# span of image vectors

#### furryvision

hi!

i have matrix
1, 1, 0, 1
2, 1, 1, -1
3, 2, 1, 0

i am told:
"a possible basis might consist of x^3 - bx - cx and 2x^3 - 3x^2 + 1. one might find the basis for the image by considering the span of the image vectors (1, 2, 3), (1, 1, 2), (0, 1, 1) and (1, -1, 0), of the standard basis of V. this should lead to the observation that the image is precisely the subspace spanned by (1, 0, 1) and (0, 1, 1)".

(a)
i've tried performing a row echelon form on the original matrix to get
1, 1, 0, 1
2, 1, 1, -1
3, 2, 1, 0
=>
1, 1, 0, 1
0, 1, -1, 3
0, 0, 0, 0

which gives me two non-zero rows (1, 1, 0, 1) and (0, 1, -1, 3). these don't make sense for any of the x^3 - bx - cx, 2x^3 - 3x^2 + 1 or (1, 0, 1), (0, 1, 1) values given in the answer.

(b)
i've also tried switching them to columns and performing row echelon form:
1, 2, 3
1, 1, 2
0, 1, 1
1, -1, 0
=>
1, 2, 3
0, 1, 1
0, 0, 0
0, 0, 0

which again gives me two non-zero rows and i'm left with (1, 2, 3) and (0, 1, 1), which again doesn't work for what they say.

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#### dwsmith

MHF Hall of Honor
hi!

i have matrix
1, 1, 0, 1
2, 1, 1, -1
3, 2, 1, 0

i am told:
"a possible basis might consist of x^3 - bx - cx and 2x^3 - 3x^2 + 1. one might find the basis for the image by considering the span of the image vectors (1, 2, 3), (1, 1, 2), (0, 1, 1) and (1, -1, 0), of the standard basis of V. this should lead to the observation that the image is precisely the subspace spanned by (1, 0, 1) and (0, 1, 1)".

(a)
i've tried performing a row echelon form on the original matrix to get
1, 1, 0, 1
2, 1, 1, -1
3, 2, 1, 0
=>
1, 1, 0, 1
0, 1, -1, 3
0, 0, 0, 0

which gives me two non-zero rows (1, 1, 0, 1) and (0, 1, -1, 3). these don't make sense for any of the x^3 - bx - cx, 2x^3 - 3x^2 + 1 or (1, 0, 1), (0, 1, 1) values given in the answer.

(b)
i've also tried switching them to columns and performing row echelon form:
1, 2, 3
1, 1, 2
0, 1, 1
1, -1, 0
=>
1, 2, 3
0, 1, 1
0, 0, 0
0, 0, 0

which again gives me two non-zero rows and i'm left with (1, 2, 3) and (0, 1, 1), which again doesn't work for what they say.

This matrix is the coefficient matrix of $$\displaystyle P_4$$?

#### HallsofIvy

MHF Helper
I think that business about the polyomials is a (probably unintentional) red herring. Just ignore that.

If $$\displaystyle \begin{bmatrix}x \\ y \\ z \\ w \end{bmatrix}$$ is in $$\displaystyle R^4$$ then $$\displaystyle \begin{bmatrix} 1 & 1 & 0 & 1 \\ 2 & 1 & 1 & -1 \\ 3 & 2 & 1 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z \\ w \end{bmatrix}= \begin{bmatrix}x+ y+ w \\2x+ y+ z- w \\3x+ 2y+ z\end{bmatrix}$$

If we let p= x+ y+ w and q= 2x+ y+ z- w, the first two components, then note that p+ q= 3x+ 2y+ z, third component. That is, any vector in the image is of the form $$\displaystyle \begin{bmatrix}p \\ q \\ p+ q\end{bmatrix}$$$$\displaystyle = \begin{bmatrix}p \\ 0 \\ p\end{bmatrix}+ \begin{bmatrix} 0 \\ q \\ q\end{bmatrix}$$$$\displaystyle = p\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}+ q\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$$

But your idea of "switching them to columns and performing row echelon form" was good (since you don't learn "column reduction"!). You just stopped too soon.

You got to $$\displaystyle \begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$

In "reduced row echelon form" we want to have "0"s above the "pivot values" also so subtract 2 times the second row from the first to get
$$\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$
where the two rows, (1, 0, 1) and (0, 1, 1), are precisely the basis vectors desired.

(Notice that your (1, 2, 3) and (0, 1, 1) also form a basis for the image.)

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