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Let $$A= \begin{bmatrix} 0 & 0 & 6 & -18 \\ 0 & 0 & -1 & 3 \\ 0 & 0 & -2 & 6 \\ \end{bmatrix} $$ Find a basis for the kernel and image of the linear transformation $T$ defined by $T(x) =Ax$.


My question is how do you handle the first 2 columns of A which are all zero's? Does that mean the top 2 values of the basis of the kernel are variables? both zero? The methods I know for calculating the basis for the kernel and image of a transformation matrix are not producing the correct answers in WebWork.

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3 Answers 3

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By elimination, you will end up with $$ \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 3 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $$ which gives you the dimension of the image of $T$ and the kernel of $T$.

Now, you can read from the matrix a basis for the image of $T$. On the other hand $$ -x_3+3x_4=0 $$ tells you how to find a basis for the kernel of $T$.

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  • $\begingroup$ Explicitly, the fact that $x_1$ and $x_2$ do not show up in the equation defining the kernel tells you that $[1 \, 0\, 0\, 0]^T$ and $[0\, 1\, 0\, 0]^T$ are in the kernel. Take these together with a vector of the form $[0\, 0\, x_3\, x_4]^T$ that satisfies the equation, and you've got your basis. $\endgroup$
    – Alex G.
    May 31, 2023 at 13:14
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Image is spanned on linearly independent columns of matrix $A$, which follows from definition of matrix of linear transformation. To find it do the row reduction of $A^T$ and its nonzero columns span the image, or equivalently, if you row reduce $A$ then columns with pivots will correspond to columns of $A$ that form image of the transformation. Kernel is set of vectors $x$ such that $Ax=0$. To find it's base you need to solve system of equations given by $Ax=0$.

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The kernel basis is:

$$ \left[ \begin{array}{cc} 0&0&1&1/3\\ \end{array} \right]^T, \left[ \begin{array}{cc} 1&0&0&0\\ \end{array} \right]^T, \left[ \begin{array}{cc} 0&1&0&0\\ \end{array} \right]^T$$

The image basis is:

$$\left[ \begin{array}{cc} 6&-1&-2\\ \end{array} \right]^T$$

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