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I know that for a set of vectors $\{ v_{1}, v_{2}, \ldots , v_{n} \} \in \mathbb{R}^{n}$ we can show that the vectors form a basis in $\mathbb{R}^{n}$ if we show that the coefficient matrix $A$ has the property $\det(A) \neq 0$, because this shows the homogeneous system has only the trivial solution, and the non-homogeneous system is consistent for every vector $(b_{1}, b_{2}, \ldots , b_{n}) \in \mathbb{R}^{n}$.

Intuitively, this concept seems applicable to all polynomials in $\mathbf{P}_{n}$ and all matrices in $M_{nn}$. Can someone validate this?

edit:

I think to make the intuition hold, $A$ must be defined as follows in $M_{nn}$:

Let $M_{1}, M_{2}, ... , M_{k}$ be matrices in $M_{nn}$.

To prove these form a basis for $M_{nn}$, we must show that $c_{1}M_{1} + c_{2}M_{2} + ... + c_{k}M_{k} = 0$ has the only trivial solution, and that every $n \times n$ matrix can be expressed as $c_{1}M_{1} + c_{2}M_{2} + ... + c_{k}M_{k} = B$.

So I believe that for $M_{nn}$, $A$ must be defined as a $n^{2} \times n^{2}$ matrix where each row vector is formed from all the $(i, j)$ entries taken from $M_{1}, M_{2}, ... , M_{k}$ (in that order.)

$\text{e.g. } A = \begin{pmatrix} M_{1_{1,1}} & M_{2_{1,1}} & ... & M_{k_{1,1}} \\ M_{1_{1,2}} & M_{2_{1,2}} & ... & M_{k_{1,2}} \\ ... & ... & ... & ... \\ M_{1_{n,n}} & M_{2_{n,n}} & ... & M_{k_{n,n}} & \end{pmatrix}$

However, I am not sure about this.

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What do you mean by $\text{det}A$ for a coefficient matrix corresponding to a set of matrices? –  Alex Becker Mar 20 '11 at 4:25
    
@Alex: editing question to reflect this. –  robjb Mar 20 '11 at 21:31

3 Answers 3

up vote 4 down vote accepted

You should be aware that for any given $n$ there's an essentially unique real vector space of dimension $n$, in the sense that any two are isomorphic (although non-canonically). For instance, the space of real polynomials of degree $\leq n$ is a real vector space of dimension $n+1$, hence isomorphic to ${\Bbb R}^{n+1}$, the space $M_n({\Bbb R})$ of square $n\times n$ real matrices is a real vector space of dimension $n^2$, hence isomorphic to ${\Bbb R}^{n^2}$, and so on. Once you prove a particular statement for ${\Bbb R}^{n}$, you proved it for ALL real vector spaces of dimension $n$.

Same thing, more in general, when you consider vector spaces over any other field than $\Bbb R$.

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Thanks! I think I will accept this answer soon, but can you confirm whether my definition of the coefficient matrix $A$ for $M_{nn}$ is correct? –  robjb Mar 22 '11 at 19:48
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@Rob: The standard way of doing it (and you may want to do that now, because you'll be using it later) is to use the entries of your matrices $M_i$ (which are really the coefficients of the linear combination that expresses $M_i$ in terms of the standard basis for the space of matrices) as the columns of your matrix, rather than the rows. This does not matter in so far as the determinant is concerned, but it will matter when you consider "change-of-basis matrices"; since it amounts to the same thing for your purposes, better go with the method that will be of use later. –  Arturo Magidin Mar 22 '11 at 20:03
    
@Rob: it is correct as long as you take as the standard basis of the space of matrices, the matrices which have a $1$ in one place and $0$ in all the others (which is what you seem to be implicitly doing). –  Andrea Mori Mar 23 '11 at 9:28

I'm not exactly sure what you would mean by the coefficient matrix of a set of vectors in $P_n$ or $M_{nn}$. If you mean e.g. let $f \in P_n$ be $a_0 + a_1x + \ldots + a_nx^n$, identify this with the vector $(a_0, a_1, \ldots, a_n)$, then consider the matrix formed by $n+1$ such vectors, then they do form a basis if and only if the determinant of the matrix $\neq 0$. The reason, of course, is that I've implicitly written down the isomorphism between $P_n$ and $\mathbb{R}^{n+1}$. The same goes for $M_{nn}$ once I make some corresponding isomorphism with $R^{n^2}$.

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You've correctly interpreted what I meant by coefficient matrix in $P_{n}$. But don't you mean they form a basis $\iff$ the matrix does NOT have determinant $0$? –  robjb Mar 20 '11 at 20:49
    
Wow sorry, you are right. –  Tony Mar 21 '11 at 2:45
    
No problem, edited to fix. –  robjb Mar 22 '11 at 19:49

In general, a square matrix over a commutative ring is invertible if and only if its determinant is a unit in that ring. (Wikipedia)

By a unit is meant an element which has a multiplicative inverse in the ring. Since zero is the only element of $\mathbf{R}$ without a multiplicative inverse, a matrix over $\mathbf{R}^n$ is invertible with nonzero determinant (note: every field is a ring).

Since the invertibility of a matrix implies that its constituent column vectors span the space, the columns of an $n x n$ matrix in $\mathbf{F}^n$ are a basis for $\mathbf{F}^n$ iff the determinant of the matrix is a unit in the ring.

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