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I am trying to find the Range and kernel of as linear transformation.

I have the following 4 x4 matrix:

$$\begin{pmatrix} 1 & 9 & 8 & 2 \\ 6 & 2 & 1 & 2 \\ 1 & 9 & 8 & 2 \\ 6 & 2 & 1 & 2 \ \end{pmatrix} \quad$$

By applying elementary row operations I obtained the matrix:

$$\begin{pmatrix} 1 & 0 & \frac{-7}{52} & \frac{7}{26} \\ 0 & 1 & \frac{47}{52} & \frac{5}{26} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \ \end{pmatrix} \quad$$

Then the range of the matrix ia given by:

$$\{ \alpha \begin{pmatrix} 1 \\ 6 \\ 1 \\ 6 \ \end{pmatrix} \quad+ \beta \begin{pmatrix} 9 \\ 2 \\ 9 \\ 2 \ \end{pmatrix} \quad \}$$

The rank is 2 and the basis for the range is given by:

$$\{ \begin{pmatrix} 1 \\ 6 \\ 1 \\ 6 \ \end{pmatrix} \quad, \begin{pmatrix} 9 \\ 2 \\ 9 \\ 2 \ \end{pmatrix} \quad \}$$

Then, the Kernel of the same matrix is given by:

$$\left(\begin{array}{cccc|c}   1 & 9 & 8 & 2 & 0\\   6 & 2 & 1 & 2 & 0 \\ 1 & 9 & 8 & 2& 0\\ 6 & 2 & 1 & 2 & 0 \\ \end{array}\right)$$

I obtained the following system of equations:

$x_1=\frac{7}{52}x_3-\frac{7}{26}x_4$

$x_2=\frac{-47}{52}x_3-\frac{5}{26}x_4$

$x_1$ arbritary and $x_4$ arbritary

then the kernel is given by:

N(A)=span $$\{ \begin{pmatrix} \frac{7}{52} \\ \frac{-47}{52} \\ 1 \\ 0 \ \end{pmatrix} \quad, \begin{pmatrix} \frac{-7}{26} \\ \frac{-5}{26} \\ 0 \\ 1 \ \end{pmatrix} \quad \}$$

I Also saw that alternatively the kernel (null space) can be given by:

N(A)=N(rref(A))= $$\{ x_3 \begin{pmatrix} \frac{7}{52} \\ \frac{-47}{52} \\ 1 \\ 0 \ \end{pmatrix} \quad+ x_4 \begin{pmatrix} \frac{-7}{26} \\ \frac{-5}{26} \\ 0 \\ 1 \ \end{pmatrix} \quad \}$$

The basis for the kernel is:

N(A)=span $$\{ \begin{pmatrix} \frac{7}{52} \\ \frac{-47}{52} \\ 1 \\ 0 \ \end{pmatrix} \quad, \begin{pmatrix} \frac{-7}{26} \\ \frac{-5}{26} \\ 0 \\ 1 \ \end{pmatrix} \quad \}$$

Nullity is 2

I am not sure if what I did is correct. I am not sure if the kernel can be given in these two different ways, and the range is well calculated (final answer is correct).

Can anyone help me on this? or point me to a website where I can find examples about how to find the kernel and range of a linear transformation?

thanks

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    If you just want to double-check the answer, you could use [Wolfram|Alpha](http://www.wolframalpha.com/input/?i=kernel+of+%28%281%2C9%2C8%2C2%29%2C%286%2C2%2C1%2C2%29%2C%281%2C9%2C8%2C2%29%2C%286%2C2%2C1%2C2%29%29) (your answer is correct). – 2017-02-26
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    So the kernel can be given in these two different ways? – 2017-02-26
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    You meant to say that $x_3$ and $x_4$ are arbitrary. – 2018-10-27

1 Answers 1

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As kennytm commented, you are correct.

A webpage with an explanation and examples that are more abstract than your question is at Kernel and Range of a Linear Transformation.

Yes, the kernel can be given in those two ways. By definition, a basis for the kernel is a linearly-independent set of vectors in the kernel that spans the kernel, and conversely, the kernel is the span (i.e., the set of all linear combinations) of the kernel's basis vectors.

Furthermore, the basis vectors can be any (two in your case) linearly-independent vectors in the kernel, so it is common to clear the denominators in any fractions. That is why Wolfram|Alpha reports different basis vectors.