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Following on from a previous question - Definition for Basis of a Subspace; it is often said that a basis of a subspace is a set of vectors that can be used to uniquely represent any vector in the subspace.

I am having some trouble understanding the meaning of the word 'uniquely' in this context

In this article - http://mathworld.wolfram.com/VectorBasis.html; it is said that 'a vector space has many different vector bases'. However in this item - https://www.quora.com/What-is-a-basis-in-linear-algebra; it is said that 'each vector in the vector space can be written exactly in one way as a linear combination of the basis vectors'.

I find the above language rather confusing as the first quote seems to suggest that there is no 'unique' way to represent any given vector - there are 'many different' bases. And then the second quote states that 'each vector in the vector space can be written in exactly one way'. These quotes seem to contradict each other.

My confusion is likely routed in a misunderstanding of what is meant by 'unique' in this context, so I seek the advice of Stack Exchange members on this topic.

Perhaps if I work through an example, users will be able to detect the error in my thinking.


$S$ is a linearly independent subset of $\mathbb R^{2}$, that spans all of the subspace:

$S = \{\begin{bmatrix}2\\5\end{bmatrix}, \begin{bmatrix}0\\6\end{bmatrix}\}$

We should be able to find any given vector in the subspace. Given the arbitrary vector $\vec x$:

$\vec x = \begin{bmatrix}x_1\\x_2\end{bmatrix}$

We can use a linear combination of the vectors in $S$ to arrive at $\vec x$:

$2(b_1) + 0(b_2) = x_1$

$5(b_1) + 6(b_2) = x_2$

By solving for $b_1$ and $b_2$, we arrive at:

$b_1 = \frac{x_1}2$

$b_2 = \frac{x_1 + x_2 * (-2 \div 5)}{-12 \div 5}$

Suppose that we want to find the vector $\vec e$:

$\vec e = \begin{bmatrix}-4\\5\end{bmatrix}$

We can use the following linear combination to get us there:

$2(\frac{-4}2) + 0(\frac{-4 + 5 * (-2 \div 5)}{-12 \div 5}) = -4$

$5(\frac{-4}2) + 6(\frac{-4 + 5 * (-2 \div 5)}{-12 \div 5}) = 5$


Hopefully I have provided enough information as for the reader to decipher my misunderstanding.

Thanks in advance.

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  • $\begingroup$ Note for example that $(1,0), (0,1), $ and $(1,1)$ are not a basis for $\Bbb R^2$ because $(2,3)$ can be written as $2(1,0)+3(0,1)$ or as $1(0,1)+2(1,1)$ $\endgroup$ – J. W. Tanner Aug 13 at 12:17
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I find the above language rather confusing as the first quote seems to suggest that there is no 'unique' way to represent any given vector - there are 'many different' bases. And then the second quote states that 'each vector in the vector space can be written in exactly one way'. These quotes seem to contradict each other.

You may be confused because you are mixing two things:

  • a basis for a subspace is not unique in the sense that a subspace can have many different bases; e.g. $\left\{ \begin{pmatrix}1 \\0 \end{pmatrix} , \begin{pmatrix}0 \\1 \end{pmatrix} \right\}$ and $\left\{ \begin{pmatrix}1 \\1 \end{pmatrix} , \begin{pmatrix}2 \\-1 \end{pmatrix} \right\}$ are two possible bases for $\mathbb{R}^2$;

  • once a specific basis is chosen, then every vector can be written in only one way as a linear combination of these basis vectors, which gives rise to its coordinates relative to that basis and we call this a unique representation with respect to that basis.


Suppose that we want to find the vector $\vec e$:

$\vec e = \begin{bmatrix}-4\\5\end{bmatrix}$

We can use the following linear combination to get us there:

What you find now is the only way to write $\vec x$ as a linear combination of the vectors in the basis $S$. We say that the coordinates of $\vec x$ with respect to the basis $S$ are unique.

However, you could choose another basis $T$ instead of $S$ and also with respect to that new basis, the vector $\vec x$ will have a unique representation.

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  • $\begingroup$ Thank you for that excellent explanation. Just one thing; when you say: "once a specific basis is chosen, then every vector can be written in only one way", just to clarify, the basis isn't so much chosen as found, is it? I am maybe being overly pedantic here, but to me it seems that we choose/find the basis by solving for $b_1$ and $b_2$, and in that way we arrive at formulas that become the constants in the linear combination. Is this a correct interpretation of what you said? $\endgroup$ – Ryan Walter Aug 13 at 12:28
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    $\begingroup$ The basis is the set of basis vectors, not the $b_1$ and $b_2$ in your notation. Concerning finding/choosing: in an exercise, you could be asked to find a basis for a subspace that is being described to you in some way. Once you're familiar with the material and that's no longer an exercise on its own, you would say you pick or choose a basis for a subspace (since it's not unique!). $\endgroup$ – StackTD Aug 13 at 12:32
  • $\begingroup$ Okay, I think I'm starting to understand what is causing the confusion. The vectors contained within the set $S$ in my example are basis vectors, and $b_1$ and $b_2$ are just constants that are used in combination with the basis vectors in order to obtain any given vector in the subspace. In this way, you can choose the basis vector as, like you mentioned, "a basis for a subspace is not unique in the sense that a subspace can have many different bases". $\endgroup$ – Ryan Walter Aug 13 at 12:46
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    $\begingroup$ Yes, I think you got it! And once a basis is chosen, these $b$'s are unique (for any given vector). $\endgroup$ – StackTD Aug 13 at 12:50
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For each basis $B$, there is one and only one way of representeing each vector as a linear combination of vectors of $B$. But, of course, if choose another basis $B^\star$, then the way of representing a vector as a linear combination of elements of $B^\star$ can be (and usually is) different from the way of representing it as a linear combination of elements of $B$.

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