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Why is it important for a matrix to be square?
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I am currently trying to self-study linear algebra. I've noticed that a lot of the definitions for terms (like eigenvectors, characteristic polynomials, determinants, and so on) require a square matrix instead of just any real-valued matrix. For example, Wolfram has this in its definition of the characteristic polynomial:
The characteristic polynomial is the polynomial left-hand side of the characteristic equation $\det(A - I\lambda) = 0$, where $A$ is a square matrix.
Why must the matrix be square? What happens if the matrix is not square? And why do square matrices come up so frequently in these definitions? Sorry if this is a really simple question, but I feel like I'm missing something fundamental.
linear-algebra<br>matrices
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edited Jun 8, 2018 at 11:24
Rodrigo de Azevedo
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asked Jun 7, 2018 at 23:45
Beneschan
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$\begingroup$<br>Note that you can't compute the determinant of a non-square matrix.<br>$\endgroup$
Arnaud Mortier
Arnaud Mortier
2018-06-07 23:53:42 +00:00
Commented<br>Jun 7, 2018 at 23:53
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$\begingroup$<br>@ArnaudMortier To be fair, I think that will just constitute another instance of the OP's question: "Why does the determinant only make sense for square matrices?"<br>$\endgroup$
Noah Schweber
Noah Schweber
2018-06-07 23:55:13 +00:00
Commented<br>Jun 7, 2018 at 23:55
$\begingroup$<br>One rather successful attempt to bring much of the machinery of square matrices to $m\times n$ matrices is the singular value theorem. It does so by doing the spectral decomposition of $A^tA$ which is also symmetric<br>$\endgroup$
N8tron
N8tron
2018-06-08 11:33:11 +00:00
Commented<br>Jun 8, 2018 at 11:33
$\begingroup$<br>Note, the definitions of eigenvectors etc that you mention don't depend on the matrix having real values. Matrices with complex values are just as useful (and of course a real matrix can be considered as a special case of a complex matrix). Also (as N8tron said) there is a very powerful concept that applies to non-square matrices, namely the "singular value decomposition" (SVD) - though to understand what it is and what it's good for, you need to learn about the concepts you mentioned for square matrices first, and the SVD may not be covered at all in a "first" linear analysis course.<br>$\endgroup$
alephzero
alephzero
2018-06-10 21:08:42 +00:00
Commented<br>Jun 10, 2018 at 21:08
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Remember that an $n$-by-$m$ matrix with real-number entries represents a linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$ (or more generally, an $n$-by-$m$ matrix with entries from some field $k$ represents a linear map from $k^m$ to $k^n$). When $m=n$ - that is, when the matrix is square - we're talking about a map from a space to itself.
So really your question amounts to:
Why are maps from a space to itself - as opposed to maps from a space to something else - particularly interesting?
Well, the point is that when I'm looking at a map from a space to itself inputs to and outputs from that map are the same "type" of thing, and so I can meaningfully compare them. So, for example, if $f:\mathbb{R}^4\rightarrow\mathbb{R}^4$ it makes sense to ask when $f(v)$ is parallel to $v$, since $f(v)$ and $v$ lie in the same space; but asking when $g(v)$ is parallel to $v$ for $g:\mathbb{R}^4\rightarrow\mathbb{R}^3$ doesn't make any sense, since $g(v)$ and $v$ are just different types of objects. (This example, by the way, is just saying that eigenvectors/values make sense when the matrix is square, but not when it's not square.)
As another example, let's consider the determinant. The geometric meaning of the determinant is that it measures how much a linear map "expands/shrinks" a unit of (signed) volume - e.g. the map $(x,y,z)\mapsto(-2x,2y,2z)$ takes a unit of volume to $-8$ units of volume, so has determinant $-8$. What's interesting is that this applies to every blob of volume: it doesn't matter whether we look at how the map distorts the usual 1-1-1 cube, or some other random cube.
But what if we try to go from $3$D to $2$D (so we're considering a $2$-by-$3$ matrix) or vice versa? Well, we can try to use the same idea: (proportionally) how much area does a given volume wind up producing? However, we now run into problems:
If we go from $3$ to $2$, the "stretching factor" is no longer invariant. Consider the projection map $(x,y,z)\mapsto (x,y)$, and think about...