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Suppose $\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}\right\}$ is a linearly dependent spanning set for a vector space $V .$ Show that each $\mathbf{w}$ in $V$ can be expressed in more than one way as a linear combination of $\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}$ . [Hint: Let $\mathbf{w}=k_{1} \mathbf{v}_{1}+\cdots+k_{4} \mathbf{v}_{4}$ be an arbitrary vector in $V$ Use the linear dependence of $\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}\right\}$ to produce another representation of $\mathbf{w}$ as a linear combination of $\mathbf{v}_{1}, \ldots, \mathbf{v}_{4} . ]$

So $\mathbf{w}$ is expressed in more than one way as a linear combination of $\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}$ and $\mathbf{v}_{4}$

Calculus 3

Chapter 4

Vector Spaces

Section 4

Coordinate Systems

Vectors

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I think this problem We're given a set off the nearly dependent vectors. We want me to be free and report. We have to solve this. We're gonna use the humans. Let's try it back to w uh, using an arbitrary number K. So W's came one of me One plus get to be too plus k for before. What we know is this We know that this be Zarley nearly depend by so they depend on each other so well, let's assume not be one is equal to, uh, some constant A to B two plus a three d three plus a Ford We pour hear Keyser arbitrating And here a czar Arbitrary. So let's write to be one in 31 in terms of B to B B m w will be, um, a one a two we two plus a three d three plus a four before plus K to be two plus three v three plus cape or we fall. So then w will be equal to a one a two plus que tu we to plus one a three plus three Well, three plus J won a four plus four me for So since case and a CZ arbitrary. And since we can, right, let's say V two in terms of We want me three Emmy for orchid Correct. Be three in terms of other valuables are directors. It means that there's more than one way Jax Prospector read, So Dan W. Can be expressed as a linear combination off their vectors, we warn me to briefly and in numerous ways.

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