Interpretation of matrices in vector spaces

Question

A matrix can be interpreted as the representation of a linear mapping between two vector spaces under their chosen bases, the Gram matrix of a bilinear form on two vector spaces, and possibly other kinds of interpretation I don't know yet?
I was wondering how to interpret a normal matrix (i.e. a square matrix $A$ s.t. $A^* A=AA^*$) in vector spaces?
What kind of linear mappings is represented as a positive definite matrix under some possibly special basis?

Thanks and regards!

What's the question in 1? That seems like a statement to me. — Qiaochu Yuan, Aug 19, 2011 at 23:03
I want to ask if there are other usual ways to interpret a matrix in vector space theory. — Tim, Aug 19, 2011 at 23:11
(2) "Another way of stating the spectral theorem is to say that normal matrices are precisely those matrices that can be represented by a diagonal matrix with respect to a properly chosen orthonormal basis of $\mathbb{C}^n$. Phrased differently: a matrix is normal if and only if its eigenspaces span $\mathbb{C}^n$ and are pairwise orthogonal with respect to the standard inner product." - Wikipedia. (3) Inner products define angles, so positive definite maps are those which never take a vector $90^\circ$ or more from itself. — anon, Aug 20, 2011 at 0:20

Qiaochu Yuan · Accepted Answer · 2011-08-20 02:10:08Z

1

A matrix always represents a linear transformation between two vector spaces, and every other use of a matrix in linear algebra is a special case of this. A bilinear form, being a bilinear map $V \times V \to k$, is the same as a linear map $V \to V^{\ast}$, for example.
Any matrix which is diagonalizable with orthogonal eigenvectors is necessarily normal. The spectral theorem says that the converse is true. Abstractly, the spectral theorem can be thought of as a statement about commutative $C^{\ast}$-algebras generated by an operator, especially in light of the Gelfand representation. This is the point of view that generalizes best to the infinite-dimensional situation.
Positive-definiteness is properly thought of as a property of a bilinear form $V \times V \to \mathbb{R}$ (or a sesquilinear form, but let's ignore this case for now). It is possible to represent such a thing with a matrix since it is the same thing as a linear map $V \to V^{\ast}$, but note that these are not the same vector space.

answered Aug 20, 2011 at 2:10

Qiaochu Yuan

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$\begingroup$ Thanks! As to 1, I think the Gram matrix of a bilinear form and the matrix of a linear transform $V\rightarrow V^*$ induced by the bilinear from are different, so the usage of a matrix as a Gram matrix of a bilinear from and the usage of a matrix as that of a linear mapping are different. As to 3, I meant to ask what kind of special linear mapping a positive definite matrix directly represent, instead of indirectly by viewing the matrix as Gram matrix of a bilinear form and then considering the linear mapping induced by the form. Anon's comment on this might be in the right direction. $\endgroup$
– Tim
Aug 20, 2011 at 2:45
1

$\begingroup$ @Tim: 1. they are the same, if you equip $V$ with a basis $e_1, ... e_n$ and $V^{\ast}$ with the dual basis $e_1^{\ast}, ... e_n^{\ast}$. 2. Positive definite matrices should not be thought of as representing linear mappings, if by this you mean maps $V \to V$. They describe bilinear forms. $\endgroup$
– Qiaochu Yuan
Aug 20, 2011 at 4:18

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