Using row reduction on coefficient matrix to prove a set whose basis has less elements than dim(V) does not span $V$

Question

Larson Edwards Falvo - Elementary Linear Algebra

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In the following exercises in Section $4.5$ Basis and Dimension, the only set that has cardinality less than the dimension of vector space is the one in #$38$

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1. I can use the fact that all bases of a vector space with dimension $n$ must have $n$ elements.

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2. I can also approach as follows:

$$2c_1 = u_1$$ $$c_1-c_2 = u_2$$ $$c_2 = u_3$$

$$\to$$

$$(c_1,c_2) = (\frac{u_1}{2}, u_3)$$

and

$$(c_1,c_2) = (u_2 + u_3, u_3)$$

Thus, there is no unique way to represent a vector in $\mathbb R^3$ using vectors in $S$ i.e. $span(S) \ne \mathbb R^3$

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3. However, I would like to prove that $S$ does not span $\mathbb R^3$ using row reduction on the coefficient matrix (the matrix consisting of the vectors in $S$).

Main Question #1: How would I do this?

I tried row reducing

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

to

$$\begin{bmatrix} 1 & 0\\ 0 & 1\\ 0 & 0 \end{bmatrix}$$

but I don't know what to do with that. Am I supposed to take the transpose?

\begin{bmatrix} 2 & 1 & 0\\ 0 & -1 & 1 \end{bmatrix}

It row reduces to

$$\begin{bmatrix} 1 & 0 & 1/2\\ 0 & 1 & -1 \end{bmatrix}$$

What do I do with that?

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Actually, in the previous section $4.4$ Spanning Sets and Linear Independence, there are some questions like these:

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However, there is no row reduction approach:

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except I guess in #$51$ where there's row reduction on the transpose (as I linked above).

For #s $9$ and $19$, I looked up the book to see if there was a similar example. There is but it involves row reduction on the augmented matrix instead of simply coefficient matrix:

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Main Question #2: How can #$51$ get away with only the coefficient matrix (albeit on the transpose) but Example 3/6 needs the augmented matrix?

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Main Question #3: How if possible to approach #s 9 and 19 using row reduction on coefficient matrix? Otherwise, why can't we use just the coefficient matrix?

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For #$9$, I would get $$\begin{bmatrix} -3\\ 5 \end{bmatrix} \to \begin{bmatrix} 1\\ 0 \end{bmatrix}$$

For #$19$, I would get $$\begin{bmatrix} -2 & 4\\ 5 & 6\\ 0 & 3 \end{bmatrix} \to \begin{bmatrix} 1 & 0\\ 0 & 1\\ 0 & 0 \end{bmatrix}$$

Where do those get me?

Answer 1

Another way to do this problem is to note that $\{\textbf{e}^1, \textbf{e}^2, \textbf{e}^3\}$ is a basis for $\mathbb{R}^3$ i.e any other basis must have the same cardinality. I think proving that result is easier and helps for future problems.

Answer 2

I think that you started in the right direction. For $(u_1, u_2, u_3) \in \mathbb{R}^3$ we have $$2c_1 = u_1$$ $$c_1 - c_2 = u_2$$ $$c_2=u_3$$

From this we obtain $\frac{u_1}{2}-u_3 = u_2$ so we can't for example get the vector $(0, 1, 0)$

Answer 3

Answer to Main Question 1:

Use (row reduced) augmented matrix

to see that

$$0c_1 + 0c_2 = 1$$

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Answer to Main Question 2:

$52$ is not asking the sets span $R^3$. It is asking if they span the same subspace. There is no augmented matrix here because there is no $b$ (as in $Ax=b$; supposedly an arbitrary vector in $\mathbb R^3$), only an $A$ (as in $Ax=b$).

Observe that

$$\{A_1x | x \in \mathbb R^3 \} = span\{(1,0,0),(0,1,0),(1,1,0)\}$$

where $A_1$ is the matrix whose columns are the elements of $S_1$

$$\{A_2x | x \in \mathbb R^3 \} = span\{(1,0,0),(0,1,0)\}$$

where $A_2$ is the matrix whose columns are the elements of $S_2$

The vectors $(1,0,0),(0,1,0),(1,1,0)$ are not linearly independent. For some reason reducing the transpose

of

$$\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{bmatrix}$$

namely

$$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 1 & 1 & 0 \end{bmatrix}$$

gives

$$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}$$

whose transpose is

$$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}$$

(itself)

where the first two columns gives 2 linearly independent vectors that also span $\{A_1x | x \in \mathbb R^3 \}$ i.e.

$$\{A_1x | x \in \mathbb R^3 \} = span\{(1,0,0),(0,1,0)\}$$

Observe now that $$\{A_1x | x \in \mathbb R^3 \} = \{A_2x | x \in \mathbb R^3 \}$$

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Answer to Main Question 3:

Using coefficient matrix only is equivalent to using augmented matrix

If there's a row of zero(es), then the set does not span the vector space.