Linearly dependent vectors, subspace and direct sum.

Question

I'm learning linear algebra, specifically linear independence and subspaces, and need help with the following problem:

Consider the vectors $\vec{x} = (1, 2, -2, 1), \vec{y} = (1, 3, -1, 4), \vec{z} = (2, 1, -7, -7) \in \mathbb{R^4}$ and let $V$ be the subspace spanned by $\vec{x}, \vec{y}, \vec{z}$. $(1)$ Show that the vectors $\vec{x}, \vec{y}, \vec{z}$ are linearly dependent and find their dependence relationship. $(2)$ Find a subspace $U$ such that $\mathbb{R^{4}} = V \oplus U$.

Since I'm having difficulties for $(2)$, I'm going to share my work for $(1)$.

$(1)$ The vector $\vec{x}, \vec{y}, \vec{z}$ are linearly dependent if the exists $\alpha_i, i= 1, 2, 3$ not all $0$ such that

$$\alpha_1 \begin{pmatrix}1\\2\\-2\\1\end{pmatrix}+\alpha_2 \begin{pmatrix}1\\3\\-1\\4\end{pmatrix}+\alpha_3 \begin{pmatrix}2\\1\\-7\\-7\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix} \tag{*}$$

$(*)$ translates to a system of linear equations. We use the Gauss-Jordan elimination method to solve it. We write down the augmented matrix and after some simple operations we find the row-reduced echelon form

$$\begin{pmatrix}1 & 1 & 2 & 0\\2 & 3 & 1 & 0\\-2 & -1 & -7 & 0\\ 1 & 4 & -7 & 0\end{pmatrix} \to \ldots \to \begin{pmatrix}1 & 0 & 5 & 0\\0 & 1 & -3 & 0\\0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}.$$

I skipped the steps to find the row-reduced echelon form to save me some time, they are very simple. There are an infinite number of solutions to the system given by $(*)$ which means that the vectors $\vec{x}, \vec{y}, \vec{z}$ are linearly dependent. From the row-reduced echelon form we see that $\alpha_1, \alpha_2$ are basic variables and $\alpha_3$ is the free variable. The dependence relationship can be written as

$$\alpha_1 = -5\alpha_3, \, \alpha_2 = 3\alpha_3, \, \alpha_3 \, \text{free}.$$

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Is my work correct for $(1)$? Besides the intermediate steps to find the row-reduced echelon form, do I need to clarify anything else? For $(2)$ I'm a bit lost. I just finished learning about vector subspaces and direct sums and everything is still not clear. I think the first step would be to find a basis for $V$ (how?). Then we know that the dimension of the subspace $U$ that we need to determine must satisfy $\dim{U} = 4 - \dim{V}$.

Any help would be appreciated.

Answer 1

Your work for part one appears to be correct, minus the typo you made (twice). The fourth coordinate of $z$ is $-7$ in the questions, and you have it as $7$ in your vector equation and augmented matrix (you clearly did the work with the other vectors though, so that's just more of a note!)

In terms of whether you need more - that would depend on your Professor, but in general I would say no. You could demonstrate that taking $a_1 = -5a_3$, $a_2 = 3a_3$ and $a_3 = a_3$ you are guaranteed to have $0$. It would look something like:

$$ 1\times(-5a_3) + 1\times(3a_3) + 2\times(a_3) = -5a_3 + 3a_3 + 2a_3 = 0a_3 = 0 \\ 2\times(-5a_3) + 3\times(3a_3) + 1\times(a_3) = -10a_3 + 9a_3 + a_3 = 0a_3 = 0 \\ -2\times(-5a_3) - 1\times(3a_3) - 7\times(a_3) = 10a_3 - 3a_3 - 7 a_3 = 0a_3 = 0 \\ 1\times(-5a_3) + 4\times(3a_3) - 7\times(a_3) = -5a_3 + 12a_3 - 7a_3 = 0a_3 = 0 $$

It is important to note further that $z = 5x - 3y$.

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For part (B) you need to find another vector space such that the Direct Sum gives you all of $\mathbb{R}^4$ We thus need $\mathbb{R}^4 = V + U$ and $V\cap U = \{0\}$

Because $V$ is linearly dependent, notably that $z = 5x - 3y$, then $V = span\{x, y\}$.

Ultimately, since you are currently dealing with a set of 2 linearly independent vectors, two more must be added such that the four together form a basis. If you add two vectors that are independent of $x$ and $y$, then by definition of linear independence you would have that $V\cap U = \{0\}$, and since you are dealing with the span of $4$ linearly independent vectors, that must form a basis for $\mathbb{R}^4$.

There are a number of ways which you can try to find these two vectors to form a spanning set. One suggestion may be to use orthogonality. This works since any non-zero, orthogonal vectors are linearly independent. Another may be to consider the following:

$$ w = \begin{bmatrix} 2 \\ 5 \\ -3 \\ 1 \end{bmatrix} $$

This vector is constructed simply by taking a linear combination $x + y$ and then differing the 4th component from what was expected.

Thus any combination of the three can be written: $$ z* = \begin{bmatrix} a + b + 2c \\ 2a + 3b + 5c \\ -(2a + b + 3c) \\ a + 4b + c \end{bmatrix} $$

Now, consider:

$$ v = \begin{bmatrix} 2 \\ 6 \\ 2 \\ 1 \end{bmatrix} $$

This matrix was derived simply by taking linear combinations of the three above, and making the system unsolvable. Specifically note that:

$$ a + b + 2c = 2 \implies a = 2 - b - 2c \\ 2(2-b-2c) + 3b + 5c = 6 \implies b = 2 - c \\ \implies -(2(2-b-2c) + (2 - c) + 3c) = -2 $$

So by taking $<2, 6, X, Y>$ where $x \neq -2$ we would arrive at an insolvable system, and thus they must be linearly independent.

So then taking $U = span\{w, v\}$ you should have that they are a basis for $\mathbb{R}^4$.

Answer 2

Both augmented matrices that you have written have rank 2, thus $x$, $y$ and $z$ span a two - dimensional subspace of $\mathbb{R^4}$, (equivalent to $\mathbb{R^2}$). Because the right matrix has more than one minor of rank 2, any two of the vectors $x$, $y$ and $z$ are linearly independent and thus form a basis in the subspace $V$. In order to construct U, also equivalent to $\mathbb{R^2}$, you need two linearly independent vectors that are not in $V$ - the simplest way to construct them is to find a vector orthogonal to two of $x$, $y$, $z$ and then find a fourth vector, orthogonal to $V$ and the newly constructed vector. This you do with the Gram-Schmidt procedure.