondingly defined as the set of all inputs which are taken to zero. $$\ker(T) = \left\{\mathbf{x}\in \mathbb{R}^5\ \big|\ A\mathbf{x} = \mathbf{0} \right\}$$ Again, there is a familiar subspace of the matrix $A$ associated with the kernel, look carefully at the definition and you should be able to figure out what it is.

By a linear transformation, they mean a function between vector spaces which satisfies $T(cx + y) = cT(x) + T(y)$. In our case, this transformation is multiplication by the matrix $A$.

The image is the set of all points in $\mathbb{R}^4$ that you get by multiplying this matrix to points in $\mathbb{R}^5$, you can find these by checking the matrix on the standard basis.

The kernel is the set of all points in $\mathbb{R}^5$ such that, multiplying this matrix with them gives the zero vector. Again you can find this in a similar way.