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I had a couple of questions about a matrix problem. What I'm given is:

Consider a linear transformation $T: \mathbb R^5 \to \mathbb R^4$ defined by $T( \vec{x} )=A\vec{x}$, where $$A = \left(\begin{array}{crc} 1 & 2 & 2 & -5 & 6\\ -1 & -2 & -1 & 1 & -1\\ 4 & 8 & 5 & -8 & 9\\ 3 & 6 & 1 & 5 & -7 \end{array}\right)$$

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

  2. Find $\ker(T)$

My questions are:

What do they mean by the transformation?

What do I use to actually find the image and kernel, and how do I do that?

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  • $\begingroup$ Most people use the definitions of im(T) and ker(T). Do you have such definitions? $\endgroup$ – rschwieb Nov 13 '12 at 17:33
  • $\begingroup$ no all I have is that statement and the reduced row echelon form of the matrix. $\endgroup$ – StealzHelium Nov 13 '12 at 17:40
  • $\begingroup$ The image is the set of points $\{Ax\}_{x \in \mathbb{R}^5}$, the kernel is the set of points $\{x | Ax=0\}$. $\endgroup$ – copper.hat Nov 13 '12 at 17:41
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    $\begingroup$ Well that explains a lot of your difficulty, if you were trying to do a problem without knowing what the words meant. $\endgroup$ – rschwieb Nov 13 '12 at 17:48
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    $\begingroup$ I wonder why did 31k people visit this question? $\endgroup$ – heptagon Sep 2 '16 at 22:32
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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:

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

(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

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

and that's that.

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  • $\begingroup$ Could you be more detailed on "arrange the vectors"? $\endgroup$ – zthomas.nc Feb 22 '17 at 18:04
  • $\begingroup$ @zthomas.nc I think during row operations the rows were switched which would lead to them not mapping out correctly $\endgroup$ – keji Mar 10 '17 at 5:40
  • $\begingroup$ @kezi nope, he just made a mistake. The vectors above the last set are the span. $\endgroup$ – user5389726598465 May 29 '17 at 8:53
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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.

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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.

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  • $\begingroup$ I'm not sure if it's a good idea to call it "points", maybe you should call them objects to be more generic. Are we sure they are points? $\endgroup$ – nbro Sep 2 '16 at 21:48

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