How to prove $I + t X$ is invertible for small enough $ | t | ?$

Question

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 $t \ge 0.$

Answer 1

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 $$ \det(I+tX) \neq 0. $$

Answer 2

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.

Answer 3

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 $b<\frac 1{\lVert A\rVert}$.

Answer 4

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.

Answer 5

$\det(A)$ 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 $I + tX \in U$.

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$\det(I + tX)$ 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.

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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.