Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X \in \text{GL}_n(\mathbb{R})$ be an arbitrary real $n\times n$ matrix. How can we prove rigorously: $$ \underset{b>0} {\exists} : \underset{|t|\le b} {\forall} : \det (I + t X) \neq 0 $$ If necessary, we could also assume that $t \ge 0.$

share|improve this question
10  
$\det(I + tX)$ is a nonzero polynomial, so it has finitely many roots. Alternately, for sufficiently small $t$ the geometric series $1 - tX + t^2 X^2 \mp ...$ converges. (This proof has the virtue of generalizing to infinite dimensions.) –  Qiaochu Yuan Jul 2 '12 at 15:12
8  
Show that $\det$ is continuous, and note that (setting $t=0$) $\det I = 1$. –  copper.hat Jul 2 '12 at 15:21
    
Just curious, but why do you use \mathtt for matrices? That's a convention I've never seen before. –  mrf Jul 3 '12 at 12:49
    
I have seen it in a couple of papers in computer science - along with mathbf for vectors. But I will try to avoid it from now on on math.stackexchange ;-) –  Hauke Strasdat Jul 3 '12 at 14:14

5 Answers 5

up vote 10 down vote accepted

Note that for $t\neq 0$ $$ \det(I+tX) = t^n\det(\frac{1}{t}I+X) = 0 $$ only if $\frac{-1}{t}$ is an eigenvalue of $X$. If $ t=0$ the statement is trivial so we exclude this case.

Since $X$ has only finitely many eigenvalues, its set of eigenvalues is bounded in magnitude. For all $t$ sufficiently small $|\frac{-1}{t}|$ will be larger than this magnitude bound, so $\frac{-1}{t}$ will not be an eigenvalue of $X$, and hence we will have $$ \det(I+tX) \neq 0. $$

share|improve this answer
1  
It's not true that $\det (tA)=t\det A$; rather, $\det (tA)=t^n \det A$. But your argument still works. –  Florian Jul 2 '12 at 15:43
    
Thanks, I have corrected it. –  nullUser Jul 2 '12 at 15:48
    
Nice one! Especially I like that it only relies on basic linear algebra. –  Hauke Strasdat Jul 3 '12 at 10:16

A matrix is invertible iff its determinant is nonzero. But determinant is a continuous function, so $GL_n$ is open as a subset of the space of all $n\times n$ matrixes, so there is a neighborhood of $I$ with no singular elements.

In particular shifting $I$ (or any other invertible matrix) by a small enough multiple of any given matrix $X$ (not necessarily invertible) does not move $I$ out of the neighborhood, leaving the matrix invertible.

share|improve this answer

Let $\lVert\cdot\rVert$ a submultiplicative norm over the set of $n\times n$ real matrices, denoted $\mathbf M_n(\Bbb R)$ (we can take $\lVert M\rVert:=\sup_{x\neq 0}\frac{\lVert Mx\rVert}{\lVert x\rVert}$, where $\lVert\cdot\rVert$ is the Euclidian norm). This norm makes $\mathbf M_n(\Bbb R)$ complete (since it's a norm over a finite-dimensional vector space over $\Bbb R$. For $A\in \mathbf M_n(\Bbb R)$ of norm $<1$, we have $$(I-A)\sum_{j=0}^{+\infty}A^j=I=\sum_{j=0}^{+\infty}A^j(I-A),$$ hence $I-A$ is invertible (the series is normally convergent, and the space being complete, convergent).

If $X=0$, the statement is obviously true, and if $X\neq 0$, we can take $b<\frac 1{\lVert A\rVert}$.

share|improve this answer

The determinant of a matrix is a polynomial (and hence continuous) in its $n^2$ entries. Taking the limit as $t\to 0$ makes all the diagonal entries tend to $1$ and all other entries to $0$, so the determinant tends to $1.$ So there exists a neighbourhood around $t=0$ such that the determinant of all those matrices has positive determinant, so are invertible.

share|improve this answer

$\det(A)$ is a polynomial function in the entries of $A$. The solution set to $\det(A) = 0$ is a closed subset, and so the set of invertible matrices is an open set. In particular, there is an open neighborhood $U$ of $I$ such that every matrix in $U$ is invertible. By choosing $t$ sufficiently small, we guarantee $I + tX \in U$.


$\det(I + tX)$ is a polynomial function of $t$, and so $\det(I + tX) = 0$ has finitely many roots, and $t=0$ is not a root. Therefore, we can find an open interval containing 0 such that $I + tX$ is invertible on that interval.


We can compute the Taylor series for $(I + tX)^{-1}$ about 0:

$$ (I + tX)^{-1} = I - t X + t^2 X^2 - t^3 X^3 + \cdots $$

It's not difficult to see that the right hand side is convergent in every component of the matrix (e.g. ratio test along with an upper bound on the entries for $X^n$) on an interval of positive radius. By multiplying through by $(I + tX)$, we can check that the sum really is the inverse.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.