There is a linear operator $f:ℝ^3\to\mathcal{P}_3(x)$ which is defined by:


I need to find core, image, rank and defect of operator $f$.

Results are:
$Im(f)=\mathcal P_3(x)$

I have correct results but I'm unsure if my theory on how to get them is correct and I would like some clarification.

So first I write above statement into canon matrix and it looks like:

$F=\begin{bmatrix} 1 && 0 && 2\\ 1 && 1 && 0\\ 0 && 1 && 1 \end{bmatrix}$

Once I simplify the matrix by multiplying first row with $-1$ and then adding it to the second row, and then multiplaying the second row with $-1$ and then adding it to the third row I get:

$F=\begin{bmatrix} 1 && 0 && 2\\ 0 && 1 && -2\\ 0 && 0 && 3 \end{bmatrix}$

I think this is enough for finding everything that I require.

Since matrix in this case can be written as independent we can say that there won't be core: $Cor(f)=\{\Theta\}$

Since there isn't any core, then defect will be zero: $d(f)=0$

Rank can be found as $r(f)=dim ℝ^3-d(f)$ and we get that rank is: $r(f)=3$

Now I'm quite uncertain on how to get the image, but my guess for this case would be that because of results from rank and core, it will stay the same as it is defined. $Im(f)=\mathcal P_3(x)$


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