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The rank of a matrix is only defined as integers. Is there some other criteria that is more quantitative.

E.g. $$A = \begin{bmatrix} 1 & 1\\ 1 & 0\\ \end{bmatrix} $$

$$B= \begin{bmatrix} 1 & 1\\ 1 & 0.999\\ \end{bmatrix} $$ $A$ is "better ranked" than $B$



I am judging the observability $O$ of a system. I have two matrices

$$ dim(A) = n\times n$$ $$ dim(C) = m\times n$$

$$O = \begin{bmatrix} C \\ C A \\ C A^2 \\ \vdots \\ C A^{n-1} \end{bmatrix}$$

The system is observable if $O$ is full ranked. What I want is to be able to differentiate between observabilities of systems

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I don't know if my answer really hit the problem but it was the first i thought about when i read the question – Dominic Michaelis Feb 24 '13 at 12:37
@DominicMichaelis I have added details. My matrix is in general not square, unfortunately. – aiao Feb 24 '13 at 12:45
Maybe you can take $O^T O$ and get a square Matrix but i don't know if that helps a lot – Dominic Michaelis Feb 24 '13 at 12:48
if you like an answer you can accept it – Dominic Michaelis Mar 21 '13 at 20:49

There is something like the condition of a matrix. This is a measurement of how strong errors in data inflict the error of your result. This is for $n\times n$ Matrixes definied which are invertible, when you have Matrixnorm. $$\operatorname{cond}(A)=\|A\|\cdot \|A^{-1}\|$$ where $\|\cdot\|$ is a matrix norm.

This is relevant for numeric as calculators often have an error of $10^{-16}$ (machine precision). So you would like to know what happens, how strong is the difference between your solution and the real solution.

As mentioned for nonsquare Matrix you can use $A^T A$ which is invertible when $A$ has maximum rank.

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I would try to build a criterion based on the singular values. See this Wikipedia article for starters. The singular values of a matrix $A$ are the square roots of the eigenvalues of $AA^T$. With your matrix $A$ we get $$ AA^T=\left(\begin{array}{cc}2&1\\1&1\end{array}\right). $$ The eigenvalues of this matrix are $$\lambda_1^2=\frac{3+\sqrt{5}}2\qquad\text{and}\qquad \lambda_2^2=\frac{3-\sqrt{5}}2.$$ While $\lambda_2$ is clearly smaller than $\lambda_1$, it is not dangerously close to zero.

With your other matrix on the other hand we get $$ BB^T=\left(\begin{array}{cc}2&1.999\\1.999&1.998001\end{array}\right). $$ The smaller eigenvalue of this matrix is very close to zero (at least in comparison to the bigger one) - sorry I don't have a CAS on this laptop yet, so can't give you the approximate values :-)

Singular values work better than eigenvalues, because the characteristic polynomial of a matrix fails to distinguish between $$ \left(\begin{array}{cc}1&100\\0&1\end{array}\right)\qquad\text{and}\qquad \left(\begin{array}{cc}1&0\\0&1\end{array}\right). $$

I don't have a definite criterion in mind. May be the Wikiarticle helps or somebody else can give a well studied solution.

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Isn't that just a Special case of the condition of a matrix ? – Dominic Michaelis Feb 24 '13 at 13:02
@Dominic: May be? I don't know. The product $AA^T$ has the benefit that it works even when $A$ is not a square matrix itself, but I don't see how to define the norm of the inverse, when the inverse does not exist. I did upvote your answer. This is just some fumbling that occurred to (and I know that we are worried about small singular values in multiantenna radio communication, when the matrix represents tha path gains from various transmitting/receiving antennas :-) – Jyrki Lahtonen Feb 24 '13 at 13:05

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