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Consider the linear transformation of $\mathbb{R}^3$ given by $Ax= (a \cdot x)a + |a|^2x$

a) What are the range (image) and kernel of A?

b) Find the matrix of A in the usual basis $e_j$

The kernel is when Ax=0. Does this mean that it is when $(a \cdot x)a = -|a|^2x$?

How does one compute the $Im(A)$ in this case and find the matrix of A in the usual basis?

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    $\begingroup$ Are you sure it's $+$? For $a\ne 0$ this has kernel$=0$ and image $=\mathbb{R}^3$. $\endgroup$ – Orest Bucicovschi Oct 5 '14 at 18:10
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$x$ is in the kernel of $A$ when $Ax=0$ which implies that $(a\cdot x)a=-|a|^2x$. But this equality can hold only if $x$ is a multiple of the vector $a$. Plugging in $x=ca$ you can see that $c=0$, and hence $$\ker A=\{0\}.$$ In other words, $A\colon \mathbb{R}^3\to \mathbb{R}^3$ is a one-to-one linear map. But any one-to-one linear map with from a finite dimensional space to itself is onto as well. Thus, $$\text{Image}(A)=\mathbb{R}^3.$$ To find the standard matrix of $A$, simply evaluate $A$ at $e_i$. More precisely, the $i$-th column of the standard matrix is $Ae_i$. For instance the first column is given by $$Ae_1=(a\cdot e_1)a+|a|^2e_1=a_1a+|a|^2e_1.$$

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  • $\begingroup$ So, since A is a linear mapping in $\mathbb{R}^3$, the matrix would simply be $\begin{pmatrix} Ae_1 & 0 & 0 \\ 0 & Ae_2 & 0 \\ 0 & 0 & Ae_3 \end{pmatrix}$? This comes out to $\begin{pmatrix} a_1a+|a|^2e_1 & 0 & 0 \\ 0 & a_2a+|a|^2e_2 & 0 \\ 0 & 0 & a_3a+|a|^2e_3 \end{pmatrix}$ $\endgroup$ – Confused_Engineer Oct 5 '14 at 18:25
  • $\begingroup$ No, The standard matrix of $A$ is the following $3\times 3$ matrix: $[Ae_1\ Ae_2\ Ae_3]$. Here $Ae_i$ is the $i$th column. I have computed in my answer what the first column of this matrix is. $\endgroup$ – EPS Oct 5 '14 at 18:28
  • $\begingroup$ Sorry, see the above edits and see if that makes more sense. The $e_1 =$ was a typo. $\endgroup$ – Confused_Engineer Oct 5 '14 at 18:30
  • $\begingroup$ You are still confused about the entries of the matrix. As I mentioned above the standard matrix is $3\times 3$. Instead of giving its entries I have given you its columns. $Ae_1$ is a column vector with $3$ entries. Indeed, $$a_1a+|a|^2e_1=a_1\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}+|a|^2\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}2a_1^2+a_2^2+a_3^2\\a_1a_2\\a_1a_3\end{bmatrix}$$. This is what the first column of the matrix is. You can compute the entries of the other columns similarly. $\endgroup$ – EPS Oct 5 '14 at 18:37
  • $\begingroup$ Alright. That makes sense. Thanks. $\endgroup$ – Confused_Engineer Oct 5 '14 at 18:43

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