# Why do mathematicians say that "let an operator be represented by a matrix" instead of operator is the matrix?

For example, look at this sentence from Perko's text on dynamical system

"It follows from Cauchy Schwarz inequality that if $T \in L(R^n)$ is represented by the matrix $A$ with respect to the standard basis for $R^n$ $_\cdots$" pg 11

What does it mean for a $T: R^n \to R^n$ to be represented by a matrix? Isn't it by definition that $T: R^n \to R^n$ is equivalent to an n by n matrix? Can someone translate exactly what it means by "represented" versus "not represented"?

• The operative phrase is "is represented by the matrix $A$ with respect to the standard basis" With respect to different bases, the matrix $A$ represents different transformations, probably completely different from the one you started with. You can't really separate a matrix from the bases that are associated with it. On the other hand, the operator $T$ is completely independent of bases. Understanding this relationship of transformations with matrices is an important step in understanding linear algebra. Commented Oct 2, 2015 at 19:39
• Dually, a given operator has different representations with respect to different bases. When the underlying vector space is just $\Bbb F^n$, which has a standard basis, we often (but not always) use that one---for general abstract vector spaces, though, we generally do not have a preferred basis. Commented Oct 2, 2015 at 19:52
• I think your question may be "Since all linear operators can be written as $x\mapsto Ax$, why do we bother denote them as $T:\mathbb R^n\to \mathbb R^n$?" In that case @rschwieb comment should be considered. As it is right now, Brian's answer does make a lot of sense. Commented Oct 2, 2015 at 20:06
• Representing $T$ as a matrix (w.r.t. some definite basis) is very useful for calculation for finite-dimensional vector spaces over $R^n$ or $C^n$. But useful vector spaces can be defined over many other types of mathematical quantities, and may be infinite-dimensional. Commented Oct 2, 2015 at 21:25
• The function $f(x)=3x$ and the number $3$ are different objects. Commented Oct 3, 2015 at 16:52

The operator $T$ is a function from $\Bbb R^n$ to $\Bbb R^n$; an $n\times n$ matrix of real numbers is not a function. If one chooses the matrix $A$ properly, one can say that the function $T(x)=Ax$ for each $x\in\Bbb R^n$, but that is far from saying that the function $T$ is the matrix $A$.

Consider the more familiar case of real-valued functions on the reals. Specifically, consider the function $f(x)=ax$: this function is certainly not the same thing as the real number $a$. The relationship between $T$ and $A$ in the first paragraph is precisely similar to that between $f$ and $a$ in this paragraph.

• So why don't these authors be more forthcoming and say let our operator mapping $R^n$ to $R^n$ be represented by $T(x) = Ax$ (or be defined as $Ax$)? Commented Oct 2, 2015 at 19:34
• @MathNewb: It’s expected that by the time you’re reading a text on dynamical systems, you’re familiar with the normal mathematical terminology and usage of basic linear algebra and other standard introductory topics. Commented Oct 2, 2015 at 19:36
• @MathNewb : Because you still have to specify in which basis multiplication by the matrix does the same thing as applying the operator. The operator maps points to points without reference to a basis or coordinates. A matrix can only be said to do so once there is a relationship between coordinates and points in the space and this relationship is provided by a basis. (Vector spaces don't come with coordinates -- so if you're thinking of vectors as lists of numbers, you've already picked a basis.) Commented Oct 3, 2015 at 21:07

The number after nine is represented in the usual base system by the sequence of digits $10$, but it isn't a sequence of digits. Properties like "two digits long" are properties of the representation, while properties like "has four factors" are properties of the number. We could represent the number in binary, as $1010$, and that would be just as valid.

Similarly, a linear transformation $T$ may be represented by a matrix $M$, but it isn't the matrix. Properties like "upper triangular" are properties of the representation, while properties like "injective" are properties of the linear transformation. In a different basis, $T$ might be represented by a matrix other than $M$, and that representation would be just as valid.

The concept of matrix can be useful in other situation than for linear operators, and there exists linear operators that can not be represented by matrices.

The matrix is just an array of elements and are added and multiplied according to certain rules. For this it is only required that there is a meaningful definition for addition and multiplication for the elements. This means that the elements does not need to be real numbers, or even complex numbers. They can for example be integers (which may not seem that alarming, but it makes a difference). There's even more "odd" things one could put in a matrix as for example integers using modulo-n operations. These later cases means that the matrix do not represent a linear operator.

Then there are linear operators that does not operate on vector spaces of finite dimension or produce values from vector spaces of finite dimensions. For example a linear operator could map from or to a vector space of sequences or even functions. These can obviously not be represented as a matrix.

As for the normal case from linear algebra 101 one should remember that a linear operator is still the same no matter which basis for the vector space(s) you use, but the matrix representing the operator will be different depending on the choice of base. Even though you may think of a vector space as having a canonical base, this is not necessarily the case (it's better to think of them as being a space in it's own right and the base you select be arbitrary and no base is "better" than another).

• Could you give an example of a linear operator which can't be represented by matrices? Commented Oct 3, 2015 at 7:58
• @Ruslan Think of linear operators on infinite-dimensional vector spaces.
– Danu
Commented Oct 3, 2015 at 10:53
• @Ruslan Take for example the Laplace or Fourier transforms. Commented Oct 3, 2015 at 11:43
• @Danu: at least physicists do often talk about matrices on infinite-dimensional Hilbert spaces. Commented Oct 3, 2015 at 14:57
• @leftaroundabout Because in physics the relevant spaces are (almost?) always separable, the analogy works quite well; but it isn't really rigorous in any case. I assume that the differences are even more obvious in the "continuous" case.
– Danu
Commented Oct 3, 2015 at 15:03