2
$\begingroup$

Given the following:

Let $T:\mathbb R^3 \rightarrow \mathbb R^3$ be a linear transformation. And let $B = ((1,1,1),(1,1,0),(1,0,1))$ be a basis of $\mathbb R^3$.

The representing matrix of the transformation is as follows: $[T]_B = \begin{bmatrix}1&0&1\\3&2&1\\2&1&1\end{bmatrix}$

Find a basis to the range ($Im T$) and the Kernel ($ker T$)

I've found the kernel by taking a generic vector from $\mathbb R^3$, applying the coordinates of the basis on it and found that $ker T = Sp{(1,0,0)}$.

I'm just not sure how to go about finding the range.

As far as I know, the steps to find it are as follows:

  1. Transpose the representing matrix.
  2. Use elementary actions to bring it to a canonical form.
  3. Restore the given matrix to a vectors via the basis coordinates.

I'm failing to understand the last part, here's what I've found: $\begin{bmatrix}1&0&1/2\\0&1&1/2\\0&0&0\end{bmatrix}$

How do I restore these vectors via the basis coordinates?

EDIT: Fixed the wrong basis B.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

The coordinates $[x,y,z]$ in the ordered basis $\{v_1,v_2,v_3\}$ represent the vector $xv_1+yv_2+zv_3$ in the standard basis $\{e_1,e_2,e_3\}$ (assuming $v_i$ are written in standard basis).

For the kernel basis, take $v\in\Bbb R^3$ in basis $B$.

$\begin{bmatrix}1&0&1\\3&2&1\\2&1&1\end{bmatrix}v=0$ gives $v=\begin{bmatrix}-k\\k\\k\end{bmatrix}$ which is the linear span of $-[1,1,1]^T+[1,1,0]^T+[1,0,1]^T=(1,0,0)$ in the standard basis.

Note that the range-space of a linear transformation is its column space (subspace spanned by column vectors) or the row-space of its transpose. You have found the linearly independent rows of $[T]_B^T$ as $[1,0,1/2]^T,[0,1,1/2]^T$ which serve as the basis vectors of the range in basis $B$. In the standard basis,

$$~[1,0,1/2]^T_S=1[1,1,1]^T+0[1,1,0]^T+\frac12[1,0,1]^T\\~[0,1,1/2]^T_S=0[1,1,1]^T+1[1,1,0]^T+\frac12[1,0,1]^T$$

$\endgroup$
2
  • $\begingroup$ Thank you for your answer, Shubham. But are you sure about my kernel basis being incorrect? To find it, I've taken a generic vector $v \in \mathbb R^3, v = (x,yz)$ applied the coordinates on it such as: $[(x,y,z)]_B = (-x+y+z, x-z, x-y)^t$ Multiply this vector by the transformation matrix, and you find $z =0, 2y+z = 0, y+z = 0$, which means the Kernel is $Sp{(1,0,0)}$ I want to say thanks again for the range of the transformation, Shubham. Also, an additional aspect of the problem which I did not add, as it's related to another subsection is that $(1,0,0) \in ker T$ $\endgroup$ Commented Jan 7, 2021 at 11:32
  • 1
    $\begingroup$ @EatayMizrachi Since you have fixed the question now, our answers match. $\endgroup$ Commented Jan 7, 2021 at 11:49

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.