Linear transforms of vector spaces has the linearity (i.e., homogeneity and superimposability), which is shared by LTI system. There are also shared concepts such as eigenvectors/eigenvalue...

I am wondering, can linear transforms (of vector spaces) be seen as LTI systems? If not, why? If yes, in particular, are there counterparts in vector spaces for the concept of Dirac $\delta$ and $h(t)$ in LTI systems, where $\delta$ represents all possible inputs (e.g., from frequency space point of view), and thus $h(t)$ characterizes the LTI system? If these two counterparts exists in Linear transforms, what's the counterpart for convolution?

A little background: I had this question recently as I am studying linear algebra (starting from some basics) ...actually I first read some books on DSP/DIP a couple months ago, and then I find out that I need to learn some basics of linear algebra (as well as other branches of math), otherwise most of the DSP/DIP literature are foreign to me...in other words, I am also new to the field of DSP/DIP...so I beg your pardon that my question may be a bit fuzzy.


  • $\begingroup$ a matrix can be seen as an LTI gain system no? $\endgroup$
    – percusse
    Commented Oct 4, 2017 at 14:59

1 Answer 1


The question of abstract nature so a rigorous answer could require more elaboration, however on the surface one can at least observe the following.

An LTI system is also a linear transform, which maps its input signal (vector) $x(t)$ or ($ x[n]$) into an output signal (vector) $y(t)$ (or $y[n]$), where both the input and output belong to the same signal space, one kind of vector space.

For such transforms and vector spaces the time invariance property can be checked easily such as $y(t-d) = \mathcal{T} \{x(t-d) \}$, but for other vector spaces such as the n-tuples or matrix spaces etc, the operation of time invariance is not trivial to check, at least intuitively. The shift of a vector (an element of the set) should be defined first.

May be the following links could help further:



  • $\begingroup$ Thanks for your answer and links (I need some time to digest the content from the links). $\endgroup$
    – bruin
    Commented Oct 5, 2017 at 5:00

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