What is the importance of the rank of a matrix? I know that the rank of a matrix is the number of linearly independent rows or columns (whichever is smaller).

Why is it a problem if a matrix is rank deficient?

Also, why is the smaller value between row and column the rank?

An intuitive or descriptive answer (also in terms of geometry) would help a lot.

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    $\begingroup$ Do you know what the column and row space of a matrix is? Do you know about vector spaces? The immediate geometric interpretation is that RankA is the dimension of the vector space spanned by the column vectors. $\endgroup$ – AnonymousCoward Feb 9 '11 at 1:55
  • $\begingroup$ I know the vector spaces are a collection of vectors that satisfies the axioms that are stated for the vector spaces. I do not understand what you mean by "vector spaces spanned by column vectors". $\endgroup$ – 0x0 Feb 9 '11 at 1:58
  • $\begingroup$ There are nice applications in graph theory when dealing with adjacency matrices. The RankA will tell you things about the corresponding graph like the number of connected components. I dont know enough about this to make it an answer though. $\endgroup$ – AnonymousCoward Feb 9 '11 at 2:01
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    $\begingroup$ If you are interested in learning the why's of linear algebra I highly recommend viewing Gilbert Strang's Linear Algebra Course and purchase his book. $\endgroup$ – Tpofofn Feb 9 '11 at 3:33

A rank of the matrix is probably the most important concept you learn in Matrix Algebra. There are two ways to look at the rank of a matrix. One from a theoretical setting and the other from a applied setting.

From a theoretical setting, if we say that a linear operator has a rank $p$, it means that the range of the linear operator is a $p$ dimensional space. From a matrix algebra point of view, column rank denotes the number of independent columns of a matrix while row rank denotes the number of independent rows of a matrix. An interesting, and I think a non-obvious (though the proof is not hard) fact is the row rank is same as column rank. When we say a matrix $A \in \mathbb{R}^{n \times n}$ has rank $p$, what it means is that if we take all vectors $x \in \mathbb{R}^{n \times 1}$, then $Ax$ spans a $p$ dimensional sub-space. Let us see this in a 2D setting. For instance, if

$A = \left( \begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array} \right) \in \mathbb{R}^{2 \times 2}$ and let $x = \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right) \in \mathbb{R}^{2 \times 1}$, then $\left( \begin{array}{c} y_1 \\ y_2 \end{array} \right) = y = Ax = \left( \begin{array}{c} x_1 + 2x_2 \\ 2x_1 + 4x_2 \end{array} \right)$.

The rank of matrix $A$ is $1$ and we find that $y_2 = 2y_1$ which is nothing but a line passing through the origin in the plane.

What has happened is the points $(x_1,x_2)$ on the $x_1 - x_2$ plane have all been mapped on to a line $y_2 = 2y_1$. Looking closely, the points in the $x_1 - x_2$ plane along the line $x_1 + 2x_2 = c = \text{const}$, have all been mapped onto a single point $(c,2c)$ in the $y_1 - y_2$ plane. So the single point $(c,2c)$ on the $y_1 - y_2$ plane represents a straight line $x_1 + 2x_2 = c$ in the $x_1 - x_2$ plane.

This is the reason why you cannot solve a linear system when it is rank deficient. The rank deficient matrix $A$ maps $x$ to $y$ and this transformation is neither onto (points in the $y_1 - y_2$ plane not on the line $y_2 = 2y_1$ e.g. $(2,3)$ are not mapped onto, which results in no solutions) nor one-to-one (every point $(c,2c)$ on the line $y_2 = 2y_1$ corresponds to the line $x_1 + 2x_2 =c$ in the $x_1 - x_2$ plane, which results in infinite solutions).

An observation you can make here is that the product of the slopes of the line $x_1 + 2x_2 = c$ and $y_2 = 2y_1$ is $-1$. This is true in general for higher dimensions as well.

From an applied setting, rank of a matrix denotes the information content of the matrix. The lower the rank, the lower is the "information content". For instance, when we say a rank $1$ matrix, the matrix can be written as a product of a column vector times a row vector i.e. if $u$ and $v$ are column vectors, the matrix $uv^T$ is a rank one matrix. So all we need to represent the matrix is $2n-1$ elements. In general, if we know that a matrix $A \in \mathbb{R}^{m \times n}$ is of rank $p$, then we can write $A$ as $U V^T$ where $U \in \mathbb{R}^{m \times p}$ and is of rank $p$ and $V \in \mathbb{R}^{n \times p}$ and is of rank $p$. So if we know that a matrix $A$ is of rank $p$ all we need is only $2np-p^2$ of its entries. So if we know that a matrix is of low rank, then we can compress and store the matrix and can do efficient matrix operations using it. The above ideas can be extended for any linear operator and these in fact form the basis for various compression techniques. You might also want to look up Singular Value Decomposition which gives us a nice (though expensive) way to make low rank approximations of a matrix which allows for compression.

From solving a linear system point of view, when the square matrix is rank deficient, it means that we do not have complete information about the system, ergo we cannot solve the system.

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    $\begingroup$ A beautiful explication, Sivaram. $\endgroup$ – Uticensis Mar 5 '11 at 0:46
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    $\begingroup$ "we cannot solve the system." - we can, but not in the usual sense... hence least squares, Tikhonov regularization, and a bunch of other fancy tricks. $\endgroup$ – J. M. isn't a mathematician Apr 29 '12 at 5:04
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    $\begingroup$ As for an application of SVD and low-rank approximations, see here. $\endgroup$ – J. M. isn't a mathematician Apr 29 '12 at 5:05
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    $\begingroup$ sorry but how do you compute the number of the entries? thank you ! $\endgroup$ – Zoe Lee Apr 17 '17 at 9:56
  • $\begingroup$ @Martin Sleziak. Where do you get the $-1$ or $-p^2$ from? $\endgroup$ – roi_saumon Oct 20 '20 at 15:17

The rank of a matrix is of major importance. It is closely connected to the nullity of the matrix (which is the dimension of the solution space of the equation $A\mathbf{x}=\mathbf{0}$), via the Dimension Theorem:

Dimension Theorem. Let $A$ be an $m\times n$ matrix. Then $\mathrm{rank}(A)+\mathrm{nullity}(A) = n$.

Even if all you know about matrices is that they can be used to solve systems of linear equations, this tells you that the rank is very important, because it tells you whether $A\mathbf{x}=\mathbf{0}$ has a single solution or multiple solutions.

When you think of matrices as being linear transformations (there is a correspondence between $m\times n$ matrices with coefficients in a field $\mathbf{F}$, and the linear transformations between a given vector space over $\mathbf{F}$ of dimension $n$ with a given basis, and a vector space of dimension $m$ with a given basis), then the rank of the matrix is the dimension of the image of that linear transformation.

The simplest way of computing the Jordan Canonical Form of a matrix (an important way of representing a matrix) is to use the ranks of certain matrices associated to $A$; the same is true for the Rational Canonical Form.

Really, the rank just shows all over the place, it is usually relatively easy to compute, and has a lot of applications and important properties. They will likely not be completely apparent until you start seeing the myriad applications of matrices to things like vector calculus, linear algebra, and the like, but trust me, they're there.

  • $\begingroup$ Did you mean "Let A be m x n matrix, where n >= m"? $\endgroup$ – Kentzo Sep 26 '18 at 22:42
  • $\begingroup$ @Kentzo:No. Why would I? $\endgroup$ – Arturo Magidin Sep 27 '18 at 2:37
  • $\begingroup$ My understanding is that otherwise $n$ as in the $m \times n$ cannot be used as $n$ in $rank(A) + nullity(A) = n$ $\endgroup$ – Kentzo Sep 27 '18 at 3:08
  • $\begingroup$ @Kentzo: The Rank-Nullity Theorem says that if $T\colon V\to W$ is a linear transformation, then $\dim(V) = \mathrm{rank}(T) + \mathrm{nullity}(T)$. An $m\times n$ matrix defines a linear transformation $\mathbb{R}^n\to\mathbb{R}^m$ via left multiplication; the nullity of this linear transformation is the nullity of the matrix $A$, and the rank of this linear transformation is the rank of $A$. So, I don’t know where you got that you must have $n\geq m$, but your “understanding” is wrong. $\endgroup$ – Arturo Magidin Sep 27 '18 at 3:59

The rank of the matrix $A$ is equal to the dimension of the Image of $A$ (which is spanned by columns of $A$), if that's a sufficient enough geometrical explanation. You can read about vector spaces here and about the image of a matrix here.

  • $\begingroup$ Can you please explain more in detail ? What do you mean by image of A ? $\endgroup$ – 0x0 Feb 9 '11 at 2:04
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    $\begingroup$ I posted a link (goo.gl/hKyvN) that explains what it is. Basically, it is the span of the columns, which means you take columns of matrix A, multiply them by all possible scalars, and get some space (which is called image of that matrix). $\endgroup$ – InterestedGuest Feb 9 '11 at 2:05

The rank of a matrix is a building stone to understanding Matrix Completion, which tackles such problems as the Netflix Prize and related issues in recommender systems. The topic of "rank" in higher dimensional space ($>2$) is an interesting topic of research.


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